1 Some Basic Concepts of Chemistry Lecture-1 1.1 Chemical Arithmetic Chemistry is the branch of physical science which deals with the study of matter, their physical and chemical properties, and their chemical composition. The main branches of chemistry are : (i) Organic Chemistry (ii) Inorganic Chemistry (iii) Physical Chemistry (iv) Analytical Chemistry (i) Organic Chemistry: It is concerned with the study of compounds of carbon except carbonate, bicarbonate, cyanides, isocyanides, carbides and oxides of carbon. (ii) Inorganic Chemistry: It deals with the study of all known elements and their compounds except organic compounds. 3
(iii) Physical Chemistry: It is concerned with the physical properties, the laws of the chemical combination and theories which are governing the reaction. (iv) Analytical Chemistry: It deals with various methods of anal y sis of chem i cal sub stances both qual i ta tive and quantitative. Some of the specialised branches are: (i) Bio-Chemistry (ii) Medicinal Chemistry (iii) Soil and Agriculture (iv) Industrial Chemistry (v) Nuclear Chemistry (vi) Polymer Chemistry
1.2 Matter and Energy Matter is anything that has mass and occupies space. All the bodies in the universe conform to this definition. Mass is the quantity of matter in a particular sample of matter whether the term weight should not be used in place of mass. Energy is defined as the capacity of doing work. Anything which has the capacity to push matter from one place to another posses energy. There are various forms of energy such as heat, light, etc. Energy is neither be created nor be destroyed.
Classification of Matter (i) Physical Classification: Matter can exist in any one of three forms Solid, liquid and Gas. In the solid state, substances are rigid. They have definite shape and fixed volume.In liquid 4
state, substances have no definite shape but possess fixed volume. In a gaseous state, substances have no definite shape and volume. Depending on temperature and pressure, a substance can exist in any one of the three forms of matter. (ii) Chemical Classification: Matter exists in nature in the form of chemical substances. A pure substance is defined as a variety of matter, which have same compositions and properties, i.e., a material containing only one substance. MATTER Homogeneous Physical Classification
Solids
Liquids
Elements
Gases
Chemical Classification Pure Substances
Compounds
Homogeneous
Heterogeneous Mixtures
Heterogeneous
(Solutions) (All Homogeneous)
Inorganic Compounds
Organic Compounds
Metals, Non-Metals and Metalloids All the elements may be classified into two groups i.e., metals and non-metals. The division is based on both physical and chemical properties. Metals are regarded as those elements which possess the following properties: (i) They are generally solids at ordinary conditions, mercury is an exception which is in liquid state. 5
(ii) They are lustrous in nature (iii) They possess high density (iv) They are good conductors of electricity and heat. (v) They possess generally high melting and boiling points.
Chemical Classification of Matter 1. Element: An element is the simplest form of a
pure
substance. It may be defined as: ''A pure substance which can neither decomposed nor built from simpler substances by ordinary physical or chemical methods''. Elements are further classified into following types: (i) Metals: These elements are generally solids and possess characteristics such as: Bright luster, hardness and ability to conduct electricity and heat. (ii) Non-metals: These elements are generally non-lustrous, brittle and poor conductors of heat and electricity. The common examples of non-metal are carbon, hydrogen oxygen, nitrogen, etc. (iii) Metalloids: These are elements which have characteristics common to both metals and non-metals. The common examples of metalloids are silicon, arsenic, bismuth, antimony, etc.
2. Compound: A compound may be defined as: A substance which is obtained by the union of two or more elements in a definite proportion, which can neither 6
decomposed nor built from simpler substances by ordinary physical or chemical methods.
Compounds are classified into two types: (i) Organic Compounds: The compounds obtained from living sources are termed organic compounds. The term organic is now applied to hydrocarbons. (ii) Inorganic Compounds: The compounds obtained from non-living sources such as rocks and minerals are termed inorganic compounds.
3. Mixture: A mixture may be defined as: A combination of two or more elements or compounds in any proportion so that the components do not lose their identity.
Mixtures are further classified into following types: (i) Homogeneous mixtures: These have the same compositions throughout the sample. The components of a mixture can be seen even under a powerful microscope. Homogeneous mixtures are also called solution. (ii) Hetrogeneous mixtures: They consist of two or more parts (called phases) which have different compositions.
Difference between a compound and a mixture: Compound 1. The constituents of a compound are always present in a fixed ratio by mass.
Mixture 1.
The constituents of a mixture may be present in any ratio
7
2. Compound is always homogeneous in nature.
2.
Mixture may or may not be homogeneous in nature.
3. The properties of a compound are different from those of its constituent elements.
3.
The properties of a mixture are midway between those of its constituents.
4. The constituents of a compound cannot be easily separated by simple mechanical means. Energy in the form of heat or light is often required.
4.
The constituents of a mixture can be easily separated by simple mechanical means.
5.
5.
Mixtures are formed as a result of a physical change.
6. Formation of a compound is 6. always accompanied by absorption or evolution of heat, light or electrical energy.
When a mixture is formed, no heat, light or electrical energy is absorbed or evolved.
7. Chemical Compounds posses 7. sharp melting and boiling points.
The melting and boiling points of mixtures are usually no sharp.
Compound are formed as a result of a chemical change.
1.3 Measurement in Chemistry ‘A unit is defined as the standard of reference chosen in order to measure a definite physical quantity in chemistry. Fundamental and Derived Units: The units of physical quantities depends on three basic units, i.e., units of mass, length and time. Since these are 8
independent on units and cannot be derived from any other units, they are called fundamental units. Thus, seven units of measurement namely, mass (Kg), length(m), time (sec), temperature (K), electric current(A), luminous intensity(L) and amount of substance are taken as basic units. All other units can be derived units. The unit of area, volume, force, work ,density, velocity, energy, etc. are derived units.
Dimensional Analysis Any calculation making use of dimensions of the different physical quantities involved is termed as dimensional analysis. A unit conversion starts with an equality between different units for the same quantity. For instance, 0.001L is equivalent to 1 mL. Representing this relationship in the form of algebraic equation: 0.001 L = 1 mL Each of these fractions are called unit factors. These ratios are more commonly called conversion factors.
Example: If the speed of light is 3×10 8 ms –1 . Calculate the distance covered by light in 2.00 ns.
Solution: Speed = 3 × 108 time = 2 .00 ns = 2 × 10 −9 Sec Distance = speed × time = 2 × 3×10 8 ×10 −9 = 6×10 −1 = 0.6 m
Example: Express each of the following in SI units: (i) Speed of Shatabdi Express, i,e., 120 miles per hr. 9
(ii) Distance between earth and sun, i.e., 93 million miles. (iii) Average between earth and sun, i.e., 5 feet 6 inches.
Solution: (i) The SI units for speed are m s −1 1 mile = 1.60 km = 1.60 × 10 3 m 1.60 × 10 3 m ∴ Conversion factor = 1 mile 1 hr = 60 × 60 s = 3.6 × 10 3 3.6 × 10 3 s ∴ Conversion factor = 1 hr 120 miles
Now, speed =
hr 120 miles
=
hr
1.60 × 10 3 m × 1mile ×
1hr
3.6 × 10 3 s
= 53.3 m s −1 . (ii)
The Si units for distance is metre (m) 1 mile = 1.60 km = 1.60 × 10 3 m Conversion factor =
1.60 × 10 3 m 1mile
Distance = 93 × 106 miles =
93 × 106 miles × 1.60 × 10 3 m 1miles
= 1.49 × 10 11 m. 10
(iii)
5 feet 6 inches = 66 inches 1 inches = 2 .54 × 10 −2 m 2 .54 × 10 −2 m Conversion factor = 1 inch Now,
66 inches =
66 inches × 5 .54 × 10 −2 m 1 inches = 1.68 m.
Seven basic physical quantities and their SI units: Physical Quantity
Unit Symbol
Length
l
Metre
m
Mass
m
Kilogram
kg
Time
t
Second
s
Electric current
i
Ampere
A
Thermodynamic temperature T
Kelvin
K
Amount of the substance
n
Mole
mol
Luminous intensity
Iv
Candela
cd
Ex-1: Does 1 gram mole of a gas occupy 22.4 L under all conditions of temperature & pressure?
Sol.
No, one gram mole of a gas occupies 22.4 L only under NTP or STP conditions, i.e., at 273K temperature and under 760 mm pressure. If these conditions are not used, then the volume is not 22.4 L.
Ex-2: The percentages of all the elements present in a compound are 92. What does it indicate ?
11
Sol.
This indicates that the compound contains oxygen also and its percentage is (100 – 92) = 8.
Ex-3: Gun powder is a mixture of sulphur, charcoal and potassium nitrate (KNO 3 ). How would you separate it into its constituents?
Sol:
Sulphur is soluble in CS2 and is insoluble in water. KNO 3 is soluble in water. Charcoal is insoluble in both water as well as CS2 .
Ex-4: Vanadium metal is added to steel to impart strength. The density of Vanadium is 5.96 g/cm 3 . Express in S.I. unit (kg / m 3 ).
Sol:
d = 5 . 96 g / cm 5 m of vanadium = 5 . 96 × 10 −3 5 . 96 g cm
−3
5 . 96 × 10 −3 kg −3 . = = 5960 m −6 3 10 m
Vol of vanadium metal = 10 −6 m 3 . And we know, d =
m v
Significant figures Number of significant figures in a physical quantity depends upon the least count of the instrument used for its measurement.
1.
Common rules for counting significant figures: Following are some of the common rules for counting significant figures in a given expression:
12
Rule 1. All non zero digits are significant. Example: x = 1234 has four significant figures. Again, x = 189 has only three significant figures.
Rule 2. All non zero digits occurring between two non zero digits are significant.
Example: x = 1007 has four significant figures. Again, x = 1.0809 has five significant figures.
Rule 3. In a number, less than one, all zero's to the right of decimal point and to the left of a non zero digit are not significant.
Example: x = 0.0084 has only two significant figures. Again, x = 1.0084 has five significant figures.
Rule 4. All zeroes on the right of the last non zero digit in the decimal part are significant.
Example: x = 0.00800 has three significant figures 8, 0,0. The zeros before 8 are not significant.
Rule 5. All zeroes on the right of the non zero digit may or may not be significant.
Example: We can write 20,000 in scientific notation as 2 × 10 4 Having 1 significant figure 2 .0 × 10 4 Having 2 significant figure 2.00 × 10 4 Having 3 significant figure
Rule 6. All zeroes on the right of the last non zero digit be come sig nif i cant, they come from a mea sur able quantity. 13
Example: Suppose distance between two stations is measured to be 3050m. It has four significant figures. the same distance can be expressed as 3.050 km or 3.050 ×10 5 cm.
2.
Rounding off: While rounding off measurements, we
use the following rules by convention:
Rule 1. If the digit to be dropped is less than 5, then the preceeding digit is left unchanged
Example: x = 7.82 is rounded off to 7.8. Again, x = 3.94 is rounded off to 3.9.
Rule 2. If the digit to be dropped is more than 5, then the preceeding digit is raised by one.
Example: x = 6.87 is rounded off to 6.9. Again, x = 12.78 is rounded off to 12.8.
Rule 3. If the digit to be dropped is 5 followed by digits other than zero, then the preceeding digit is raised by one.
Example: x = 16.351 is rounded off to 16.4. Again, x = 6.758 is rounded off to 6.8.
Rule 4. If the digit to be dropped is 5 followed by zeroes, then the preceeding digit is left unchanged, if it is even.
Example: x = 3.250 is Rounded off to 3.2 Again x = 12.650 is rounded off to 12 .6
Rule 5. If the digit to be dropped is 5 or 5 followed by zero, then the preceeding digit is raised by one.
Example: x = 3.750 is rounded off to 3.8. Again, x = 16.150 is rounded off to 16.2. 14
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Questions Q.1: Express the following numbers upto three significant figures. (i) 306.35
(ii) 0.0038816
(iii) 1.78975×10 4
(iv) 0.25400
(v) 2.65986×10 3 Ans: (i) 306.35 = 306 (ii) 0.0038816 = 3.88×10 −3 (iii) 1.78975×10 4 = 1.79×10 4 [rounded off according to rule vi (b)] (iv) 0.25400 = 2.54×10 −1 (v) 2.65986×10 3 = 2.66×10 3 [rounded off according to rule vi (b)]. 15
Q.2: The mass of a sample of iron metal is 5.932 g. If the density of iron is 7.8 g/cm 3 , what is its volume? Ans: Volume = Mass = 5.932 Density 7.8 g cm −3 The smallest number of significant figures in this calculation is two. Hence the result has been rounded off to two significant figures. Q.3: "The star of India" sapphire weighs 563 carats. If one carat is equal to 200 mg, what is the weight of the gemstone in grams? Ans: Weight one carat = 200 mg ∴ Q.4
Weight of 563 carats =
Express the following in scientific notation: (i) 0.0048
(ii) 234,000
(iv) 500.0
(v) 6.0012
Ans. (i) 4.8×10 −3 (iv) 5.000×10 2 Q.5
200 × 563 = 112.6 g . 1000
(ii) 2.34×10 5
(iii) 8008
(iii) 8.008×10 3
(v) 6.0012×10 0
How many significant figures are present in the following ? (i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Ans. (i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5
Q.6
Round up the following upto three significant figures : (i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Ans. (i) 34.2 (iii) 0.0460
16
(ii) 10.4 (iv) 2810
Q.7
How many significant figures should be present in the answer of the following calculation ? (i)
0.02856 × 29815 . × 0.112 0.5785
(ii) 5×5.364
(iii) 0.0125 + 0.7864 + 0.0215 Ans.
Q.8
(i)
The least precise term has three significant figures (i.e., in 0.112). Hence, the answer should have three significant figures.
(ii)
Leaving the exact number (5), the second term has four significant figures. Hence, the answer should have four significant figures.
(iii)
In the given addition, the least number of decimal places in the term is 4. Hence, the answer should have four significant.
Why are the atomic masses of most of the elements fractional ?
Ans. Atomic masses of most of the elements are fractional because most of elements occur in nature as a constant mixture of isotopes. The atomic masses of the isotropes actually the average relative masses of the isotopes depending on their abundance. Q.9
Express the following in the scientific notation : (i) 0.0048
(ii) 234,000
(iv) 500.0
(v) 6.0012
Ans. (i) 4.8 × 10 −3 (iv) 5.00 × 10 2
(ii) 2.34 × 10 5
(iii) 8008
(iii) 8.008 × 10 3
(v) 6.0012 × 10 0
Q.10 How many significant figures are present in the following ? (i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
(ii) 3
(iii) 4
(v) 4
(vi) 5
Ans. (i) 2 (iv) 6
17
Q.11 Round up the following upto three significant figures : (i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Ans. (i) 34.2 (iii) 0.04597
(ii) 10.4 (iv) 281
Q.12 How many significant figures should be present in the answer of the following calculations
Ans.
