Stoichiometry and the Mole Concept Notes

Stoichiometry and The Mole Concept Symbols and Formulae A symbol is used to represent one atom of an element. The symbol can be a two letter, the first of which must be a capital. Examples are H for hydrogen and Li for Lithium. An integer in front of a symbol indicates the number of that atom. Examples are 3H represents represents three atoms of hydrogen, 5Cu represents represents five atoms of copper. The formula of a substance shows the relative numbers of atoms or ions that have combined together. Examples are as follow: The formula represents molecule of nitrogen Substance is an element

e.g

N2

e.g Substance is a covalent compound

H2 O e.g

Substance is an ionic compound

Na 2O

one

One molec molecule ule of nitro nitrogen gen contai contains ns two nitrogen atoms 2N2 repr epresen esentt two two mole molecu cule les s of nitrogen. Two molecules of nitrogen contain a total of four N atoms The formula represents one molecule of water One molecule of water contains two hydr hydrog ogen en atom atoms s and and one one oxyg oxygen en atom 2H2O repr epresen esentt two two mole molecu cule les s of water that contain a total of four H atoms and two O atoms Na2O represents one unit of sodium oxide One unit unit of sodiu sodium m oxide oxide contai contains ns + 2two Na ions and one O ion 2Na2O represents two units of sodium oxide that contain four Na + ions and two O2- ions

Writing Formulae of Simple Compounds The valency of an element is needed to be able to write the formulae of a compound. The valency of an element is a number which shows its combining power and in ionic compounds, the valency of an ion is equal to the charge 1

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POSITIVE ION H Hydrogen + Na Sodium + K Potassium 2+ Ba Barium 2+ Ca Calcium 2+ Mg Magnesium 3+ Al Aluminium + Ag Silver (I) 2+ Cu Copper (II) 2+ Fe Iron (II) +

2+

Pb

Zn2+ Ni2+ Fe3+ Cr3+

VALENCY 1 1 1 2 2 2 3 1 2 2

Lead (II)

2

Zinc Nickel (II) Iron (III) Chromium (III)

2 2 3 3

NEGATIVE ION F Fluoride Cl Chloride Br Bromide I Iodide 2S Sulphide 2O Oxide OH Hydroxide NO3 Nitrate NO2 Nitrite Hydrogen HCO3 -

carbonate

CH3CO OSO32SO42CO32PO43-

VALENCY 1 1 1 1 2 2 1 1 1 1

Ethanoate

1

Sulphite Sulphate Carbonate Phosphate

2 2 2 3

In writing formula of ionic compounds note that: • •

Ionic compounds are made up of positive and negative ions In an ionic compound, the total positive charges must equal the total negative charges. In other words, the total charges in an ionic compound must equal zero

Example 1 Sodium chloride

Magnesium oxide

Method A Formula of Na+ ions Valency of 1 ions Simplest ratio 1 of combining ions Formula NaCl 2+ Formula of Mg ions Valency of 2 ions Simplest ratio 1 of combining ions Formula MgO

Method B -

Cl 1 1

O22 1











Calcium nitrate

Sodium sulphate

Iron (II) phosphate

Formula of ions Valency of ions Simplest ratio of combining ions Formula Formula of ions Valency of ions Simplest ratio of combining ions Formula Formula of ions Valency of ions Simplest ratio of combining ions Formula

Ca2+

NO3-

2

1

1

2

Ca(NO3)2 Na+ SO421

2

2

1

Na2SO4 Fe PO432+

2

3

3

2

Fe3(PO4)2

Worked Example: 1. Iron Iron (III (III)) oxide xide is a chem chemic ical al comp compou ound nd foun found d in an iron iron ore called haematite (a) (a) Give Give the the sym symbo boll of all all the the elem elemen ents ts in in iron iron (II (III) I) oxi oxide de

(b) (b)

Give Give the the for formu mula la of all all the the ions ions in in iron iron (II (III) I) oxi oxide de

(c) (c)

What What is the the for formula mula of iron iron (III (III)) oxi oxide de? ?

In writin writing g the formu formula la of coval covalent ent compou compounds nds,, the valenc valency y of an elem elemen entt is the the numb number er of cova covale lent nt bond bonds s whic which h it can can form form with with hydrogen atom.

