Chapter 1-8 Worksheet Solutions

Worksheet 1.1: Solutions

Precision, accuracy and significant figures No.

Answer

1

B and D.

2

a N

3

Averaging a set of results reduces the effects of random errors associated with taking measurements.

4

a 3

5

a 1.407

6

a 7.80

7

a 6

8

a 6.7

9

a 44.6

10

Measured quantities have an inherent uncertainty.

b P

c B

b 4

c 3

× 10

× 10

2

b 5.005

b 6.01

3

× 10

d 4

× 10

b 6.000

2

b 0.30

× 10

2

× 10

3

c 9.83

c 9.800

× 10

× 10

1

2

d 7.5

d 6.00

× 10

–3

× 10

-4

3

–1

or 3.0 × 10

c 2.0

d 4.4

× 10

2

b 358.2

722

= 60.2 g (The answer should be given to 3 significant figures. The 12 is an exact 12 number, and is therefore not relevant to the significant figure count.)

Page 1 © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008. This page from the Chemistry Dimensions 2, Teacher’s Resource may be reproduced for classroom use.


Calculations involving gases and solutions No. 1

Answer a

n(C6H12O6) =

5.00

m

=

M

180.0

mol

n(O) = 6 × n(C6H12O6) = 6 × 8.36

m

n(H2O2) =

b

M

=

5.00 180.0

= 0.167 mol

mol

34.0

n(atoms) = 4 × n(H2O2) N (atoms) (atoms) = n × N A = 4 × 2

3

4 5

7

3.0 23

V 1 T 1

pV = = nRT and and n =

c(NaCl) =

n

=

m M m

=

V M × V c1 × V 1 1.2 × 1.0

∴ 0.79

=

c2

0.67

pV = = nRT and and n =

a b

= 117 mL

0.090

V 1 =

2

= 4.4 × 10 kPa 2

–1

V 2 T 2

∴

V 2 =

∴ pV = =

=

V 1 × T 2 T 1

=

600 × 325 260

= 750 mL

62.0 × 8.31 × 397 mRT RT ∴ V = = = = 22.1 L pM 210 × 44.0 M m

70.2 58.5 × 1.0

= 1.2 M

= 1.79 L

L of water must be added. m

M The gas is methane.

9

0.30 0.30 × 35.0 5.0

23

× 6.02 × 10 = 2 × 10 g mol

= constant ∴

V 2 =

8

=

0.49 × 8.31 × 327

–22

M = = 4 × 10

T 6

c2

23

× 6.02 × 10 = 5.92 × 10

(H2O) added = 117 – 35 = 82 mL V (H

pV = = nRT ∴ p =

V

34.0

c1 × V 1

c1 × V 1 = c2 × V 2, ∴ V 2 = ∴

8.36

c2 × V 2 c1

=

∴ pV = =

1.50 × 250 5.00

mRT 2.06 × 8.31× 8.31 × 300 –1 RT ∴ M = = = = 16 g mol pV 20 × 16 M m

= 75 mL

1.50% m/v means 1.5 g of ammonia in 100 mL of solution We require x g of ammonia in 150 mL. 150 ∴ x = 1.50 × = 2.25 g 100 Page 1 © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008. This page from the Chemistry Dimensions 2, Teacher’s Resource may be reproduced for classroom use.


Calculations involving gases and solutions No. 10

11

Answer

p × V = = constant ∴ p1V 1 = p2V 2 ∴ V 2 =

pV = =

mRT

mRT

∴ M = =

=

p1 × V 1 143.8 × 687 = = 496 mL p2 199

0.778 × 8.31 × 299

pV M 99.8 × 122 × 10 The halogen must be bromine (Br 2).