(i)
0.02856 × 29815 . × 0.112 0.5785
(ii)
5×5.364
(iii)
0.0125 + 0.7864 + 0.0215
(i)
The least precise figure (0.0112) has 3 significant figures. Therefore, the answer should have three significant figures.
(ii)
The second figure (5.364) has 4 significant figures. Therefore, the answer should be reported upto four significant figures. The exact figure (5) is not considered in this case.
(iii)
In this case, the least precise figures (0.0125 and 0.0215) have 3 significant figures. Therefore, the answer should be reported upto three significant figures.
Q.13 When do zeros present in a number become insignificant ? Ans. The zeros written to the left of the first non-zero digit in a number are insignificant. For example, in the number 0.014, both the zeros are insignificant. Q.14 Is velocity a basic or derived quantity according to SI system ? Ans. Velocity is a derived quantity and it depends upon two basic quantities i.e., distance and time. The SI units of velocity are : Velocity = Distance / Time = ms–1
18
Q.15 Why is air not always regarded as homogeneous mixture ? Ans. Air which contains certain suspended particles such as dust particles is a hetrogeneous mixture and not a homogeneous mixture. Q.16 Statements given below pertain either to an element or mixture or compound. Identify the statement which corresponds to (a) an element (b) mixture (c) compound. (i)
The properties of the reactants are entirely different from the properties of the products of their chemical reactions.
(ii)
The constituents retain their individual chemical identity.
(iii)
The pure substance which cannot be subdivided into two or more substances by any chemical means.
Ans. (i)
The statement corresponds to a compound.
(ii)
The statement corresponds to a mixture.
(iii)
The statement corresponds to an element light.
Q.17 What do you understand by limnochemistry ? Ans. Chemistry of water reservoirs like rivers, lakes, etc. is called Limnochemistry. In this branch, we study the variation of pH with seasons and temperature. Moreover, we also study reactions occurring in aqueous medium of water reservoir. Q.18 What do you understand by Phytochemistry ? Ans. Photochemistry is the branch of chemistry which involves the study of chemical composition of various parts of plants.
19
Lecture-2 1.4 Laws of Chemical Combinations 1. Law of Conservation of Mass This law was stated by the french chemist ‘‘antoine Laurent lavoisier (1774)". This law states that: During any physical or chemical changes, the total mass of the products remains equal to the total mass of the reactants. Lavoisier showed that when mercuric oxide was heated it produced free mercury and oxygen. The sum of masses of mercury and oxygen was found to be equal to the mass mercuric oxide Heat
Mercuric oxide → Mercury + Oxygen 100 g
92.6 g
7 .4 g
Law of conservation of mass is also known as law of indestructibility of matter.
Example: Is law of conservation of mass always valid ? Solution No, it is not valid for nuclear reactions. In these
reactions, a certain amount of mass gets converted into energy known as nuclear energy. Therefore, mass is not conserved and it does not remain constant.
2. Law of Constant composition or Definite Proportions This law deals with the composition of chemical compounds. It was discovered by the french chemist, Joseph Proust (1799). This law states that: A chemical compound always contains same elements combined together in same proportion by mass. 20
It implies that in a chemical compound the elements are present in fixed proportion not i.e., arbitrary ratio by mass. For example, pure water obtained from different sources such as, river, well, spring, sea, etc., always contains hydrogen and oxygen combined together in the ratio 1: 8 by mass. Similarly, carbon dioxide can be obtained by different methods such as: (a) burning of carbon, (b) heating limestones (c) the action of dilute hydrochloric acid on marble pieces. It can be shown experimentally that different samples of carbon dioxide contain carbon and oxygen in the ratio 3 : 8 by mass.
Example : When is the law of definite proportions not obeyed ? Solution : The law of definite proportions is not obeyed when the element exists in isotopic forms. For example, in HCl the elements H and Cl may be in the ratio 1 : 35 and 1 : 37 by mass in case 35 C and 37 Cl isotopes are considered.
3. Law of Chemical Combination Law of Multiple Proportions: When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other, bear a simple ratio to one another. Compound of Sulphur and Oxygen : Like carbon, the element, sulphur, also forms two oxides, sulphur dioxide and sulphur trioxide. In sulphur dioxide, 32 parts by mass of sulphur combine with 32 parts by mass of oxygen but in case of sulphur trioxide, 32 parts by mass of sulphur combine with a fixed mass of sulphur combine with 48 parts by mass of 21
oxygen. Therefore, the masses of oxygen which combine with a fixed mass of sulphur (32 parts) in the two oxides are 32 and 48 respectively. These bear a simple ratio of 32 : 48 or 2 : 3 to each other.
Example : Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is M 3 O 4 , find that of the second.
Solution: In the first oxide, oxygen = 27.6, metal = 100 — 27.6 = 72.4 parts of mass. As the formula of the oxide is M 3 O 4 , it means 72.4 parts by mass of metal=3 atoms of metal In the second oxide, oxygen = 30.0 parts by mass and metal = 100—30= 70 parts by mass. But 72.4 parts by mass of metal = 3 atoms of metal So, 70 parts by mass of metal = (3/72.4)×70 atoms of metal = 2.90 atoms of metal Also, 27.6 parts by mass of oxygen = 4 atoms of oxygen So, 30 parts by mass of oxygen = (4/27.6)×30 atoms of oxygen = 4.35 atoms of oxygen Hence, ratio of M : O in the second oxide = 2.90 : 4.35 = 1 : 1.5 = 2 : 3 Therefore, formula of the metal oxide is M 2 O 3 .
22
4. Law of Reciprocal Proportions The ratio of the masses of two elements A and B which combine separately with a fixed mass of the third element C is either same or some simple multiple of the ratio of the masses in which A and B combine directly with each other. The elements H and O combine separately with the third element S to form H 2 S and SO 2 and they combine directly with each other to form H 2 O as shown in figure.
The masses of H and O which combine with the fixed mass of S, viz, 32 parts are 2 and 32, i.e., they are in the ratio 2 : 32 or 1 : 16. When H and O combine directly to form H 2 O, the ratio of their combining masses is 2 : 16 or 1 : 8. The two ratios are related to each other as
1 1 : = 1 : 2 i.e., they 16 8
are simple multiple of each other.
Example: Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and 11.10% of hydrogen. Show that these data illustrate the law of reciprocal proportions. 23
Solution: In NH 3 , 17.65 g of H combine with N=82.35 g Therefore, 1 g of H combine with N=
82 .35 g = 4.67 g 17 .65
In H 2 O, 11.10 g of H combine with O=88.90 g Therefore, 1 g of H combine with O=
88. 90 g = 8.01g 11.10
Therefore, ratio of the masses of N and O which combine with fixed mass (=1 g) of H = 4.67 : 8.01 = 1 : 1.72. In N 2 O 3 ratio of masses of N and O which combine with each other = 36.85 : 63.15 = 1 : 1.71. Thus, the two ratios are the same. Hence, it illustrates the law of reciprocal proportions.
Ex am ple : Carbondioxide con tains 27.27% of car bon, carbondisulphide contains 15.79% of carbon and sulphurdioxide contains 50% of sulphur. Show that the data is in agreement with law of reciprocal proportions.
Solution: In CS 2 C : S mass ratio is 15.79 : 84.21
24
15.79
C
27.27
CS2 84.21 S
50
CO2
CS2
72.73 O 50
15.79 parts of carbon combine with sulphur = 84.21 ∴
27.27 parts will combine with S =
84.21 × 27 .27 = 145 .434 15 . 79
hence, ratio of S : O is 145.434 : 72.73 i.e., 2 : 1 In SO 2 the ratio of S : O is 1 : 1 Since, the ratio of S : O is a simple whole number ratio, therefore law of reciprocal proportions is proved.
5. Gay Lussac’s Law of Gaseous Volumes When gases react together, they always bear a simple ratio to one another and to the volumes of the products, if these product are also gases, provided all measurement of volumes are done under similar conditions of temperature and pressure. Consider, for illustration, the following example: Combination between hydrogen and chlorine. One volume of hydrogen and one volume of chlorine always combine to form two volumes of hydrochloric acid as:
25
H2
1vol. 1vol.
+
Cl2
1vol. +
2HCl 2vol.
1vol.
2vol.
6. Avogadro’s Hypothesis/ Law/ Principle It may be stated as follows: ''Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.'' On applying Avogadro’s law to the reaction between hydrogen and chlorine, Hydrogen + Chlorine → Hydrochloric acidgas 1 Vol.
1 Vol.
2 Vol.
Applying Avogadro’s law n moleculesof + n molecules → 2 n moleculesof hydrogen
of chlorine
hydrochloricacidgas
1 molecule + 1 molecule → 2 molecules 1 1 molecule + molecule → 1 molecule 2 2 It means that one molecule of hydrochloric acid gas contains
1 2
1 molecule of chlorine. Since a 2 molecule may contain more than one atoms, it is possible to divide it into atoms. Thus, Avogadro’s law is not in conflict with Dalton’s Atomic Theory.
molecule of hydrogen and
26
Ex. Which law co-relates the mass and volume of a gas ? Sol. It is the Avogardo’s Law and states that equal volumes of
all gases under similar conditions of temperature and pressure contain equal number of moles (or molecules).
Add to your Knowledge Avogadro’s law is also helpful in developing the relationship between (a) Molecular mass and vapour density (b) Mass and volume of gas (a) Relationship between molecular mass and vapour density By definition Vapour density (V.D.) =
=
=
=
Density of gas Density of hydrogen Mass of some volume of gas at S.T.P. Mass of same volume of H 2 at S.T.P. Mass of N molecules of gas Mass of N molecules of H 2 Mass of 1 molecule of gas Mass of 1 H 2 molecule
27
= = Thus,
Mass of 1 molecule of gas 2 × Mass of H atom 1 × Molecular mass of gas 2
Mol.mass=2 × V.D.
(b) Relationship between mass and volume of gas Mol. mass
= 2 × V.D. =
=
2 × mass of 1 L of gas at S.T.P.(1atm) Mass of 1 L of H 2 at S.T.P. 2 × Mass of 1 L of gas at S.T.P. 0.089 g
= 22 .4 × MAss of 1 L of gas at S.T.P. = Mass of 22.4 L of gas at S.T.P. Thus,
Mass of 22.4 L of gas at S.T.P.
= Molar mass (in g mol −1 )
1.5 Dalton’s Atomic Theory The main points of this theory are as follows: 1.
Matter is made up of extremely small invisible particles called atoms.
2.
Atoms of the same element are identical in all respects, i.e., size, shape, and mass.
28
3.
Atoms of different element have different masses, sizes and also possess different chemical properties
4.
Atoms of the same or different elements combine together to form compound atoms (now called as molecules).
5.
When atoms combine with one another to form compound atoms (molecules), they are in simple whole number ratios, such as 1 : 1, 2 : 1, 2 : 3 and so on.
6.
Atoms of two elements may combine in different ratios to form more than one compound. For example, sulphur combines with oxygen to form sulphur dioxide and sulphur trioxide, the combining ratios being 1 : 2 and 1 : 3 respectively.
7.
An atom is the smallest particle that takes part in a chemical reaction. In other words, whole atoms, rather than fractions of atoms take part in a chemical reaction.
Explanation of the Laws of Chemical Combination by Dalton’s Atomic Theory: 1.
Law of Conservation of Mass. Matter is made up of atoms (postulate 1) which can neither be created nor destroyed (postulate 8). Hence, matter can neither be created nor destroyed.
2.
Law of Constant composition. It follows directly from postulate 5.
3.
Law of Multiple proportions. As follows directly from postulate 6.
4.
Law of Reciprocal proportions. As atoms combine with each other in simple multiple of each other. 29
Limitations of Dalton’s Atomic Theory The main drawbacks of Dalton’s Atomic Theory are : (i) It could explain the laws of chemical combination by mass but failed to explain the law of gaseous volumes. (ii) It could not explain why atoms of different elements have different masses, sizes, valencies etc. (iii) It could not explain, why do atoms of the same or different elements combine to form molecules? (iv) It could not explain, what is the nature of binding force between atoms and molecules which accounts for the existence of matter in three states, i.e., solids, liquids and gases. (v) It makes no distinction between the ultimate particles of an element or a compound.
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Questions Q.1
Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below :
1 Pa = 1 Nm −2
If the mass of air at sea level is 1034 g, calculate the pressure in pascal. Ans. Pressure = Force = Mass × Accleration of gravity Area Area Mass of air = 1034 g = 1.034 kg Acceleration of gravity = 9.806 ms −2 Area = 1 cm 2 = 10 −4 m 2 Q.2
The following data is obtained when dinitrogen and dioxygen react to gether to form different compounds : Mass of dinitrogen
14 g
14 g
28 g
28 g
Mass of dioxygen
16 g
32 g
32 g
80 g 31
Ans. By keeping 14 g as the fixed mass of dinitrogen (N 2 ), the ratios by mass of dioxygen (O 2 ) combining with 14 g of dinitrogen are : 16 : 32 : 16 : 40 or 2 : 4 : 2 : 5. Since this ratio is simple whole number, the data obeys the Law of Multiple proportions. Q.3
If the speed of light is 3.0×10 8 ms −1 , calculate the distance covered by light in 2.00 ns.
Ans. Distance travelled by light in 1 s = 3.0 × 10 8 m Distance travelled by light in 2.0 × 10 −9 s = (3.0 × 10 8 m)×(2.0 × 10 −9 s) / (1s) = 0.600 m. Q.4
Convert the following into basic units (i) 28.7 pm
Ans.
(i)
(ii) 15.15 µs
(iii) 25365 mg.
1 pm = 10 −12 m 28.7 pm = (10 −12 m)× (28.7 pm) / (1.0 pm) = 2.87×10 −11 m
(ii)
1µs = 10 −6 s 15.15 µs = (15.15)× (10 −6 s) / (1.0 µs) = 1.515×10 −5 s
(iii)
1 mg = 10 −6 kg 25365 mg = (25365 mg)× (10 −6 kg) / (1mg) = 2.5365×10 −2 kg
Q.5
We breathe in fresh air in the morning walk ? Is it pure as well ?
Ans. Fresh air may be regarded as pure by an ordinary person but not by a chemist. Actually fresh air is the mixture of number of gases like oxygen, nitrogen, noble gases, carbon dioxide, water vapours etc. It is therefore, not pure from the angle of a chemist.