Relative Molecular Mass (M r) By definition, Mr of a substance is defined as the average mass of a











For example; 1. Relative elative molecula molecularr mass mass of sulphuric sulphuric acid acid The formula of sulphuric acid = H 2SO4 Sulphuric acid contains: 2 Hydrogens 1 Sulphur 4 Oxygens

2 x Ar of H = 2 x 1 = 2 1 x Ar of S = 1 x 32 = 32 4 x Ar of O = 4 x 16 = 64 Therefore Mr = 2 + 32 + 64 = 98

2. Relative elative molecular molecular mass mass of hydrated hydrated sodium sodium carbonate carbonate The formula of hydrated sodium carbonate = Na2CO3 . 10H2O Hydrated sodium carbonate contains: 2 Sodiums 2 x Ar of Na = 2 x 23 = 46 1 Carbons 1 x Ar of C = 1 x 12 = 12 3 Oxygens 3 x Ar of O = 3 x 16 = 48 10 Water 10 x (Mr of H2O) molecules 10 x (2 x Ar of H + 1 x Ar of O) 10 x (2 x 1 + 1 x 16) 10 x (18) = 180 Therefore Therefore Mr = 46 + 12 + 48 + 180 = 286

Test Your Understanding Using the Ar found in your periodic table, find the Mr of the following substances: Calcium hydroxide Ca(OH)2 Sodium oxide

Na2O

Ammonium sulphate Calcium hydrogen Ca(HCO3) carbonate 2 Nitrogen trichloride Hydrogen sulphide











The Percentage Composition of Elements in a Molecule If we know the chemical formula of a substance, we can calculate its percentage composition by mass, mass, without doing as experiment. The rules are: 1. Write down down the chemical chemical formul formula a of the substance substance 2. Find out its Mr 3. Express Express the total total mass mass of each elemen elementt as a percentag percentage e of the total relative molecular mass For example; 1. Percenta Percentage ge compositi composition on of sulphuri sulphuric c acid Chemical formula H2SO4 Relat elativ ive e mole molecu cula larr 98 mass 2 Percentage ×100 =2.04% % of hydrogen = composition 98 % of sulphur = % of oxygen =

32 98 64 98

×100 =32.65% ×100 =65.31%













2. Percentage Percentage composition of hydrated hydrated sodium sodium carbonate crystals Chemical formula Na2CO3 . 10H2O Relat elativ ive e mole molecu cula larr 286 mass 46 Percentage ×100 =16.08% % of sodium = composition 286 % of carbon = % of oxygen = % of water =

12 286 48 286

180 286

×100 =4.20%

×100 =16.78%

×100 =62.94%

The percentage composition of pure substances does not change. For example water molecule will always contain 11.11% hydrogen and 88.89% oxygen whether its in New York, New Delhi, New Zealand or the Moon (it never varies)

Test Your Understanding What What is the percen percentag tage e compos compositi ition on of the eleme elements nts,, carbon carbon and hydrogen, in the hydrocarbon molecules listed below? a) Methane (CH4)

b) Ethane (C2H6)





















The Emperi Emperical cal Formula ormula:: The Fo Formu rmula la from from the Percen Percentag tage e Composition By defini definitio tion, n, the empiri empirical cal formul formula a is the simpl simplest est formu formula la which which shows the relative numbers of the atoms of the different elements present. The The empir empirica icall formu formula la can be deter determi mined ned once once the percen percentag tage e or mass of each element is known. The rules are: 1. Divide the percentage or mass of each element by its Ar

2. Divide by the smallest smallest number to convert to the simplest simplest ratio 3. The number number of atoms atoms of the different different elements elements is the empiri empirical cal formula For example: Given that a molecule contain 88.89% oxygen and 11.11% hydrogen, what is its empirical formula?

Divide % by Ar Simplest ratio Empirical formula

H 11.11% ÷ 1 = 11.11 11.11 ÷ 5.55 = 2 H2

O 88.89% ÷ 16 = 5.55 5.55 ÷ 5.55 = 1 O

Test Your Understanding Calculate the empirical formula of the hydrocarbon molecules listed below using the percentage composition given a) 75 75% % C, C, 25% 25% H

b) 80 80% % C, C, 20% 20% H











THE MOLE What is the mole? It is a term in chemistry to describe an amount of atoms, ions and molecules. It enables chemists to count these particles by weighing.