12 13

−3

–1

= 159 g mol

600 ppm = 600 mg in 1.0 L = 600 000 μg in 1.0 L = 600 000 ppb n(Pb(NO3)2) = c(Pb(NO3)2) =

m

3.5

=

M 331.2 n 0.0106

= 0.0106 mol

= = 0.177 M V 0.060 0.177 × 10 c × V c2 = 1 1 = = 0.059 M V 2 30 14

c2 × V 2

c1 =

V 1

=

0.108 × 300 30

= 1.08 M

n(Na2CO3) = c × V = = 1.08 × 0.030 = 0.0324 mol m(Na2CO3) = n × M = = 0.0324 × 106.0 = 3.4 g 15

2.0% m/v means 2.0 g of glucose in 100 mL of solution. s olution. ∴ in 50 mL of solution, s olution, 0.50 × 2.0 g Total m(glucose) = (0.50 × 2.0) + (0.50 × 6.0) = 4.0 g 4.0 g of glucose in a volume of 300 mL gives a concentration of

16

p1V 1 T 1

17

c2 =

=

p2V 2 T 2

c1 × V 1 V 2

=

∴

V 2 =

p1V 1T 2

16 × 50 100

T 1 p2

=

750 × 190 × 273 298 × 760

4.0 300

×100 =

1.3% m/v

= 172 mL

= 8% m/v

8% m/v = 8 g/100 mL = 80 g L –1 = 80 000 mg L –1 = 8.0 × 104 ppm 18

a

b

2.33 g in 398 mL n(AgNO3) =

m

∴ 2.33

=

×

100 398

in 100 mL = 0.585% m/v

2.33

mol 169.9 M n 2.33 –2 c(AgNO3) = = = 3.45 × 10 M 169.9 × 0.398 V Page 2 © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008. This page from the Chemistry Dimensions 2, Teacher’s Resource may be reproduced for classroom use.


Stoichiometry No. 1

Answer

n(SO2) =

V

50.0

=

24.5

V M

= 2.04 mol

n(H2SO4) = n(SO2) 2 m(H2SO4) = n × M = 2.04 × 98.1 = 2.00 × 10 g 2

n(H3PO4) =

m

=

1000

= 10.2 mol

M 98.0 n(NO2) = 5 × n(H3PO4) = 51.0 mol 3 m(NO2) = n × M = 51.0 × 46.0 = 2.35 × 10 g 3

n(NaHCO3) = c × V = 2.86 × 0.0299 = 0.0855 mol n(H2SO4) = c(H2SO4) =

4

a b

5

2

× n(NaHCO3) = 0.0428 mol

n

=

0.0428 0.0238

V

= 1.80 M

2NaN3(s) → 2Na(s) + 3N2(g) 75 m n(NaN3) = = M 65.0 3 n(N2) = × n(NaN3) 2 nRT 3 75 × 8.31 × 303 = 43 L V (N2) = = × 2 65.0 × 101.3 p

m(Br 2) = density × volume = 3.12 × 10.0 = 31.2 g 31.2 m = 0.195 mol n(Br 2) = = M 159.8 n(NaOH) = c(NaOH) =

6

1

n(H2SO4) =

10 4

n V m

× n(Br 2) = 0.488 mol =

0.488 0.239

=

= 2.04 M

19.56

= 0.1994 mol

M 98.1 n(NH3) = 2 × n(H2SO4) = 2 × 0.1994 = 0.3988 mol 0.3988 × 8.31 × 360 nRT = = 3.94 L V (NH3) = 2.99 × 101.3 p



Stoichiometry No.

Answer

7

MgCO3(s) → MgO(s) + CO 2(g) 100 m = 2.48 mol n(MgO) = = M 40.3 n(CO2) = n(MgO) = 2.48 mol V (CO2) = n × V M = 2.48 × 22.4 = 55.6 L

8

a b

9

CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l) 10.0 m = 0.100 mol n(CaCO3) = = M 100.1 n(HNO3) = c × V = 0.500 × 0.100 = 0.0500 mol 0.100 mol of CaCO 3 will react with 0.200 mol of HNO 3, ∴ HNO3 is the limiting reagent. 1 n(CO2) = × n(HNO3) = 0.0250 mol 2 V (CO2) = n × V M = 0.0250 × 24.5 = 0.613 L m

6.57

= 0.0877 mol 74.9 n(NaOH) = c × V = 0.779 × 0.250 = 0.195 mol 1 As: 3 NaOH, ∴ 0.0877 mol of arsenic would react with 0.263 mol of sodium hydroxide ∴ NaOH is the limiting reagent. n(As) =