32
Lecture-3 1.6 Atomic and Molecular Mass The atomic mass of an element can be defined as the number which indicates how many times the mass of one atom of the element is heavier in comparison to the mass of one atom of hydrogen. Atomic mass of an element also can be defined as the number which indicates how many times the mass of one atom of the element is heavier in comparison to 1/12 th part of the mass of one atom of carbon-12( 12 C). A = Atomic mass of an element Massofoneatom of the element
= 1 th part of themassof oneatom of carbon − 12 12 A=
Massof oneatom of the elements Massof oneatom of carbon − 12
× 12
Ex:
Calculate the molar mass of glucose (C 6H 12 O6 ) and the number of atoms of each kind in it.
Sol:
Molecular mass of glucose (C 6H 12 O6 ) = 6 × ( 12 .011u) + 12 × ( 1.008u) + 6 × ( 16.00u) = 180.162 u Calculation of number of atoms of each kind 1 mole of glucose (C 6H 12 O6 ) = 6 moles of carbon + 12 moles of hydrogen +6 moles of oxygen 33
Hence, Atoms of carbon
= 6 × 6.02 × 10 23 = 36.12 × 10 23
Atoms of hydrogen
= 12 × 6.02 × 10 23 = 72 .24 × 10 23
Atoms of oxygen oxygen = 6 × 6.02 × 10 23 = 36.12 × 10 23
Atomic Mass Unit:
The quantity of 1/12th mass of an atom of carbon-12 ( 12 C) is known as the atomic mass unit and is abbreviated as amu. The actual mass of one atom of carbon-12 is 1.9924×10 −23 g or 1.9924×10 −26 kg.
(i) Atom: The smallest particle of an element that can take part in chemical change but generally cannot exist freely as such.
(ii) Molecule: The Smallest particle of a substance (element or compound) which has free or independent existence and possess all characteristics properties of the substance.
(iii) Molecular Mass: Molecular mass of a substance may be defined as: The average relative mass of its molecule as compared to the mass of an atom of carbon(C 12 ) taken as 12.
Ex-1: Atomic mass of Ca is 40 gm/mole, means 1 mole of Ca & having mass = 40 g.
Sol:
In 1 mole number of Ca = 6.023 × 10 23 . So, 6.023 × 10 23 atoms of Ca having mass = 40 g.
34
1 atom of Ca having mass = = 40
1 23 g 6.023 × 10
40 g 6.023 × 10 23
= 40 × 1.67 × 10 −24 g = 40 amu
1 is constant for all, 6.023 × 10 23 means 1.67 × 10 −24 gm 40 AMU means, mass of one atom of Ca So, 1 AMU = 1.67 × 10 −24 gm = 1.67 × 10 −27 kg
Ex-2: Find out which is having minimum mass. (a) (b) (c) (d)
Sol:
10 mole H 2 10 g H 2 10 amu H 2 All same
(c) i.e., 10 × 1.67 × 10 −24 gm
(iv) Vapour density (v.d): It is the density of a gas or vapour w.r.t. H 2 at constant pressure and temperature and given as mol. wt . of gas v . d. =
2 mol. wt . of H 2
2 let, mol. wt. of gas = M mol. wt. of H 2 = 2 35
M v . d.= 2 2 2 M v . d.= 2 it is unit less
Ex-3: Find out v.d. of O 3 , O 2 and CO 2 . Sol: v . d. =
Mol. wt . 2
48 = 24 2 32 v . d. of O 2 = = 16 2 44 v.d. of CO 2 = = 22 2 v . d. of O 3 =
Ex-4: V.d. of gas is 40. Find out molecular weight. Sol: V.d. =
Mol. wt . 2
Mol. wt. = 2 × v . d. = 2 × 40 = 80
Ex-5: v.d. of sulphur vapour is 64. Find out formula of sulphur vapour.
Sol:
We know that sulphur vapour is represented asS , S2 , S4 , S6 , S8 . First of all find out molecular wt. of sulphur vapour.
36
Mol. wt . 2 mol. wt . 64 = 2 mol. wt. = 2 × 64 = 128 mol. wt. = 128 Mol. wt. = No. of sulphur × wt. of one S Let, formula is S x 128 = 32 × x or, x = 128 / 32 = 4 S 4 = Formula of vapour v . d. =
Ex-6: If v.d. of gas is 10(O 2 = 4). Find out molecular weight of the gas.
Sol:
v.d. of gas is 10. When we assume that v.d. of oxygen is 4.
But, Real v.d. of O 2 is 32/2 = 16 mean, real v.d. = 16 given v.d. is = 4 We have to find out the factor between actual and given v.d. give v.d. ×F = Real v.d. Re al v . d. F= given v . d 16 =4 4 means, for the given gas, Real v.d. = F × given v.d. Real v.d. = 4 × 10 = 40 So, mol. wt. = 2 × v . d. = 2 × 40 = 80 F=
37
1.7 Empirical Formula and Molecular Formula Empirical formula: The formula which gives the simplest whole number ratio of the atoms of various elements present in one molecule of the compound is called empirical formula.
Example : Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass (At. mas Fe=56, O=16). Ans. Fe2 O 3 Element Symbol %age Iron
Fe
Oxygen
O
69.9
30.1
At. Rel. no. mass of atoms 56
16
Simplest ratio
Whole no. ratio
699 56
1.248 1.248
1×2
=1.248
=1
30.1 16
1881 1224
=1.881
=1.50
=2 1.5×2 =3
Molecular formula: The formula which gives the actual
number of various elements present in one molecule of the compound is called molecular formula. Percentage Composition: Mass %age of an element =
Massof that element in thecompound × 100 Molar massof thecompound
Mass %age of H in H 2 O =
2 × 1.008 × 100 = 11.18% 18.02
Mass %age of O = 100 − 11.18 = 88. 79% 38
1.8 Stoichiometry of Chemical Reaction When chemical equations is written in the balanced form, it gives quantitative relationships between the various reactants and products in terms of moles masses, molecules and volumes. This is called stoichiometry (Greek word meaning ‘ to measure an element’). The coefficients of the balanced chemical equation are called stoichiometric coefficients. For example, a balanced chemical equation along with the quantitative information conveyed by it is given below: CaCO 3
+
1mole 40 × 12 + 3 × 16 = 100 gm
2HCl → CaCl 2 + H 2 O + CO 2 1moles 1moles 1mole 1mole 2( 1 + 35 .5 ) 40 + 2 × 35 .5 12 + 16 × 32 = 72 g = 111g = 18g
The problems involving these calculations may be classified into the following different types: 1. Volume: Volume Relationships, i.e.,volume of one of the reactants or products is given and the volume of the other is to be calculated.The general method of calculations for all the problems of the above types consists of the following steps: (1) Balance the reaction (2) Find out cofficient (3) Calculate according to cofficient
39
Note:
Molal solution means m = 1 Deci. molal solution means m =
1 10
Centi. molal solution means m =
1 100
2. Atom: It is the smallest part of the compound which is neutral and having following properties. (1) It is having three fundamental particle. (a) Protons (b) Neutrons (c) Electrons (2) Generally represented as z X A . Where Z = No. of protons (P) or atomic No. A = P + N = Atomic mass 3. Molecule: It is the combination of atoms and having neutral in nature.
Example: O 2 , N2 , Cl 2 , CO 2 4. Atomic Weight: It is the weight of 6.023 × 10 23 atoms of the given.
Example: If weight of 3.0115 × 10 20 atoms is 20mg then find out atomic weight of the atom.
Solution: Since 3.0115 × 10 20 atoms having wt. = 20 mg 20 × 10 −3 gm 1 atom is having weight = 3.0115 × 10 20 Hence, 6.023 × 10 23 atoms are having weight 20 × 10 −3 23 . = × 6 023 × 10 3.0115 × 10 20 = 20 × 10 −3 × 2 × 10 3 = 40 gm/mole 40
5. Molecular Weight: It is the weight of 6.023 × 10 23 molecule of the given.
Example: If Weight of 0.06023 × 10 18 molecules of a gas is 1 µg , then find out the molecular weight of the given gas.
Solution: Since, 0.06023 × 10 18 molecule of gas is having Weight = 1µg 1 × 10 −6 gm 1 molecule is having weight = 0.06023 × 10 18 Hence, weight of 6.023 × 10 23 molecule of gas 1 × 10 −6 23 = 18 × 6.023 × 10 0.06023 × 10 = 1 × 10 −6 × 10 7 = 10 gm/mole
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Questions Q.1
What will be the mass of one
12
C atom in g ?
Ans. 1 mole of 112 C atoms = 6.022×10 23 atoms = 12g Thus, 6.022×10 23 atoms of
12
C have mass = 12g
1 atom of 12 C will have mass = 12 / 6.022×10 23 g = 1.9927×10 −23 g. Q.2
Which one of the following will have largest number of atoms ? (i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g Cl 2 (g)
(Atomic masses : Au = 197, Na = 23, Li = 7, Cl = 35.5 amu) Ans. (i) (iii) (iii) (iv) Q.3
1 1 = mol = × 6.02 × 10 23 atoms 197 197 1 1 1gNa = mol = × 6.02 × 10 23 atoms 23 23 1 1 1gLi = mol = × 6.02 × 10 23 atoms 7 7 1 1 2 1gCl2 = mol = × 6.02 × 10 23 molecules = × 6.02 × 10 23 atoms 71 71 71 1gAu
How many moles of methane are required to produce 22g CO 2 (g) after combustion ?
Ans. The balanced chemical equation for the combustion of methane is; CH 4 +
2O 2
→
CO 2
+
1 mole
2H 2 O 1 mole 12 + 2 ×16 = 44g
Thus, to produce 44 g of CO 2 , CH 4 required = 1 mole Therefore, to produce 22g of CO 2 , CH 4 required = (1/44)×22 mole = 0.5 mole. Q.4
How much copper can be obtained from 100g of copper sulphate (CuSO 4 )?
Ans. Molecular mass of CuSO 4 = Atomic mass of Cu + Atomic mass of S + 4× Atomic mass of O = 63.5 + 32 + 4 ×16 = 159.5 u. Gram molecular mass of CuSO 4 = 159.5 g Now, 159.5 g of CuSO 4 have Cu = 63.5 g ∴ 42
100 g of CuSO 4 have Cu = (63.5 g) × 100 g / 159.5 g = 39.81 g
Q.5
Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.
Ans. Empirical formula of the oxide of iron = Fe 2 O 3 Molecular formula of the oxide of iron = n× Empirical formula = 1× (Fe 2 O 3 ) = Fe 2 O 3 (Since there in no common factor in Fe 2 O 3 , therefore n = 1 ). Q.6
Calculate the average atomic mass of chlorine from the following data: Isotope
% Natural Abundance
Atmic mass
35
Cl or Cl–35
75.77
34.9689 u
37
Cl or Cl–37
24.23
36.9659 u
Ans. Average atomic mass of Chlorine = 75.77 × 34.9689 u+24.23 × 36.9659 u (75.77 + 24.23) = Q.7
2649.59 u + 895.68 u = 35.45 100
Nitrogen and hydrogen react to form ammonia according to the reaction N 2 (g) + 3H 2 (g) → 2NH 3 (g) If 1000 g of H 2 react with 2000 g of N 2 , (i)
Will any of the two reactants remain unreacted ? If yes, which one and what would be its mass ?
(ii)
Calculate the mass of ammonia (NH 3 ) which will be formed.
Ans. N 2 (g) 28g
+ 3H 2 (g)
→
3×2 = 6g
2NH 3 (g) 2×17 = 34 g
According to available data, 28 g of N 2 require H 2 = 6 g 2000 g of N 2 require H 2 = 6 g × (2000 g) / (28 g) = 428.6 g But H 2 actually available = 1000 g This means that H 2 is in excess and will remain unreacted. (i)
Mass of H 2 that remains unreacted = 1000 – 428.6 = 571.4 g
(ii)
Mass of NH 3 formed may be calculated as follows : 6 g of H 2 will form NH 3 = 34 g 428.6 g H 2 will form NH 3 = (34 g)×(4280.6 g) /(6.0 g) = 2428.8 g 43
Q.8
Which of the following has largest number of atoms ? (i) 1 g of Au
(ii) 1 g of Na
(iii) 1 g of Li
(iv) 1g of Cl 2
Ans. (i)
197 g of Au have atoms = 6.022×10 23 (Gram atomic mass of Au = 197 g) (1g) 1 g Au has atoms = 6.022 × 10 23 × ∴ = 3.06 × 10 21 atoms (197g)
(ii)
23 g of Na have atoms = 6.022×10 23 (Gram atomic mass of Na = 23 g) (1g) 1 g Na has atoms = 6.022 × 10 23 × ∴ = 2.62 × 10 22 atoms (23g)
(iii)
7 g of Li have atoms = 6.022×10 23 (Gram atomic mass of K = 39 g) (1g) 1 g Li has atoms = 6.022 × 10 23 × ∴ = 8.60 × 10 22 atoms (7g)
(iv)
71 g of Cl 2 have molecules = 6.022×10 23 (Gram molecules mass of Cl 2 = 71 g) 71 g of Cl 2 have atoms = 2×6.022 ×10 23 1 g of Cl 2 has atoms = 2 × 6.022 × 10 23 ×
∴
(1g) = 167 . × 10 22 atoms (71g)
Thus, 1 g of lithium (Li) has the largest number of atoms. Q.9
What will be mass of one 12C in g ?
Ans. 6.022×10 23 atoms of carbon have mass = 12 g 1 atoms of carbon have mass =
(12g) × (1atom) (6.022 × 10
23
atom)
= 1993 . × 10 −23 g
Q.10 Use the data given in the following table to calculate the molar mass of naturally occurring argon, Isotope
Isotopic molar mass
Abundance
36
Ar
35.96755 g mol −1
0.337 %
38
Ar
37.96272 g mol −1
0.063 %
40
Ar
39.9624 g mol −1
99.600 %
Ans. Molar mass of argon in the average molar mass and may be calculated as:
44
=
(0.337) × (35.96755) + (0.063) × (99.6) × (39.9624) (00.337 + 0.063 + 99.600)
=
12.121 × 2.3916 + 3980.26 = 39.947 g mol –1 100
Q.11 A welding fuel gas contains carbon and hydrogen only, Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g water and no other products. A volume of 10.0 L (measured at NTP) of this welding gas is found to weight 110.6 g. Calculate (i) empirical formula (ii) molar mass (iii) molecular formula of the gas. Ans. Step-1. Calculation of mass percent of carbon and hydrogen. CO 2 ≡ C 44g
12g
3.38 g =
12 g × 3.38 g = 0.9218g of C 44 g
H 2 O ≡ 2H 18g
2g
0.690 g =
2g × 0.69 g = 0.0766 g of H 18g
Total mass of fuel gas = 0.9218 + 0.0766 = 0.9984 g Percentage of carbon =
Mass of carbon × 100 Mass of fuel gas
=
0.9218 g × 100 = 92.33% 0.9984 g
Percentage of hydrogen = =
Mass of hydrogen × 100 Mass of fuel gas
0.0766 g × 100 = 7.67% 0.9984 g
Step-2. Determination of empirical formula of fuel gas.