Definition of the mole: The amount of substance which contains the Avogadro Number of particles. The Avogadro Number is 6.02 X 1023 (or 602 000 000 000 000 000 000 000)

Definition of Avogadro Number: The number of atoms in 12g of the carbon-12 isotope Substances Drawing Pins

Unit

Number

Gross

144

Football boots Pair

2

Eggs Dozen

12

Playing cards Pack

52











Moles of Atoms: Relative Atomic Mass in Grams If we weigh 12 g of carbon, this would contain 6.02 x 1023 atoms of carb carbon on.. In the sam same way, way, so would would 23 g of sodi sodium um,, or 27 g of aluminium, aluminium, or 16g of oxygen, or 1 g of hydrogen. The mass of one mole of atoms is its relative atomic mass in grams Element Hydrogen Carbon Oxygen Aluminium Calcium Silver

Relative atomic mass 1 12 16 27 40 108

Mass Mass of 1 mol mole e 1g 12 g 16 g 27 g 40 g 10 8 g

Num Number ber of atom atoms s in 1 mole 6.02 x 1023 6.02 x 1023 6.02 x 1023 6.02 x 1023 6.02 x 1023 6.02 x 1023

Moles of Molecules: Relative Molecular Mass in Grams If we weigh 44 g of carbon dioxide, this would contain 6.02 x 1023 molecules of carbon dioxide. In the same way, so would 18 g of water, or 32 g of oxygen gas, or 2 g of hydrogen gas. mole of molecu molecules les is its relative relative molecular molecular The The mass mass of one mole mass in grams Molecule

Carbon dioxide (CO2) Water (H2O) Oxygen (O2) Hydrogen (H2)

Relative Molecular mass 44

Mass Mass of 1 mol mole e

Num Number ber of atom atoms s in 1 mole

44 g

6.02 x 1023

18 32 2

18 g 32 g 2g

6.02 x 1023 6.02 x 1023 6.02 x 1023













In all chemical calculations, gas volumes will be measured at room temperature and pressure (r.t.p). Room temperature is taken as 25oC (298K) and room pressure as one atmosphere (760 mm of mercury).











Molecule

Carbon dioxide (CO2) Chlorine (Cl2) Oxygen (O2) Hydrogen (H2)

Relative Molecular mass 44

Mass of 1 mole 44 g

Number of Volume atoms in 1 occupied mole at r.t.p 23 6.02 x 10 24 000 cm3

71 32 2

71 g 32 g 2g

6.02 x 1023 6.02 x 1023 6.02 x 1023

Calculations with Moles Formula A:

mass mass in grams grams Numberof moles = Ar or or Mr of atom or molecule molecule Example: How many moles are there in (a)88 g of carbon dioxide? (b)64 g of oxygen molecules? Solution: (a)Mr (a) Mr of carbon dioxid dioxide e molecule = 44

24 000 cm3 24 000 cm3 24 000 cm3











(c) 64 g of methane methane gas?











Formula B:

Mass in grams = Number of moles x Ar or Mr of substance Example: How many grams are there in (a)10 moles of water? (b)0.25 (b) 0.25 moles of oxygen atoms? Solution: (a) Mr of water molecules H2O = 18

Mass in grams = 10 x 18 = 180 g (b)Mr (b) Mr of oxygen atoms O = 16 Mass in grams = 0.25 x 16 = 4 g

Test Your Understanding How many grams are there in (a) 5 moles of carbon dioxide CO2 ?