=

M

n(H2) produced =

1 2

× n(NaOH) = 0.0975 mol

m(H2) = n × M = 0.0975 × 2.0 = 0.195 g 10

a b

Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l) m 8.00 n(Ba(OH)2) = = = 0.0467 mol 171.3 M n(HCl) = c × V = 1.886 × 0.120 = 0.226 mol 0.0467 mol of Ba(OH) 2 would react with 0.0934 mol of HCl, n(BaCl2) = n(Ba(OH)2) = 0.0467 mol 0.0467 n = = 0.389 M c(BaCl2) = V 0.120

HCl is in excess

∴



Determining the molecular formula of a gas

No

Answer

1

Na 22.8

:

B : O 21.5 55.7 Mole ratio : : 23 10.8 16.0 0.99 : 1.99 : 3.48 1 : 2 : 3.5 2 : 4 : 7 Empirical formula is Na2B4O7.

2

C

:

As 75.7

:

H : O 0.112 9.57 × 10 21 Mole ratio 0.0556 : : 1.0 6.02 × 10 23 0.0556 : 0.111 : 0.0159 3.5 : 7 : 1 7 : 14 : 2 Empirical formula is C7H14O2.

3

Mole ratio

O 24.3

: 74.9 16.0 1.01 : 1.52 2 : 3 Empirical formula is As2O3. EFM = 197.8

MF is (As2O3)a, where a =

RMM EFM

=

395.6 197.8

= 2

Molecular formula is As4O6. 4

C: H :O 40.0 6.67 53.3 Mole ratio : : 12.0 1.0 16.0 3.33 : 6.67 : 3.33 1 : 2 : 1 Empirical formula is CH2O. EFM = 30

MF is (CH2O)a, where a =

RMM EFM

=

60 30

= 2

Molecular formula is C2H4O2. 5

It absorbs water and so allows the mass of hydrogen in the sample to be determined.

6

It absorbs carbon dioxide and so allows the mass of carbon in the sample to be determined. Page 1 © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008. This page from the Chemistry Dimensions 2, Teacher’s Resource may be reproduced for classroom use.


Determining the molecular formula of a gas No 7

Answer

m(C) in sample = m(H) in sample =

12.0 44.0 2.0 18.0

× 43.54 = 11.9 g × 22.26 = 2.47 g

C:H 11.9 2.47 Mole ratio : 12.0 1.0 0.99 : 2.47 1 : 2.5 2:5 Empirical formula is C2H5. 8

9

10

The sample must vaporise at a temperature below that of the steam (below 100°C).

M =

mRT pV

=

0.16 × 8.31 × 373 101.3 × 46 × 10

m(C) in sample = m(H) in sample =

12.0 44.0 2.0 18.0

−

3

= 106 g mol –1

× 27.83 = 7.590 g × 11.38 = 1.264 g

m(O) in sample = 18.99 – 7.590 – 1.264 = 10.14 g

C : H : O 7.590 1.264 10.14 Mole ratio : : 12.0 1.0 16.0 0.633 : 1.264 : 0.634 1 : 2 : 1 Empirical formula is CH2O. mRT 6.21 × 8.31 × 473 = = 89.99 g mol –1 M = 2 pV 1.25 × 10 × 2.17 EFM = 30

MF is (CH2O)a, where a =

RMM EFM

=

90 30

= 3

Molecular formula is C3H6O3. 11

Mass spectrometry



Gravimetric analysis No.

Answer

1

It was necessary to dry it so that water did not add any mass to the compound. Grinding it helped it to dissolve.

2

n((NH4)2SO4)

=

m =

M

1.00 132.1

= 7.57 × 10 –3 mol

3

Between steps 5 and 6, additional filtration with washing of the residue would have been required to remove the insoluble components.

4

BaCl2(aq) + (NH4)2SO4(aq) → BaSO4(s) + 2NH4Cl(aq) If the fertiliser was pure: –3 n(BaCl2) needed = n((NH4)2SO4) = 7.57 × 10 mol V (BaCl2)

=

n =

c

7.57 × 10 -3 0.200

= 0.0379 L = 37.9 mL

5

The precipitating agent should be in excess to ensure complete precipitation.

6

The supernatant liquid was tested with BaCl 2 solution. No precipitate was observed.