Element no. ratio C
%
Atomic mass
Gramn atoms (moles)
Atomic ratio (molar ratio)
Simplest whole
92.33
12
92.33 = 7.69 12
7.69 =1 7.69
1
Empirical formula of the fuel gas = CH. Step-3. Calculation of molecular mass of the fuel gas. 10.0 L of the fuel gas at NTP weigh = 11.6 g 45
10.0 L of the fuel gas at NTP weigh = ∴
116 . × 22.4 = 25.98 g 10.0
Molecular mass of the fuel gas = 25.98 g = 26.0 g = 26 amu
Step-4. Calculation of molecular formula of the gas. Empirical formula mass = 12 + 1 = 13 amu Molecular formula = 26 amu n= ∴
Molecular mass 26 = =2 Empirical formula mass 13
Molecular formula = n × Empirical formula =2 × CH = C 2 H 2
Q.12 What is the difference between the mass of a molecule and molecular mass ? Ans. Mass of a molecule is that of a single molecule also known as its actual mass. But molecular mass is the mass of Avogardo’s number (6.022×10 23 ) of molecules. Q.13 The average atomic mass of chlorine is 35.5 amu. Do we come across a sample of the element with atomic mass 35.0 amu ? Ans. No, it is not possible. The fractional atomic mass of an element is its average mass and not the actual mass. In this case, the element chlorine exists as two isotopes with atomic mass 35 amu and 37 amu respectively in the ratio of 3 : 1. The average comes out to be fractional i.e., 35.5 amu. Q.14 Are the atomic masses of the elements their actual masses ? Ans. No, atomic masses of the elements are not actual masses. These are only relative masses because the actual masses are very small. For detail, consult text part. Q.15 Give an example of a molecule in which the ratio of the molecular formula is six times the empirical formula. Ans. The compound is glucose. Its molecular formula is C 6 H12 O 6 while empirical formula is CH 2 O. Q.16 How many oxygen atoms are present in 96 amu of ozone ? Ans. 96 amu of ozone (O 3 ) contain
46
96 amu = 6 atoms of oxygen. 16 amu
Lecture-4 1.9 Mole Concept > > >
It is denoted by ‘n’ 1 mole of a gas at NTP is having volume 22.4 L 1 mole of a compound is having weight equal to it’s molecular weight. So on the basis of different concept mole having following types.
Example: 20 gm of He sample is having moles. Solution:
So,
n= n=
weight of sample atomic weight of He 20 =5 4
Mole of molecule sample Mole(n) =
weight of sample Molecular weight
Example: 880 gm sample of CO 2 gas is there. Find out moles of the gas.
Solution:
Mole(n)= n=
weight of sample Molecular weight 880 = 20 44
If number of particles, ions atom or molecules are given then: Mole(n) =
Number of Particles No.(6.023 × 10 23 ) 47
Example: Find out moles of 12.04 × 10 24 atoms of Na. Solution:
12 .04 × 10 24 1 n= = 2 × 10 = 20 moles. 23 6.023 × 10 If volume at NTP is given for a gas then moles =
volumeof gasat NTP 22 .4 L
Example: Find out moles of 448 L of O 2 gas at NTP. Solution: Moles(n)= 448 = 20 22 .4
Example: How many atoms of oxygen are present in 300 g of CaCO 3 ?
Solution: Molecular mass of CaCO 3 in grams = 100 g Now, 1 mole of CaCO 3 contain = 3 mole of O atoms. or
100 g of CaCO 3 contain = 3 × 6.02 × 10 23 Oxygen atoms ∴ 300 g of CaCO 3 contain O atoms 3 × 6.023 × 10 23 = × 300 = 54.207 × 10 23 100 = 5.4207 × 10 24 Oxygen atoms.
Note:
Alternative short method is givenin video lecture
Example: Chlorophyl, the green colouring material of plants contains 2.68% of magnesium by mass. Calculate the number of magnesium atoms in 5.00 g of this complex.
48
Solution: Mass of magnesium in 5.00 g of complex is =
2 .68 × 5 .00 = 0.134 g. 100
Gram atomic mass of magnesium = 24 24 g of magnesium contain = 6.02 × 10 23 atoms 0.134 g of magnesium would contain 6.02 × 10 23 = × 0.134 = 3.36 × 10 21 atoms 24 Therefore, 5.00 g of the given complex would contain 3.36 × 10 21 atoms of magnesium. Note:
Alternative short method is given in video lecture
Types of Problems (1) Problem related with mass and the volume of gas sample. Example: Find out volume of H 2 at NTP if 48g of sample is there. Solution: Moles of H 2 Moles (n)= n= from (i) n =
Weight of sample Molecular weight Volumeof H 2 at NTP 22 .4
...(i) ...(ii)
48 = 24 from 48 gm 2 49
So, n = 24 = V= 24 × 22 .4
V 22 .4 or
V = 537 .6 Litre
Example: If 448 litres of CO 2 gas sample is there at NTP then, find out mass of CO 2 sample.
Solution: Moles of the CO 2 at NTP n=
448 = 20 22 .4
mass of the CO 2 gas weight of CO 2 n= Molecular weight
or
n = 20
or
n=
x 44
x = 880 gm
(2) Problems related with number of atoms or molecules Example: If 300 gm CaCO 3 is there then find out number of atoms in CaCO 3
Solution: To find number of atoms in CaCO 3 first of all. Find out No. of atoms in one CaCO 3 then number of CaCO 3
Toal atom = No. of CaCO 3 × No. of atoms in one CaCO 3 Moles of CaCO 3 (n) = n= Number of CaCO 3 =? 50
Weight of CaCO 3 Molecular weight of CaCO 3
300 100
or
n=3
n= 3=
Number of CaCO 3
N0(6.023 × 10 23 )
Number of CaCO 3 N0
Number of CaCO 3 = 3N0 Number of atoms in one CaCO 3 = 5 CaCO 3 = Ca + C + 3.0 1
1
3=5
So, total number of atoms in CaCO 3 = No. of CaCO 3 × No. of atoms in one CaCO 3 = 3N0 × 5 = 15 N0 = 15 × 6.023 × 10 23
or
9.0345 × 10 24
(3) Problems related with charge on the ions and number of electrons and protons Example: If 540 gm of Al +++ is there then find out number of electrons and charge on Al +++ sample.
Solution: One Al +++ is having charge = 3e− (+ve charge) in Al number of electrons = 13 So, in Al +++ number of electrons = 10 First of all find out number of Al +++ Moles(n)=
Weight 540 = 27 27
Moles(n) = 20
51
Q
Moles(n) =
No. of Al +++ N0
No. of Al +++ = 20N0 Total charge = No. of Al +++ × charge of 1 Al +++ = 20N0 × 3 electron = 60N0 e− = 60 × 6.023 × 10 23 × 1.6 × 10 −19 = 578.2 × 10 4 coulomb = 5 . 78 × 106 coulomb No. of electron in one Al +++ = 10 So, total electron = No. of Al +++ × electron in 1 Al +++ 20N0 × 10 = 200N0 .
Example: If a gas sample of CO −− 3 at NTP is having 120 gm weight, then find out following: (a) Volume of gas sample at NTP. (b) Number of atoms in the sample. (c) Number of electrons. (d) Number of protons. (e) Number of charge.
Solution: First of all find out moles of the substance and sample.
n= 52
Weight of sample Molar mass
n=
120 60
n = 20
(a) Volume at NTP, We know that, moles(n) = 2=
V 22 .4 L V 22 :4
V = 2 × 22 .4 = 44.8 Litre (b) Moles(n) =
Number of CO 3−− N0
Number of CO 3−− = 2N0 (N0 = 6.023 × 10 23 ) In this number of atoms = Total number of CO 3−− is having atoms = 4 = 2N0 × 4 = 8N0 because one CO 3−− is having atoms = 4 (c) In one CO 3−− Total number of electrons In
C
=
6
electrons
In
O
=
8
electrons
So. total electrons in, CO 3−− = 6 + 8 × 3 + 2 (free negative charge) 53
Total electrons = 32 in one CO 3−− Number of CO 3−− = 2N0 So, total number of electrons = 2N0 × 32 = 64 N0 (d) Number of protons, In one CO 3−− Total number of protons In C = 6 In O = 8 So. total electrons in,
protons protons
CO 3−− = 6 + 8 × 3 = 6 + 24 = 30 Total number of protons = total number of CO 3−− × protons in one CO 3−− = 2N0 × 30 = 60N0 (e) Charge on one CO 3−− q = 2 e− where e− = 1.6 × 10 −19 coulomb total charge = total number of CO 3−− Q = 2 N × 2 e− = 4 e−N0 total charge. 54
1.10 Limiting Reagent Many a time, the reactions are carried out when the reactants are not present in the amounts as required by a balanced chemical reaction. In such situations one reactant is in excess over the other. The reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction Hence the reactant, which gets consumed, limits the amount of product formed and is therefore, called the limiting reagent.
Example: If a reaction is as, A Ratio is
+
1
2B
→
2
C 1
So, if we use 40 moles of A then 80 moles of B should be there to produce 40 moles of C
Example: For, 2A + B → 2C, in the reaction if 40 moles of A reacts with 60 moles of B then find out limiting reactant.
Solution: Given moles for the reaction 2A +
B → 2C
40
60
0
initial.
If we use 1 mole of B then 2 moles of A should be. Means if we use 60 moles of b then A should be 120. Which is 40, therefore, A is less in amount hence, A is limiting Reagent So, reaction is according to it's amount. In this reaction proper ratio for the production is, 55
2A +
B → 2C
2
1
2
So, 2X
X
2X
40
20
40
B is used only 20 so remaining B is = 60 - 20 = 40.
Finding formula of hydrocarbon on the basis of percentage. Hydrocarbons are of four types mainly. 1.
Alkane
2.
Alkene
3.
Alkyne
4.
Benzene ring
Alkane:
General formula C n H 2n + 2 Where, n = 1, 2, 3,........
if n = 1 we get, CH 4 mol. wt. = 16 So, general mol. wt. of hydrocarbon, C n H 2n + 2 atomic weight of C = 12, Atomic weight of H = 1. = 12 n + 1( 2 n + 2) = 12 n + 2 n + 2 = 14n + 2
Alkene:
General formula C n H 2n where, n = 1, 2, 3,........ So, general mol. wt. of hydrocarbon C n H 2n = 12 n + 1( 2 n) = 14n
56
Alkyne:
General formula C n H 2n − 2 where, n = 1, 2, 3,....... So, general mol. wt. of hydrocarbon C n H 2n − 2 = 12 n + 1( 2 n − 2) = 12 n + 2 n − 2 = 14n − 2
Example: If in a compound percentage of carbon is 80% and hydrogen is 20% then find out formula of hydrocarbon. If molecular weight is 30.
Solution: C = 80%
H = 20%
% of mass
moles
mole ratio
C
80
80/12
1
H
20
20/1
3
So, empirical formula is CH 3 Mol. wt. = 30
Empirical wt. = 15
f =
mol. wt . Emperical wt .
f =
30 =2 15
Molecular formula = Empirical formula × f = CH 3 × 2 = C 2 H 6
57
Alternative Method For a hydrocarbon molecular weight of the hydrocarbon may be 14n + 2 14n 14n — 2 So, mol. wt. should be according to these three. 14n + 2 = 30
or
14n = 28
n=2
14n = 30 14n — 2 = 30
or
14n = 32
means, only first equation is satisfy the mol. wt. So, it is alkane having n = 2 i.e., C n H 2n + 2 or C 2 H 6 . Mole concept used in the reaction By using POAC
1.11 Principle of Atoms Conservation In this POAC, we assume that moles of the atoms in reactant as well as in product side are conserved.
Example: CaCO 3 → CaO + CO 2 For the given reaction moles of Ca in CaCO 3 are same as that of Ca in CaO. First of all find out relation between moles of Ca and moles of CaCO 3 . In
58
CaCO 3 → Ca
+
C+
3O
1
1
1
3
X
X
X
3X
Moles of Ca = moles of CaCO 3 Moles of C = moles of CaCO 3 Moles of O = 3 × moles of CaCO 3 In
CaO
→ Ca
+
O
1
1
1
X
X
X
Moles of CaO = moles of Ca Moles of caO = moles of O In
CO 2 →
C
+
2O
1
1
2
X
X
2X
Moles of C = moles of CO 2 Moles of O = 2 × moles of CO 2 For CaCO 3 → CaO +
CO 2
If we conserve Ca then moles of Ca in CaCO 3 = moles of Ca in CaO nCa = nCaCO 3 nCa = nCaO So,
nCaCO 3 = nCaO Conserving C ⇒ Moles of C in CaCO 3 = Moles of C in CO 2 59
nC = nCaCO 3 | Means,
nC = nCO 2
nCaCO 3 = nCO 2
Conserving O ⇒ Moles of O in CaCO 3 = Moles of O in CO 2 + moles of O in CaO n 0 = 3 × nCaCO 3
|
n 0 = 2 nCO 2
|
n 0 = nCaO
So, 3 × nCaCO 3 = 2 × nCO 2 + 1 × nCaO
Example: If 400 gm CaCO 3 is heated then find out wt. of CaO and CO 2 formed.
Solution: CaCO 3 → CaO + CO 2 For Ca
nCa in CaCO 3 = nCa in CaO 1 × nCaCO 3 = 1 × nCaO X 400 = 100 56 X = 224 g
For C
nC in CaCO 3 = nC in CO 2 1 × nCaCO 3 = 1 × nCO 2 y 400 = 100 44 X = 176 g
60
Combustion Reaction In the combustion reaction POAC is applied.
Example. In the combustion of methane, why is methane regarded as the limiting reagent ?
Sol. Methane (CH 4 ) is regarded as the limiting reactant
because air or oxygen is always present in excess. The amounts of CO 2 and H 2 O formed in the reaction depend upon the amount of methane only. Therefore, it is regarded as the limiting reagent.
Example: CH 4 is having combustion then find out the relation between moles of all.
CH 4
+
So,
For,
O2 → CO 2 C
+ H2O
If POAC is there. 1 × nCH 4 = 1 × nCO 2
For,
H
If POAC is there. 4 × nCH 4 = 2 × nH 2 O
For,
O
If POAC is there. 2 × nO 2 = 2 nCO 2 + 1 × nH 2 O
Example: If 20 moles of a hydro carbon on combustion gives 40 mols of CO 2 and 40 moles of H 2 O, then find out formula of hydro carbon.