Formula C:

Number of particles = Number of moles x Avogadro Number Example How many particles are there in (a)16 g of oxygen molecules? (b)11 (b) 11 g of carbon dioxide molecules? Solution (a)Number (a) Number of moles of oxygen molecules

= Mass / Mr

= 16 / 32 = 0.5 moles Number of particles

= 0.5 x 6 x 1023

= 3 x 1023 molecules (b)Number (b) Number of moles of carbon dioxide dioxide molecules

= Mass / Mr

= 11 / 44 = 0.25 moles Number of particles

= 0.25 x 6 x 1023













Formula D:

Volume of gas = Number of moles x molar gas volume Example What volume (at r.t.p) would the following gases occupy? (a)2 moles of oxygen gas? (b)0.1 mole of hydrogen gas? (Molar Gas Volume at r.t.p = 24 dm3) Solution (a) Volume of oxygen gas = 2 x 24 = 48 dm3 (b) Volume of hydrogen gas = 0.1 x 24 = 2.4 dm3











Moles and Equations Stoichiometry is the study of the quantitative composition composition of chemical subs su bsta tanc nces es and and quan quanti tita tati tive ve chan change ges s that that occu occurr in chem chemic ical al reactions. Consider a car manufacturer uses liquid hydrogen as fuel. He may have have to dete determ rmin ine e how how much much fuel fuel is nece necess ssar ary y for for a part partic icul ular ar journey. When hydrogen burns in oxygen it forms water. The balanced chemical equation is: 2H2 (g) + O2 (g)



2H2O (g)

In the reaction, reaction, 2 hydrogen molecules react with one oxygen molecule to form 2 water molecules. If we use moles, then 2 moles of hydrogen gas react with one mole of oxygen gas to form two moles of steam . A balanced chemical equation shows the stoichiometric ratio and this gives the exact amount of reactant and product.











Mole Ratio:

No No of moles moles of oxygen oxygen No No of moles moles of water water

=1 2 1

No of moles of oxygen =   x No of moles of water  2 1

Mass Mass of water water

No of moles of oxygen =   x Mr of water water  2 1

54

No of moles of oxygen =   x   = 1.5 moles  2   18  In the reaction, 1.5 moles of oxygen or 48g of oxygen [1.5 x 32] are required.











maximum limits of impurity: chloride 0.02% sulphate 0.02% sodium 0.05%











Percentage Yield In any reaction, the amount of product produced experimentally is alwa always ys less less than than prod produc uced ed theo theorretic etical ally ly (by (by calc calcul ulat atio ion n from from equation). This can be expressed in terms of percentage percentage yield

Percentage yield =

erimental value

exp  theoreti ×100% cal value value   theoretical

Example: Calc Calcul ulat ate e the the per percent centag age e yiel yield d if, if, on heat heatin ing g 50 50g g of lime limest ston one e (CaCO3), 21g of lime (CaO) were obtained. Calcium carbonate



CaCO3 (s)

CaO (s)



calcium oxide + carbon dioxide + CO2 (g)













Moles and Solution What is a solute? solvent? and solution? Solute: Solv Solven ent: t: Sol Solution tion::

a substance that dissolves in a liquid the the liq liqui uid d in in whi which ch the the sol solut ute e diss dissol olve ves s the the res resul ulti ting ng mixtur xture e

SOLUTE

+

SOLVENT



SOLUTION











Concentration of Solutions It is necessary for a chemist to try to work out the number of moles of solute in a certain volume of solution. For example how much sodium hydroxide is contained in 200 cm3 of 2 mol dm-3 sodium hydroxide solution? This can be calculated as follows: 1000 cm3 of 2 mol dm-3 contains 2 moles of NaOH 1 cm3 of 2 mol dm-3 contains

2 1000

mole of NaOH

2 × 200











To find the concentration of the unknown acid, the following acid and alkali relationship is required: MV

MV

  =   n   n  acid alkali where V= n= equation

M = molarity or concentration inn mol dm-3 volume in cm3 number of moles shown in the chemical











Test Your Understanding What What volu volume me of 2 mol mol dm-3 sul sulphu phuri ric c acid acid would would be requi require red d to neutralize (a) 25 cm3 of 4 mol dm-3 sodium hydroxide?











Further Further Molar Calculations from Equation Problem 1 Calculate how much carbon dioxide gas would be evolved if 50 g of chalk (calcium carbonate) were dissolved in excess acid. You may assume that the reaction takes place at room temperature and pressure (r.t.p). Molar gas volume = 24 000 cm3 The equation for the reaction is;

Stoichiometry and the Mole Concept Notes

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