Solution: Let the hydro carbon is C x H y. If combustion reaction is as,
C x Hy +
O2 → CO 2 + H 2 O 61
So,
For,
C
If POAC is there x × nC xH y = 1 × nCO 2
For,
H
y × nC x H y = 2 × nH 2 O
x and y are number of C and H used in the hydrocarbon.
For calculation nC xH y = 20
nCO 2 = 40
nH 2 O = 30
X × 20 = 1 × 40
or
X=2
y × 20 = 2 × 40
or
y= 4
So, formula of hydrocarbon is C 2 H 4
Note: If in place of moles we are using volume then. This volume is equivalent to moles.
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Questions Q.1
In three moles of ethane (C 2 H 6 ), calculate the following : (i)
No. of moles of carbon atoms
(ii)
No. of moles of hydrogen atoms
(iii)
No. of molecules of ethane
Ans. (i)
1 mole of C 2 H 6 has moles of carbon atoms = 2 moles 3 moles of C 2 H 6 have moles of carbon atoms = 2× 3 = 6 moles
(ii)
1 mole of C 2 H 6 has moles of hydrogen atoms = 6 moles 3 moles of C 2 H 6 have moles of hydrogen atoms = 6× 3 = 18 moles
(iii)
1 mole of C 2 H 6 has molecules = 6.022 × 10 23
3 moles of C 2 H 6 have molecules = 6.022 × 10 23 × 3 =1.81×10 4 molecules Q.2
Identify the limiting reactant if any in the following reaction mixtures ? A + B → AB 2 (i)
300 atoms of A + 200 molecules of B 2
(ii)
100 atoms of A + 100 molecules of B 2
(iii)
5 moles of A + 2.5 moles of B 2
(iv)
2.5 moles of A + 5 moles of B 2
(v)
2 moles of A + 3 moles of B 2
Ans.
A
+
1 atom, 1mole
B
AB 2
→
1 molecule, 1 mol
1 molecule, 1 mol
In the light of the above information, let us find the limiting reactant if any in all the cases. (i)
1 atom of A will react with molecules of B 2 = 1 300 atoms of A will react with molecule of B 2 = 300 But the molecules of B 2 actually available = 200 ∴
(ii)
B 2 is the limiting reactant.
1 atom of A will react with molecules of B 2 = 1 100 atoms of A will react with molecule of B 2 = 100 The molecules of B 2 actually availiable = 100 ∴
(iii)
There is no limiting reactant in this case.
1 mole of A will react with moles of B 2 = 1 5 moles of A will react with moles of B 2 = 5 63
But moles of B 2 actually availiable = 2.5 ∴ (iv)
B 2 is the limiting reactant.
1 mole of A will react with moles of B 2 = 1 2.5 moles of A will react with moles of B 2 = 2.5 But moles of B 2 actually availiable = 2.5 This shows that 5 moles of A can react whereas only 2.5 moles of A are actually available ∴
(v)
A is the limiting reactant.
1 mole of A will react with moles of B 2 = 1 2 moles of A will react with moles of B = 2 But the moles of B 2 actually availiable = 3
∴ Q.3
A is the limiting reactant.
Calculate the number of atoms present in: (i) 52 moles of He
Ans. (i)
(ii) 52 amu of He
(iii) 52 g of He
1mole of He contain atoms = 6.022 × 10 23 52 moles of He contain atoms = 6.022 × 10 23 ×52 = 3.13 × 10 25 atoms
(ii)
Atomic mass of He = 4 amu ; 4 amu is the mass of He atoms = 1 52 amu is the mass of He atoms = 1/4 × 5 = 13 atoms
(iii)
Gram atomic mass of He = 4 g ; 4 g of He contain atoms = 6.022 × 10 23
52 g is He contain atoms =6.022 × 10 23 /4 = 7.83 × 10 24 atoms Q.4
In the combustion of methane, why is methane regarded as the limiting reactant ?
Ans. Methane (CH 4 ) is regarded as the limiting reactant because air or oxygen is always present in excess. The amounts of CO 2 and H 2 O formed in the reaction depend upon the amount of methane only. Therefore, it is regarded as the limiting reactant. Q.5
Why is limiting reactant so named ?
Ans. Limiting reactant i.e., the reactant present in fixed amount is so named because it limits the participation of other reactants even if present in excess in a particular reaction. Q.6
Is the number of molecules in one mole of a gas at 100°C and 500 mm pressure equal to or less than the Avogardo’s number ?
Ans. Avogardo’s number of molecules (particles) are present in one mole of a gas i.e., molecular mass of a gas expressed in grams. Any change in temperature and pressure has no influence on the number of particles. 64
Lecture-5 1.12 Equivalent weight and Equivalents Equivalent Weight (E): It is the amount of the given element or metal which reacts with 1 gm of hydrogen.
Example: Suppose we have to find equivalent weight of
oxygen. Then combination of hydrogen and oxygen will give H 2 O.
Solution: From this formula, 2 gm of hydrogen reacts with 16 gm oxygen. 16 1 gm of hydrogen will reacts with = 8 gm oxygen. 2 Means Equivalents weight of oxygen is = 8.
Example: If we have to find equivalent weight of Al then, Solution: Al reacts with hydrogen to give AlH 3 . From this, 3 gm H reacts with = 27 gm Al 27 1 gm H reacts with = = 9 gm Al 3 Means, Equivalent weight of Al = 9
Equivalent Weight (New Concept) For metals equivalent weight =
Example:
atomic weight max imum ch arg e
Al
E=
27 = 9 3
Ca
E=
40 = 20 2 65
For elements equivalents weight =
Example:
atomic weight max imum ch arg e
O2
E=
16 =8 2
or
E=
32 =8 4
molecular weight
For Acid
E=
Example:
H 2 SO 4
E=
M 2
HCl
E=
M 1
Number of H + replaced
E=
M 3
H 3 PO 4
E=
M 2
H 3 PO 3
E=
M 1
H 3 PO 2
In case of a reaction, NaOH + H 2 SO 4 → NaHSO 4 + H 2 O Find out Equivalent weight of H 2 SO 4 E=
66
Molecular weight No. of H + replaced
Here H 2 SO 4 is having 2H + but number of H + replaced is one. So,
E=
Molecular weight 1
For, NaOH + H 3 PO 4 → Na2 HPO 4 + H 2 O Find out Equivalent weight of H 3 PO 4 . So,
E=
Molecular weight Number of H + replaced
In this number of H + replaced are 2H + E=
Molecular weight 2
Equivalent Weight of the Compound or Salt (E) For Compound Equivalent weight E=
Molecular weight of compound Charge on cation × NUmber of cation
Example: Find out equivalent weight of (a)
CaCO 3
(c)
Al 2( SO 4 )3
(b) Na2 SO 4
Solution: (a) For CaCO 3 E= Q
Molecular weight F
F = X ×y 67
X, Charge on cation (Ca ++ ) = 2 y, Number of cations (Ca++ ) = 1 So,
E=
100 100 = = 50 1×2 2
(b) For Na2 SO 4 E= Q
Molecular weight F F = x ×y
x, Charge on cation (Na+ ) = 1 y, Number of cations (Na+ ) = 2 So,
E=
M 1×2
or
E=
M 2
(c) For Al 2 (SO 4 ) 3 E=
Molecular weight F
Q
F = x ×y
x, Charge on cation (Al +++ ) = 3 y, Number of cations (Al +++ ) = 2 So,
E=
M 2×3
or
E=
M 6
Note: If compound is ionic then equivalent weight of compound (AB)= Ecation + Eanion
68
Example: If Al 2 O 3 is there, find out equivalent wt. of Al 2 O 3 . Solution:
E Al 2 O 3 = E Al + E0 = 9+8 = 17
or
E Al 2 O 3 =
Molecular weight Charge on cation × Number of cations
Molecular weight of Al 2 O 3 = 27 × 2 + 16 × 3 = 54 + 48 = 102 X = Charge on cation i.e., + 3 (Al +++ ) y = Number of cations i.e., 2 (2Al) So,
F = x ×y E Al 2 O 3 =
or
F = 3×2 =6
102 = 17 6
Finding equivalent of the given : Equivalence of the given =
Weight of the given E
Example: Find out equivalence of 200g CaCO 3 Solution: Equivalence. = E=
weight of the given
Molecular weight F
E Where, F = x × y
x = Charge on cation = +2 y = Number of cations = 1 69
E= Euivalence =
100 = 50 2
wt . 200 = =4 50 E
If a reaction is there A + 2B → 3C then, equivalence of A = Eq. of B = Eq. of C Eq . ofA = Eq. of B =
wt . EB
Eq . of C =
wt . EC
wt . EA
[Eq.= Equivalence]
Example: If 400 gm Ca reacts with oxygen to give CaO then find out weight of O 2 used.
Solution: Reaction is, Ca + O 2 → CaO Let weight of O 2 used is x. In this Equivalence of Ca = Equivalence of O 2
70
Equivalence of Ca =
wt . ECa
Equivalence of Ca =
400 20
ECa = 40/2 = 20
Equivalence of O 2 =
wt . EO 2
Equivalence of O 2 =
x 8
From the concept, Equivalence of Ca = Equivalence of O 2 400 x = 20 8 or
x = 20 8
or
x = 160 gm
wt. of O 2 used is 160 gm.
Relation between Equivalence and Moles If x gram of a given is there. Then, moles n=
wt . of substance atomic wt ./ molecular wt .
Equivalence (eq.)= If,
Wt . of substance Equivalent(E)
....(i) ....(ii)
(II) Eq . atomic wt . / molecular wt . = E (I) n Eq . Atomic wt . molecular wt . = = n E E
For, Acid,
71
E=
Molecular wt . No. of H * replaced
For, Base, E=
Molecular wt . No. of OH − replaced
For, Metal / Element, E=
atomic wt . Maximum ch arg e
In case of compound,
Example: Find out equivalent wt. of Al 2 (SO 4 ) 3 Solution:
E= E=
For,
Molecular wt . Ch arg e on cation × NO. of cations
Mol. wt . Mol. wt . = 3×2 6
Al 2 (SO 4 ) 3 E=
Atomic wt ./ molecular wt . F
F = No. of H + for acid = No. of OH − for base = Ch arg e on metal So,
72
F=
Atomic wt ./ molecular wt . E
Eq . atomic wt . Molecular wt . = = n E E
So,
Eq . = F or n
F=
Eq . n
Example: 20 moles of Ca is having...........Equivalence. Eq . = F or n
Solution:
Eq. =2 20
Where, F = No. of charge on Ca i.e., = 2 Eq.= 40
∴
Example: Find out No. of Eq. of 30 moles of H 3 PO 4 . Solution: Eq . = F
or
n
Eq. =3 30
Where, F = 3 ∴
Eq.= 90
Note: If for a reaction, moles are given then find out. equivalence first. Then calculate,
Example: 20 moles of Ca reacts with O 2 then find weight of O 2 used.
Solution: Ca + O 2 → CaO. Find out equivalence first. Eq . =F n
for Ca F= 2
Eq . =2 n
Eq. = 2 × 20 = 40 73
Eq. of Ca = Eq. of O 2 40 = Eq . of O 2 40 =
wt . x = E 8
x = 320 gm
Example: If 20 moles of Al reacts with 20 moles of O 2 then find out.
(a) Moles Al 2 O 3 formed (b) Weight and name of remainder.
Solution: Al + O 2 → Al 2 O 3 Eq . =F n
for Al
Eq. =3 20
Eq. = 3 × 20 = 60
F=3
Eq.= 60 Eq . =F n
for O 2 F = 4
Eq . =4 n
Eq. = 4 × 20 = 80 Eq.= 80
74
Means,
Al
+
O 2 → Al 2 O 3
Eq. used
60
80
0
Final
0
20
60
(a) Eq. of Al 2 O 3 = 60 Eq . =f n E= F=6
f= ? Mol. Wt . No. of cations × ch arg eon cations
For Al 2 O 3 E=
Mol. wt . M = 2×3 6
Eq . = 6, n
So,
60 =6 n
n = 10
Remainder is O 2 and remaining Equivalence is 20. For O 2 Eq . = 20 = 20 =
wt . E
wt. 8
Wt. = 160 gm.
1.13 Reactions in Solutions A majority of reactions in the laboratories are carried out in solutions. The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways. 1. Mass per cent or weight per cent (w/w %) 2. Mole fraction 3. Molarity 75
4. Molality Let us now study each one of them in detail (i) Mass per cent : It is obtained by using the following relation: Mass per cent =
mass of solute massof solution
× 100
Problem: A solution is prepared by adding 2g of a substance A to 18g of water, calculate the mass per cent of the solute.
Solution: Mass per cent of A =
mass of A mass of solution
=
2g × 100 2 gof A + 18gof water
=
2g × 100 20g
× 100
= 10% (ii) Mole Fraction: It is the ratio of number of moles of a particular component to the total number of mole of the solution. If a substance A dissolves in substance B and their number of moles are n A andn B respectively then the mole fraction of A and B are given as Mole fraction of A = = 76
No.of moles of A No. of moles of solution nA n A + nB
Mole fraction of B = =
No.of moles of B No. of moles of solution nB n A + nB
(iii) Molarity: It is the most widely used unit and is denoted by M. It is defined as the number of oles of the solute in I litre of the solution. Thus, Molarity (M) =
No. of moles of solute volume of solution in litres
Suppose we have 1 M solution of a substance say NaOH and we want to prepare a 0.2 M solution from it. 1 M NaOH means 1 mol of NaOH present in 1 litre of the solution for 0.2 M solution we require 0.2 moles of NaOH in 1 litre solution. Hence, we have to take 0.2 moles of NaOH and make the solution to 1 litre. Now how much volume of concentrated (1M) NaOH solution be taken which contains 0.2 moles of NaOH can be calculated as follows: If 1 mole is present in 1 L or 1000 mL then 0.2 mol is present in
1000 mL 1 mol
× 0.2 mol
= 200 mL Thus, 200 mL of 1 M NaOH are taken and enough water is added to dilute it to make it 1 litre. 77
In fact for such calculations, a general formula, M1 × V1 = M2 × V2 where M and V are molarity and volume respectively can be used in this case, M1 is equal to 0.2; V1 = 1000mL and M2 = 1.0 ; V2 is to be calculated substituting the value in the formula: 0.2 M × 1000mL = 1.0 M × V2 ∴
V2 =
0.2 M × 1000mL = 200 mL 1.0 M
Problem: Calculate the molarity of NaOH in the solution prepared by dissolving its 4g in enough water to form 250 mL of the solution.
Solution: Since molarity (M) =
No. of moles of solute volume of solution in litres
=
Mass of NaOH / Molar mass of NaOH 0.250L
=
4g / 40g 0.1 mol = 0.250L 0.250L
= 0.4 mol L–1 = 0.4 M
Note: that molarity of a solution depends upon temperature because volume of a solution is temperature dependent.
78
(iv) Molality: It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m. Thus, molality (m)=
No. of moles of solute Mass of solvent in kg
Note: Molality is temperature independent Problem: The density of 3 M solution of NaCl is 1.25 g mL –1 . Calculate molality of the solution.
Solution: M=3 mol L –1 Mass of NaCl in 1 L solution
= 3×58.5 =175.5g mass of
1L solution
= 1000× 1.25 =1250g
(since density
= 1.25 g mL –1 ) mass of
water in solution
= 1250–175.5 =1074.5g
Molality =
=
No. of moles of solute Mass of solvent of kg 3 mol 1.074.5 kg
= 2 . 79 m
79
Relation between Moles fraction and Molality Example: If moles fraction of urea in water is 0.2 find out molality of urea solution.
Solution: We know that molality m=
n(molesof solute) wt . of solvent (kg)
If moles of solute are ‘n’ and moles of solvent are ‘N’ then, mole fraction of urea X=
n n +N
n= moles of urea N = moles of water Now, given is, x = 0.2 0.2 =
n n +N n = 5n 0.2
SO,
n +N =
or
n +N = 5n
or
N = 4n
For molality, m=
n wt . of solvent(kg)
Let the wt. of solvent or water is y Then,
80
.....(i)
moles
N=
y mol. wt .
mol. wt. = 18 for H 2 O N= or
y 18
y = 18 N gm
Therefore, molality, m=
n m = 80 1000 N = 4n
We know that m=
n 18 ×
n wt . of solvent(kg)
4n 1000
=
1000 55 .55 = = 14 18 × 4 4
Example: If molality of urea in water is 2m. Find out mole fraction.
Solution:
m=
n(molesof solute) wt . of solvent(kg)
Mole fraction x =
n n +N
Let the weight of solvent is 1 kg. So, m=
n 1 81
or
....(i)
n=m=2
Weight of solvent is 1kg, therefore, Moles
N=
wt . 1000 = mol. wt . 18
2 2 x = 1000 = + 2 57 .5 18
Hence,
Example: 600 gm of urea is dissolved in 100 litre of solution. M=
n V
Solution: n = 600 = 10 V = 100L 60
Moles of urea, Mol. wt. of urea = 60 M = 10 / 100
So,
M = 0.1
Example: 20% urea solution is there having density 2g/ml. Find out molality, molarity of solution.
Solution: 20% urea solution means, 20 g of urea in 100g of solution.
Hence, 80g of solvent Therefore, Molality m = n=
82
n(molesof solute) Wt . of solvent (kg) 20 1 = 60 3
wt. of solvent = 80g or
80 kg 1000
20 1 3 = 100 = 4.16 m = 60 = 80 8 24 1000 1000 Molarity of solution is M=
n density of solution V
20 M = 60 50 1000
d=
mass V
M=
1 20 × 3 1
V=
M=
20 3
V=
massof solution d 100 = 50 ml 2
Molar solution = 1m decimolar solution = ( 1 / 10) m centi molar solution = ( 1 / 100)n
83
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Questions Q.1
Calculate the mass percentage of different elements present in sodium sulphate (Na 2 SO 4 ).
Ans. Molecular mass of Na 2So 4 = 2 × Atomic mass of Na + Atomic mass of S + 4 × Atomic
mass of O
= 2×23 + 32 + 4×16 = 46 + 32+ 64 = 142 u. The percentage of different elements present can be calculated as: Percentage of sodium (Na)=
=
Mass of Na × 100 Molecular mass of Na 2So 4 46 × 100 142
= 32.39% Percentage of sulphur(S) =
=
Mass of S × 100 Molecular mass of Na 2SO 4 32 × 100 142
= 25.54% Percentage of oxygen(O) =
=
Mass of O × 100 Molecular mass of Na 2SO 4 64 × 100 142
= 45.07% Q.2
Calculate the mass of sodium acetate (CH 3 COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0 g mol −1.
Ans. Molar mass of CH 3 COONa = 82 g mol −1 (given) Molarity of solution = 0.375 M = 0.375 mol L −1 Volume of solution = 500 mL = 500/1000 = 0.5 L
85
Molarity of solution(M) = (0.375 molL–1) =
Mass of CH 3COONa / Molar mass Volume of Solution in litres
W (82gmol –1) × (0.5 L)
W =(0.375molL–1) × (82 g mol −1) × (0.5 L) = 15.375 g. Q.3
Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g mL −1 and the mass percent of nitric acid in it being 69%.
Ans. Mass of solution = 100 g Density of solution = 1.41 g mL −1 Volume of solution =
Mass of solution 100 g = = 70.92 mL Density of solution 1.41gML−1
Concentration of HNO 3 in moles per litre means molarity 69 g Mass of HNO 3 63 g mol −1 Molar Mass of HNO 3 Molarity(M) = = 70.92 Volume of solution in litre L 1000 = 15.44 mol L−1 =15.44 M Q.4
What is the concentration of sugar (C12 H 22 O11) in mol L –1 if 20 g of it are dissolved in enough water to make final volume upto 2L ?
Ans. The concentration in mol L −1 means molarity (M). From the available data, it can be calculate as : Mass of sugar = 20 g Molar mass of sugar (C12 H 22O11) = 12×12 + 22×1 + 16×11 = 342 g mol −1 Volume of solution in litre = 2 L Molarity of solution =
= 86
Mass of sugar / Molar mass Volume of solution inlitres 20g −1
(342mol ) × (2L)
= 0.029 mol L−1 = 0.029M.
Q.5
If the density of methanol is 0.793 kg L –1, what is its volume needed for making 2.5 L of its 0.25 M solution ?
Ans. Step-1. Calculation of mass of methanol (CH 3 OH) Molar mass of methanol (CH 3 OH) = 12 + 4×1 + 16 = 32 g mol −1 Volume of solution = 2.5 L Molarity of solution =
Mass of methanol / Molar mass Volume of solution in litre
0.25mol L–1 =
W 32 g mol −1 × 2.5L
W = 0.25 mol L–1 × 32 g mol –1 × 2.5 L = 20 g Step-2. Calculation of volume of methanol Mass of methanol = 20 g = 0.002 kg Density of methanol = 0.793 kg L −1 Volume of methanol = Q.6
0.002kg Mass = = 0.0025 L Density 0.793kgL–1
A sample of drinking water was found to be severely contaminated with chloroform CHCl 3 , supposed to be carcinogen. The level of contamination was 15 ppm (by mass) (i) Express this in percent by mass (ii) Determine the molality of chloroform in the water sample.
Ans. (i)
Calculate of percent by mass 15 ppm level of contamination means that 15 parts or 15 g of chloroform (CHCl 3 ) are present in 10 6 parts or 10 6 g of the sample i.e., water. ∴
(ii)
Mass percent =
15 g 6
10 g
× 100 = 15 . × 10 −3
Calculate of molarity of the solution Mass of chloroform = 1.5 × 10–3 g Molar mass of chloroform (CHCl 3 ) = 12 + 1 + (3×35.5) = 119.5 g mol −1 87
Mass of sample i.e., water = 100 g . × 10 −3 g 15 Mass of chloroform Molar mass of chloroform 119.5 g mol –1 Molarity of solution (m) = = 100 Mass of solvent in kg kg 1000 = 125 . × 10 −4 mol kg −1 = 125 . × 10 −4 m Q.7
How are 0.50 mol Na 2 CO 3 and 0.50 M Na 2 CO 3 different ?
Ans. 0.50 mol Na 2 CO 3 represent concentration in moles. 0.50 mol Na 2 CO 3 represent concentration in moles/litre (molarity). Q.8
If 10 volumes of dihydrogen react with five volumes of dioxygen gas, how many volumes of water will be produced ?
Ans. 2H 2 (g)
+
O 2 (g) → 2H 2 O(g)
2 vol
1 vol
10 vol 5 vol
10 vol
2 vol
10 volumes of water vapours will be produced Q.9
Calculate the molarity of a solution of ethanol in water in which mole fraction of ethanol is 0.04.
Ans. Mole fraction of etahnol (X B ) may be given as : XB =
nB n A +nB
Here, n B = no. of moles of ethanol ; n A = no. of moles of water According to available data,
X B = 0.04,n A = 0.04 =
1000g (18gmol —1)
= 55.55 mol
nB 55.55+nB
2.222+(0.04)nB = nB
88
or
nB +0.04nB = 2.222 or 0.96nB = 2.222
or
nB =
2.222 = 2.31 0.96
2.31 moles of ethanol are dissolved in 1000 g (or 1000 mL) of water or 1000 mL of the solution. In this case, the volume of solution is considered to be the same as that of the solvent i.e., water. In other words, the solution is regarded as dilute solution, Molarity of solution = 2.31 M. Q.10 Calcium carbonate reacts with aqueous HCl to give CaCl 2 and CO 2 according to the reaction: CaCO 3 (s) + 2HCl (aq) → CaCl 2 (aq) + CO 2 (g) + H 2 O (l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl ? Ans. Step-1. Calculation of the mass of HCl present Molarity of HCl solution = 0.75 M = 0.75 mol L −1 Volume of HCl solution = 25 mL = 25/1000 = 0.025 L Molarity of solution (M)=
Mass of HCl / Molar mass Volume of solution in litres
(0.75mol L–1) =
Mass of HCl (36.5 g mol –1) × (0.025 L)
Mass of HCl = (0.75 g mol L −1)×(36.5 g mol −1)×(0.025 L) = 0.684 g Step-2. Calculation of the mass of CaCO 3 reacted CaCO 3 (s) + 2HCl (aq) → CaCl 2 (aq) + CO 2 (g) + H 2 O (l) (1 mol)
(2 mol)
40+12+48 2×36.5 = 100 g
= 73 g
73 g of HCl require CaCO 3 for reaction = 100 g 0.684 g of HCl require CaCO 3 for reaction =
100 g × 0.684 g = 0.94 g 73 g
Q.11 Out of 1 M H 2 SO 4 and 1N H 2 SO 4 , which is more concentrated and why ? Ans. 1 M H 2 SO 4 solution is more concentrated than 1 N acid solution. Actually 1M solution of acid isd prepared by dissolving 98 g of the acid (1 gram mole) per litre of the solution while 1 N solution of acid is prepared by dissolving 49 g 89
of the acid (1 gram equivalent) in the same volume of the solution. Since 1 M solution has a greater amount of solute dissolved in it, this solution is more concentrated than 1 N solution of the acid. Q.12 What is the effect of temperature on molarity of a solution ? Ans. Molarity of a solution normally decreases with rise in temperature. Q.13 What is the significance of N/10 NaOH solution ? Ans. It means that 0.1 gram equivalent (4 g) of NaOH is dissolved per litre of the solution. Q.14 What are SI units of molarity ? Ans. SI units of molarity : mol dm −3 . Q.15 Are the molar volumes of CO 2 and SO 2 different ? Ans. No, these are the same. In fact, molar volume is the volume occupied by 22.4 dm 3 of a gas at STP.
90
Summary 1. Precision refers to the closeness of set of values obtained by identical measurement while accuracy refers to the closeness of a single measurement to its true value. 2. Significant figures 3 All non-zero digits are significant 3 The zeros to the right of the decimal point or zeros between two non-zero digits are significant. 3 Zeros to the left of the first non-zero digit in a number or zeros at the beginning of a number are not significant. 3 In addition or subtraction, the result should be mentioned in the same number of decimal place as that of the term with the least decimal places. 3 In multiplication and division, the result should be mentioned in same number of significant figures as the least precise term used in calculation. 3. Units for measurement— 3 CGS system—Length (centimeter), mass (gram), time (second) 3 MKS system—Length (meter), mass (kilogram), time (second) 3 SI system—The SI has 7 basic units from which all other units are derived called derived units. Length
meter (m)
Mass
kilogram (kg)
Time
second (s)
Temperature
kelvin (K)
Electric current
ampere (A)
Luminous intensity
candela (Cd)
Amount of substance
mole (mol)
3 Area (m 2 ), volume (m 3 ), density (kg mm −3 ), velocity (ms −1), acceleration (ms −2 ), molar mass (kg mol −1), molar volume (m 3 mol −1), molar concentration (mol m −3 ) are se of the examples of derived units. 4. Conversion factors 1L
= 1000 mL
1J
= 1Nm =1kgm 2s -2
1 Cal
= 4.184 J
1 eV
=1.602 × 10 -19 J 91
1 eV/atom
= 96.5 kJ mol −1
1 amu
= 931.5016 MeV
1 kilowatt hour
= 3600 kJ
1 horse power
= 746 watt
1J
=10 7 erg
1 esu
= 3.3356 × 10 -10 C
1 dyne
=10 -5 N
1 atm
=101325 Pa
1 bar
=1 × 10 5 Nm -2
1 L-atm
=101.3 J
1 year
= 3.1536 × 10 7 s
1 debye (D)
=1 × 10 -18 esu cm
1 mole of a gas
= 22.4 L at STP
1 mole of a substance
= N0 molecules
1 g-atom
= N0 atoms
1 pm
=10 -12 m
1Å
=10 -10 m
1 nm
=10 -9 m
5. Atomic mass and its determination average mass of an atom average mass of an atom 3 Atomic mass = 1 × mass of an atom of C12 12 3 Average atomic mass—(If an element exists in two isotopic forms having atomic masses ‘a’ and ‘b’ in the ratio m : n) ( m × a) + ( n × b) = m+ n 3 Gram Atomic Mass (GAM)—Atomic mass of an element when expressed in grams. mass of an element (g) 3 Number of gram atoms = GAM GAM 3 Mass of one atom of an element (g) 6.02 × 10 -23 3 Methods of determination of atomic mass (i) Dulong and petits method— 92
Atomic mass =
6.4 specific heat (cal)
This law is applicable to solid elements only except Be, B, C and Si. (ii) Vapour density method— Atomic mass = equivalent mass × valency Valency of metal whose chloride is volatile 2 × vapour density of chloride = equivalent mass of meatl+35.5 (iii) Specific heat method—It is suitable only for gases. Atomic mass of a gaseous element molecular mass = atomicity (C p / C
y
= γ =1.66 for monoatomic, 1.40 for diatomic and 1.33 for triatomic gases)
6. Molecular mass and its determination mass of one molecular of the substance Molecular mass = 1 × mass of one atom of C12 12 3 Actual mass of 1 molecule = molecular mass ×166 . × 10 −24 g 3 Gram molecular mass (GMM)—Molecular mass of an element or compound when expressed in gram. mass of substance (g) 3 Number of gram molecules = GMM 3 Gram molar volume volume of 1 mole of any gas at STP 22.4 L molecular mass in gram 3 Density of a gas at NTP 22400 mL 3 Number of atoms in a substance = number of GMM × 6.02 × 10 23 × atomicity 3 Methods of determination of molecular mass (i) Diffusion method —
r1 = r2
M2 M1
(ii) Vapour density method — Molecular mass = 2 × vapour density
93
(iii) Victor Meyer method—It is based on Dalton’s law of partial pressure and Avogadro’s hypothesis. 3 22400 ml of vapours of a substance molecular mass of that substance 7. Equivalent mass and its determination atomic mass 3 Equivalent mass of an element = valency 3 Equivalent mass of an acid = 3 Equivalent mass of a base = 3 Equivalent mass of a salt =
molecular mass basicity
molecular mass acidity
formula mass total positive or negative charge
3 Equivalent mass of an oxidising agent =
formula mass total change in oxidation number
3 Equivalent mass of common oxidising agent changes with the medium of the reaction. 3 Methods of determination of equivalent mass (i) Hydrogen displacement method— mass of metal Equivalent mass of metal = × 1008 . mass of H 2 displaced mass of metal × 11200 volume of H 2 displaced at STP
or Equivalent mass of metal = (ii) Oxide formation method— Equivalent mass of metal = =
mass of metal × 8 mass of O 2 mass of metal × 5600 vol of O 2 at STP in mL
(iii) Chloride formation method— mass of metal × 35.5 mass of chlorine mass of metal × 11200 = vol of Cl2 at STP in (mL)
Equivalent mass of metal =
(iv) Metal displacement method— eq mass of metal added mass of metal added = mass of metal displaced eq mass of metal displaced 94
=
W1 E1 = W2 E2
(v) Neutralization method (for acids and bases) w Equivalent mass of acid or base = (where w is the mass of acid or base (in V×N g), V is the volume of base or acid in liter required for neutralization and N is normality of base or acid. (vi) Double decomposition method AB + CD → AD ↓ +CB Mass of compound AB eq mass of A + eq mass of B = Mass of compound AD eq mass of A + eq mass of D or
eq mass of salt ( E1) mass of salt taken ( w1) = mass of ppt obtained ( w2 ) eq mass of salt in ppt ( E2 )
(vii) Volatile chloride method Equivalent mass of metal =
2 × VD of chloride − 35.5 Valency
(viii) Silver salt method (for organic acids) 108 × mass of silver metal salt Equivalent mass of an acid = − 107 mass of Ag 8. Percentage composition and molecular formula mass of element Percentage of element = × 100 molecualr mass Molecular formula = n × empirical formula molecular formula mass where, n = empirical formula mass
95
Name of Topics with Related Questions
3 Physical Quantities and Their Measurements: Q.1.13, Q.1.14,
Q.1.15, Q.1.15, Q.1.22, Q.1.27. 3 Laws of Chemical Combination: Q.1.21. 3 Significant figures and scientific notation: Q.1.16, Q.1.18,
Q.1.19, Q.1.20. 3 Related with atomic mass molecular mass and percentage: Q.1.1,
Q.1.2, Q.1.9, Q.1.32. 3 Mole Concept: Q.1.4, Q.1.7, Q.1.10, Q.1.23, Q.1.28, Q.1.30,
Q.1.33. 3 Chemical Equation and Stoichiometry: Q.1.24, Q.1.26, Q.1.34,
Q.1.35, Q.1.36. 3 Problem related with combustion: 3 Problem related with Concentration of Soluton: Q.1.5, Q.1.6,
Q.1.11, Q.1.12, Q.1.17, Q.1.25, Q.1.29. 3 Empirical and Molecular Formulae: Q.1.3, Q.1.8.
96
NCERT Solved Questions 1.1
Calculate the molecular mass of: (i) H 2 O
Ans.
(ii) CO 2
(iii) CH 4
(i)
Molecular mass of H 2 O = 2 (1 amu) + 16 amu =18u
(ii)
Molecular mass of CO 2 = 12 amu + 2 x 16 amu = 44u
(iii)
Molecular mass of CH 4 = 12 amu + 4 (1 amu) = 16u
1.2.
Calculate the mass per cent of different elements present in sodium sulphate (Na 2 SO 4 ).
Ans.
Mass % of an element Mass of that element inthecompound × 100 = Molecular mass of the compound Molecular mass of Na 2 SO 4 = 2(23.0) + 32.0 + 4 x 16.0 = 142 g mol –1 46 Mass per cent of sodium = × 100 = 32.39% 142 32 Mass per cent of sulphur = × 100 = 22.54% 142 64 Mass per cent of oxygen = × 100 = 45.07% 142
1.3
Determine the empirical formula of an oxide of iron which has 69.9 % iron and 30.1% dioxygen by mass. (Atomic mass : Fe = 55.85 amu, O = 16•00 amu).
Ans: Element Symbol % by mass
Atomic mass
Moles of the element (Relative no. of moles)
Simple ratio
Simplest whole number molar ratio
Iron
Fe
69.9
55.85
69.9 = 1.25 55.85
1.25 =1 1.25
2
Oxygen
O
30.1
16.00
30.1 = 1.88 16.00
1.88 = 1.5 1.25
3
∴
Empirical formula = Fe 2 0 3 . 97
1.4.
Ans:
Calculate the amount of carbon dioxide that could be produced when (i)
1 mole of carbon is burnt in air.
(ii)
1 mole of carbon is burnt in 16 g of dioxygen.
(iii)
2 moles of carbon are burnt in 16 g of dioxygen.
The balanced equation for the combustion of carbon in dioxygen or air is C (s ) 1 mole
(i)
+
O2 (g) 1 mole (32g)
→ CO2 (g) 1 mole (44g)
In air, combustion is complete. Hence, 1 mole of carbon on combustion produces CO 2
(ii)
As only 16 g of dioxygen is available, it is the limiting reactant because 16 16g O2 means mols of O2 = = 0.5 32 If 0.5 mole of O2 is used then 0.5 moles of CO2 will be wt x so, moles n= = 0.5 = mol.wt 44 ⇒
(iii)
x = 22
Here again, dioxygen is the limiting reactant. Therefore, CO 2 produced from 16 g dioxygen 16 or nO2 = 0.5 Since nC = 2 ,nO2 = 32 In this 0.5 O2 will be consumed to give 0.5 moles of CO2 x nCO2 = 44 x 0.5 = 44 x = 22
1.5.
Calculate the mass of sodium acetate (CH 3 COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol –1 .
Ans.
0.375 M aqueous solution means that 1000 ml of the solution contain sodium acetate 0.375 mole ∴500 ml. of the solution should contain sodium acetate 0.375 = mole 2
98
Molar mass of sodium acetate = 82.0245 g mol –1 0.375 mole × 82.0245 g mol –1 = 15.380 g. ∴Mass of sodium acetate required = 2 n moles OR molarity M= = V volume 0.375 =
Given M = 0.375 V = 500 ml = 0.5 lt
h 0.5
n = 0.5 × 0.375 =
0.375 2
wt 0.375 = mol.wt 2 0.375 wt = × 82.0245 = 15.38 gm 2
moles n =
1.6.
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g ml –1 and mass per cent of nitric acid in it being 69%.
Ans.
Mass percent of 69% means that 100g of nitric acid solution contains 69 g of nitric acid. Molar mass of nitric acid (HNO3 ) = 1 + 14 + 48 = 63g mol –1 ∴
Moles in 69 g HNO 3 n=
69 g 63 gmol –1
= 1.095 mole
Volume of 100 g nitric acid solution density = So volume = ∴
mass volume
mass 100g = = 70.92 mL = 0.07092L density 1.41gmL–1
Conc. of HNO 3 in moles per litre 1.095 mole = = 15.44 M 0.07092 L
* Alternatively, (shortcut) mass percent × density × 10 moles = molarity = litre m. mass 69 × 1.41 × 10 = = 15.44 M 63 99
1.7.
How much copper can be obtained from 100 g of copper sulphate (CuSO 4 ) ? (Atomic mass of Cu = 63.5 amu)
Ans.
1 mole of CuSO 4 contains 1 mole (1 g atom) of Cu Molar mass of CuSO 4 = 635 + 32 + 4 × 16 = 159.5 g mol –1 Thus, Cu that can be obtained from 159.5 g of CuSO 4 = 63.5 g ∴Cu that can be obtained from 100 g of CuSO 4 63.5 = × 100g 159.5 = 39.81 g 1 mole of CuSO 4 will give 1 mole of Cu CuSO 4 → Cu + S + 40 So moles of CuSO 4
n=
mass 100 = nol.wt 159.5
n moles of CuSO 4 will give n moles of Cu so, n = moles of Cu = moles of CuSO 4 100 = moles of Cu 159.5 wt = 63.5 100 wt = 63.5 × = 39.812 159.5 1.8.
Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9% and 30.1% respectively. Given that the molar mass of the oxide is 159.8 g mol −1 (Atomic mass: Fe = 55.85, o = 16.00 u.
Ans.
Calculation of empirical formula
Element
Per centage
At. Mass
Relative no. of Moles
Simple ratio
Simples t Whole no. Ratio
Fe
69.9
55.85
69.9 = 1.25 55.85
1.25 =1 1.25
1×2=2.0
O
30.1
16.00
30.1 = 1.88 16.00
1.88 = 1.5 1.25
1.5×2 =3.0
100
Empirical formula = Fe 2 O 3 . Empirical formula mass of Fe 2 O 3 = 2 x 55.85 + 3 x 16.00 = 159.7 g mot –1 Molecular Formula = n × Empirical formula Molar mass 159.8 =1 ⇒ n= 159.7 Empirical formula mass (nis emperical factor)
⇒ ∴
Hence. molecular formula is same as empirical formula, Fe2 O3 .
⇒
OR Let the formula be Ax By moles of A x nA x = ⇒ = moles of B y nB y Let Formula is Fe x Oy nFe x 69.9 x x 2 so, = ⇒ = ⇒ = so formula is Fe2 O3 55.85 y y 3 n0 y 30.1 16 1.9.
Calculate the atomic mass (average) of chlorine using the following data: Isotope
Ans.
35
Cl(1)
37
Cl(2)
% Natural Abundance
Atomic Mass
75.77
34.9689
24 .23
36.9659
Fractional abundance of 35 Cl = 75.77, atomic mass = 34.9689 Fractional abundance of 37 Cl = 24.23, atomic mass = 36.9659 ∴Average atomic mass = (.75777) (34.9689 amu) + (0.2423) (36.9659 amu) = 35.45
1.10.
In three moles of ethane (C 2 H6 ), calculate the following: (i)
Number of moles of carbon atoms
(ii)
Number of moles of hydrogen atoms
(iii)
Number of molecules of ethane. 101
Ans.
(i)
1 mole of C 2 H6 contains 2 moles of carbon atoms
∴
3 moles of C 2 H6 will contain C-atoms = 6 moles
(ii)
1 mole of C 2 H6 contains 6 moles of hydrogen atoms
∴
3 moles of C 2 H6 will contain H-atoms = 18 moles
(iii)
1 mole of C 2 H6 contains Avogadro’s number, i.e., 6.02 × 1023 molecules
∴
3 moles of C 2 H6 will contain ethane molecules = 3 × 6.02 × 1023 = 18.06 × 1023 molecules.
or moles =
No.of given No.(6.023 × 1023 ) =
1.11.
3=
No.of molecules 6.02 × 1023
No.of molecules 18.06 × 1023
What is the concentration of sugar (C 12 H22 O11 ) in mol L−1 if its 20 g are dissolved in enough water to make a final volume up to 2 L ?
Ans.
Molar mass of sugar (C 12 H22 O11 ) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol −1 20 g No. of moles in 20 g of sugar = = 0.0585 mole 342 mol −1 Volume of solution = 2 L (Given) Moles of solute 0.0585 mol Molar concentration = = Volumeof sol inL 2L = 0.0293 mol L−1 = 0.0293 M Formula concentration (molarity) =
moles n = volume V
1.12.
If the density of methanol is 0.793 Kg L−1 , what is its volume needed for making 2.5 L of its 0.25 M solution?
Ans.
Molar mass of methanol (CH3 OH) = 32 gmol −1 = 0.032 Kgmol −1 1 liter of 0.25 M CH3 OH contains moles of CH3 OH = 0.25 ∴
102
2.5 L of 0.25 M CH3 OH will contain moles
= 0.25 x 2.5 = 0.625 Mass of 0.625 moles of CH3 OH = 0.625 × 32 = 20 g The density of CH3 OH is 0.793 KgL−1 or 0.793 g cm −3 Hence, volume of methanol required 20 g M or = = d 0.793 gcm−2 = 25.22 mL 1.13.
Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal, is as shown below. 1Pa = 1Nm−2 If mass of air at sea level is 1034 g cm−2 , calculate the pressure in pascal.
Ans.
Pressure is the force (i.e., weight) acting per unit area But weight = mg ∴
Pressure = Weight per unit area =
Conversion =
1034 × 10 −3 Kg × 9.8ms −2 10
−4
2
m
1034 g × 9.8 ms −2 cm2 = 1.013 × 10 5
N m2
= 1.013 × 10 5 Pa (Pascal) 1.14.
What is the SI. unit of mass? How is it defined?
Ans.
S.I. unit of mass is Kilogram (Kg). It is defined as the mass of platinum iridium (Pt-Ir) cylinder that is stored in a air tight jar at international Bureau of weight and measured in Frame.
1.15.
Match the following prefixes with their multiples: Prefixes
Ans.
Multiplies
(i)
micro
10 –6
(ii)
deca
1010
(iii) mega
106
(iv) giga
109
(v)
10 −15
femto
micro = 10 −6 ,deca = 10 ,mega = 106 ,giga = 109 , femto = 10 −15 103
1.16.
What do you mean by significant figure?
Ans.
The total number of digits in a number including the last digit whose value is uncertain is called significant figure.
1.17.
A sample of drinking water was found to be severely contaminated with chloroform, CHCl3 , supposed to be carcinogen. The level of contamination was 15 ppm (by mass).
Ans.
(i)
Express this in percent by mass
(ii)
Determine the molarity of chloroform in the water sample.
(i) ∴ (ii)
15 ppm means 15 parts in million (106 ) parts 15 % by mass = 6 × 100 = 15 × 10 −4 = 1.5 × 10 −3 % 10 Molar mass of chloroform (CHCl3 ) = 12 + 1 + 3 × 25.5 = 119.5 mol −1
100 g of the sample contain chloroform = 1.5 × 10 −3 g ∴
1000 g (1 Kg) of the sample will contain chloroform
1.5 × 10 −3 ⇒ × 1000 = 1.5 × 10 −2 g 100 moles of solute molarity = k / t .of solvent (Kg) n=
1.5 × 10 −2 = 1.255 × 10 −4 119.5
wt of solution = wt of solute + wt of solution Wt of solute = wt of solution — wt of solution = 1000 − 1.5 × 10 −2 = 1000 gm 1 Kg 1.255 × 10 −4 = 1 = 1.255 × 10 −4 M 1.18.
Ans. 104
Express the following in scientific notation: (i) 0.0048
(ii) 234, 000
(iv) 500.0
(v) 6.0012.
(i) 4 .8 × 10 −3
(ii) 2.34 × 10 5
(iii) 8008 (iii) 8.008 × 103
(iv) 5.000 × 102 1.19.
(v) 6.0012 × 10°
How many significant figures are present in the following? (i) 0.0025
(ii) 208
(iv) 126,000
(iii) 5005
(v) 500.0
(vi) 2.0034
Ans.
(i) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5.
1.20.
Round up the following upto three significant figures:
Ans.
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
(i) 34.2
(ii) 10.4
(iii) 0.0460 1.21.
Ans.
(iv) 2810.
The following data were obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen
Mass of dioxygen
(i)
14g
16g
(ii)
14g
32g
(iii)
28g
32g
(iv)
28g
80g
(a)
Which law of chemical combination is obeyed by the above experimental data? Give its statement.
(b)
Fill in the blanks in the following conversions:
(i)
1 Km =..... mm =..... pm
(iii)
1 mL =..... L =..... dm3
(a)
Fixing mass of dinitrogen as 28 g, the masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides. These are in the ratio 2 : 4 : 2 : 5 which is a simple whole number ratio. Hence, the given data obey the law of multiple proportions. 1000 m 100 cm 10 mm 1Km = 1Km × × × = 106 mm 1 Km 1m 1 cm
(b)
1Km = 1Km × (ii)
(ii)
1 mg =..... Kg =..... ng
1000 m 1 pm × −12 = 1015 pm 1 Km 10 m 105
(iii)
1mg = 1 mg ×
1 Kg 1g × = 10 −6 Kg 1000 mg 1000 g
1mg = 1 mg ×
1g 1 ng × −9 = 106 ng 1000 mg 10 g
1 mL = 1 mL ×
1L = 10 −3 L 1000 mL
1mL = 1cm3 = 1cm3 ×
1 dm × 1 dm × 1 dm = 10 −3 dm3 10 cm × 10 cm × 10 cm
1.22.
If the speed of light is 3.0 × 108 ms −1 , calculate the distance covered by light in 2.00 ns.
Ans.
Distance covered = Speed × Time = 3.0 × 108 ms −1 × 2 ns = 3.0 × 108 ms −1 × 2 ns
10 −9 s 1 ns
= 6.0 × 10 −1 m = 0.6m 1.23.
In the reaction, A + B2 → AB2 , identify the limiting reagent, if any, in the following mixtures
Ans.
(i)
300 atoms of A + 200 molecules of B
(ii)
2 mol A +3 mol B
(iii)
100 atoms of A + 100 molecules of B
(iv)
5 mol A + 2.5 mol B
(v)
2.5 mol A + 5 mol B
(i)
According to the given reaction, 1 atom of A reacts with 1 molecule of B.
∴
200 molecules of B will react with 200 atoms of A and 100 atoms of A will be left unreacted. Hence,
B is the limiting reagent while A is the excess reagent.
106
(ii)
According to the given reaction, 1 mol of A reacts with 1 mol of B
∴
2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant.
(iii)
No limiting reagent because equal moles are reacting.
(iv)
2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.
1.24.
(v)
2.5 mol of A will react with 2.5 mol of B. Hence, B is the limiting reagent which is consumed completely
(vi)
2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + 3H2 (g) → 2NH3 (g) (i) Calculate the mass of ammonia produced if 2000gm dinitrogen reacts with 1000 gm dihydrogen (ii) Will any of the two reactants remain unreacted ? (iii) If yes, which one and what would he its mass ?
Ans.
(i) According to given equation, 28 g of N2 reacts with 3 mol of H2 i.e., 6 g of 112 ∴
2000 g of N2 will react with H2 6 = × 200g = 428.6 g. 28
Amount of H2 given = 1000g Thus, N2 is the limiting reagent while H2 is the excess reagnt As 28 g of N2 will produce NH3 = 2 × 17 = 34 g ⇒
2000 g of N2 will produce NH3 34 = × 2000 g = 2428.57 g 28
(ii) H2 will remain unreacted. (iii) Mass of H2 left unreacted = 1000 g − 428.6 g = 571.4 g 1.25
How are 0.50 mol Na2 CO3 and 0.50 M Na2 CO3 different?
Ans.
Molar mass of Na2 CO3 = 2 × 23 + 12 + 3 × 16 = 106gmol −1 0.50 mol Na2 CO3 means 0.50 × 106 g = 53 g 0.50 M Na2 CO3 means 0.50 mol, i.e., 53 g Na2 CO3 are present in 1 liter of the solution.
1.26. Ans.
Convert the following into basic units: (i) 28.7 pm (ii) 15.15 µ s (iii) 25365 mg 10 −12 m 28 . 7 pm = 28 . 7 pm × = 2.87 × 10 −11 m (i) 1 pm
107
10 −6 m = 1.515 × 10 −5 s 1µ s
(ii)
15.15µ s = 15.15 µ s ×
(iii)
25365 mg = 25365 mg ×
1g 1 Kg × 1000 mg 1000 g
= 2.5365 × 10 −2 Kg 1.27.
Which one of the following will have largest number of atoms? (i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl 2 (g)
(Atomic masses: Au = 197, Na = 23, Li = 7, Cl = 35.5 U) mass Ans. We know moles = atomic mass / mol.mass moles = (i)
Noof atoms / molecules No (6.023 × 1023)
1 gAu =
1 1 mol = × 6.02 × 1023 atoms 197 197 = 3.055 × 1021 atoms
(ii)
1 gNa =
1 1 mol = × 6.02 × 1023 atoms 23 23 = 2.617 × 1022 atoms
(iii)
1 1 1 gLi = mol = × 6.02 × 1023 atoms 7 7 = 8.6 × 1022 atoms
(iv)
1 gCl2 =
1 1 mol = × 6.02 × 1023 molecules 71 71 2 = × 6.02 × 1023 atoms 71
Thus 1 g of Li has the largest number of atoms. 1.28.
Ans.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. n(C 2 H5OH) xC 2 H 5OH = = 0.040 (Given)x 2 = 0040 The aim is to find n(C 2 H5OH) + n(H2 O) number of moles of ethanol in 1 L of the solution which is nearly = I L of water .
108
Since mole fraction of C 2 H5OH is very less means water is more so mass of solution is govern by mass of H2 O In this way solvent volume of = mass of solvent 1000 g No. of moles in I L of water = = 55.55 moles 18 gmol −1 Substituting n(H2 O) = 55.55 in eqn (i), we get n(C 2 H2 OH) − 0.040 or 0.96 n(C 2 H5OH) n(C 2 H5OH) + 55.55 Since
nC 2 H 5OH <<< nH2 O (55.5) nC 2 H 5OH + 55.55 = 55.55
Hence
nC 2 H 5OH = 55.55 × 0.04 = 2.2220 = 2.22 moles 2.22 So molarity = = = 2.22 M volume 1 1.29.
What will be the mass of one 12 C atom in g?
Ans.
1 mol of 12 C atoms (12 gm) = 6.022 × 1023 atoms ∴
1 atom of 12 C will have mass 12 = g 6.022 × 1023 = 1.99 × 10 −23 g.
1.30.
How many significant figures should be present in the answer of the following calculations? 0.02856 × 298.15 × 0.112 (i) 0.5785 (ii) 5.×5.364 (iii) 0.0125 + 0.7864 + 0.0215
Ans.
1.31.
(i) The least precise term has 3 significant figures (i.e., in answer should have 3 significant figures.
0.112). Hence, the
(ii)
Leaving the exact number (5), the second term has 4 significant figures. Hence, the answer should have 4 significant figures.
(iii)
In the given addition, the least number of decimal places in the term is 4. Hence, the answer should have 4 significant.
Use the data given in the following table to calculate the molar mass of naturally occurring argon. 109
Isotope
Ans.
Isotopic molar mass
Abudance
36
Ar
35.96755 gmol −1
0.337
38
Ar
37.96272 gmol −1
0.063
40
Ar
39.9624 gmol −1
99.600
Molar mass of Ar = Σ
Isotopic molor mass × Persentage 100
= 35.96755 × 0.00337 + 37.96272 × 0.00063
+39.96924 × 0.99600 100
= 39.948 gmol −1 . 1.32.
Calculate the number of atoms in each of the following : (i) 52 moles of He (ii) 52 u of He
Ans.
(iii) 52 g of He
(i)
mol of He = 6.022 × 1023 atoms
∴
52 mol of He = 52 × 6.022 × 1023 atoms = 3.131 × 1025 atoms no.of He moles = 6.023 × 1013 no.of He 52 = 6.023 × 1023
or
No of He = 52 × 6.023 × 1023 = 3.131 × 1025 (ii)
∴
1 atom of He u of He 4 u of He = 1 atom of He 1 52 u He = × 52 atoms = 13 atoms 4
(iii) 1 mole of He = 4 g = 6.022 × 1023 atoms ∴
52 g of He =
6.022 × 1023 × 52 atoms 4
= 7.8286 × 1024 1.33.
110
atoms
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at S.T.P.) of this welding gas is found to weight
11.6 g. Calculate (i) empirical formula (ii) molar mass of the gas, and (iii) molecular formula. Ans.
Mass of carbon in 3.38 g CO2 =
12 × 3.38 g = 0.9218 g 44
Mass of hydrogen in 0.690 gH2 O 2 = × 0.690 g = 0.0767 g 18 As compound contains only C and H, therefore, total mass of the compound = 0.9218 + 0.0767g = 0.9985g % of C in the compound =
0.9218 × 100 = 92.32 0.9985
=
0.0767 × 100 = 7.68 0.9985
% of H in the compound
Calculation of Empirical Formula Element
% by mass
Atomic mass
Moles of the element
Simplest molar ratio
Simplest whole no. molar ratio
C
92.32
12
1
1
H
7.68
1
92.32 = 7.69 12 7.68 = 7.68 1
1
1
∴
Empirical formula = CH
or Calculation of Molar Mass 10.0 L of the gas at S.T.P. weight = 11.6 g ∴
22.4 L of the gas at S.T.P. will weight 11.6 = × 22.4 = 25.984 g 10.0
Thus, molar mass = 26 gmol −1 Empirical formula mass of CH = 12 + 1 = 13 111
Calculation of molecular Formula Molecular mass 26 ∴ n= = =2 E .F . mass 13 ∴ 1.34.
Molecular formula = (CH)2 = C 2 H2
Calcium carbonate reacts with aqueous HCl according to the reaction : CaCO3 (s ) + 2HCL(aq) → CaCl2 (aq) + CO2 (g) + H2 O (I) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl ?
Ans.
Step 1. To calculate mass of HCl in 25 mL of 0.75 M HCl 1000 mL of 0.75 M HCl contain HCl = 0.75 mol = 0.75 × 36.5 g = 27.375 g ∴25 mL of 0.75 HCl will contain HCl 27.375 = × 25 g = 0.6844 g. 1000 Step 2. To calculate mass of CaCO 3 reacting completely with 0.9125 g of HCl CaCO3 (s ) + 2HCl(aq) → CaCl2 (aq) + CO2 (g) + H2 O(l) 2 mol of HCl, i.e., 2 × 36.5 g = 73 g HCl react completely with CaCO 3 = 1 mol = 100 g ∴
1.35.
0.6844 g HCl will react completely with CaCO 3 100 = × 0.6844 g = 0.9375 g. 73
Chlorine is prepared in the laboratory by treating manganese dioxide (Mn0 2 ) with aqueous hydrochloric acid according to the reaction. 4 HCl(aq) + MnO2 (s ) → 2H2 O(I) + MnCl2 (aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 u)
Ans.
1 mol of MnO2 , i.e., 55 + 32 = 87 gMnO2 react with 4 moles of HCl, i.e., 4 × 36.5 g = 146 g of HCl. ∴
112
5.0 g of MnO2 will react with HCl 146 = × 5.0 g = 8.39 g. 87
Additional Exercise Very Short Answer Questions 1. What is the basic SI unit for (i) Temperature
(ii) Electric current
2. Give one example each of homogeneous and heterogeneous mixture. 3. What is the number of hydrogen atoms in 60 a.m.u. ethane? 4. What is the value of gram molecular volume at S.T.P.? 5. 6 L of N 2 and 6 L of H 2 at S.T.P. are allowed to react completely to form ammonia. What is the volume of ammonia formed at S.T.P. ? 6. What is molecular mass of heavy water? 7. What is difference between molality and molarity? 8. Briefly explain precision and accuracy with their difference. 9. Under what conditions the zeros in a number are significant. 10. What is a limiting reagent.
Short Answer Qusestions 1. Study the number of significant figures in each of the following: (i) 2.56 ×10 3
(ii) 256
(iii) 5000
(iv) 0.00256
(v) 0.0320
(vi) 13.560.
2. A Block of wood having the dimensions 10cm × 5cm × 20cm weights 1.5 kg. What is the density of wood, expressed in grams per cubic centimetre? 3. What is the mass of 1.50 litre of mercury? The density of mercury is 13.6 g cm –3 . 4. Classify the following substances as elements, compounds and mixtures. In case of mixtures clearly indicate whether the mixture is homogeneous or heterogeneous. (i) Gasoline
(iii) Distilled water
(v) Air
(vi) Sand
(vii) Diamond
(viii) Wood.
(ii) Tap water
(iv) Milk
5. State and explain the law of multiple proportions and law of definite proportion. 6. What are the various postulates of Dalton’s atomic theory? 7. State the following terms: (i) Atomic mass (ii) Gram atomic mass (iii) Atomic mass unit.
113
8. Define atomic mass unit. What is its mass in terms of grams? 9. Can two different compounds have same empirical formula? Illustrate your answer with two examples. 10. Define molarity. What are its units? What is the effect of temperature on molarity of a solution?
mmm
114