7

Heat helps larger crystals of BaSO4 form so it is better to wash it in hot water than cold. Washing removes any ions adsorbed onto the surface of the precipitate.

8

This means that the precipitate was dried, weighed, then dried and weighed again. If the masses are the same then constant mass has been achieved and there is no more water present.

9

n((NH4)2SO4) m((NH4)2SO4)

= n(BaSO4) =

1.63

m =

M

233.4

–3

= 6.98 × 10 mol

–3

= n × M = 6.98 × 10 × 132.1 = 0.923 g 0.923 % m/m ((NH4)2SO4) = × 100 = 92.3% 1.00 10

• • • • •

Errors in weighing Incomplete dissolution of fertiliser Incomplete precipitation Losses due to filtration Not completely drying the precipitate



Gravimetric analysis problems No. 1 2

Answer

Neither sodium nor nitrate ions form any insoluble compounds. a i b

AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s) ii AgNO3(aq) + KCl(aq) → KNO3(aq) + AgCl(s) Because chloride ions are in both compounds, the chloride concentration is measured correctly. The amount of chlorine is not equal to the amount of sodium or to the amount of potassium, therefore neither of these two metals can be analysed using this procedure.

3

Water for rinsing should be used sparingly. The more water used, the more precipitate that will dissolve and be lost.

4

There are many possible answers. For example: Mg(NO3)2 → MgCO3(s) AgNO3 → AgI(s) CuSO4 → Cu(OH)2(s)

5

a b

6

7

2–

Barium sulfate can form very fine particles. These particles will pass through normal filter paper. a b

8

2+

Ba (aq) + SO4 (aq) → BaSO4(s) By simply removing the water, the positive ion is caught up in the remaining solid as well as the sulfate ion, and any other ionic substances that are in the fertiliser.

a b

AgNO3(aq) + LiCl(aq) → LiNO3(aq) + AgCl(s) n(AgNO3) = c × V = 2.0 × 0.025 = 0.050 mol 0.02 mol of AgNO 3 reacts with 0.02 mol of LiCl, hence AgNO 3 is in excess. There is no problem with adding excess silver nitrate. In fact, it is important to add an excess amount to ensure all the chloride is precipitated. C8H15Cl3 + 3AgNO3(aq) → 3AgCl(s) + … m 0.478 = 0.00333 mol = n(AgCl) = 143.4 M 1 × 0.00333 = 0.00111 mol n(C8H15Cl3) = 3 m(C8H15Cl3) = n × M = 0.00111 × 217.5 = 0.24 g 0.24 % purity = × 100 = 12% 2.0



Gravimetric analysis problems No.

Answer

9

a b

10

a b

2FeCl3(aq) + …→ Fe2O3 + … 0.644 m = 0.0040 mol = n(Fe2O3) = 159.6 M n(FeCl3) = 2 × n(Fe2O3) = 0.0080 mol m(FeCl3) = n × M = 0.0080 × 162.3 = 1.3 g 1.3 % purity FeCl 3 = × 100 = 65% 2.0 Mass will be too high, ∴ the % P will be overestimated. Precipitate will be lost, ∴ the % P will be underestimated.



Revision of acids and bases No.

Answer

1

a

2

a

3

a

2KHCO3(s) + H2SO4(aq) → K 2SO4(aq) + 2H2O(l) + 2CO2(g) b Fe2O3(s) + 6HNO3(aq) → 2Fe(NO3)3(aq) + 3H2O(l) c Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)

b 4

a b c

5

2+

–

H2O Ca(OH)2(s) ⎯ ⎯ ⎯ → Ca (aq) + 2OH (aq) + – b H2SO4(l) + H2O(l) → H3O (aq) + HSO4 (aq) – + 2– HSO4 (aq) + H2O(l) → H3O (aq) + SO4 (aq) c Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l)

a b

2–

2–

S and O H2SO4 and H3PO4 –

+

HCl/Cl and NH4 /NH3 – – 2– H2O/OH and OH /O H2CO3/HCO3 – /CO32– and H2SO4/HSO4 – /SO42– ( H 2CO3 decomposes to form H 2O and CO2.) HSO4 – (aq) + OH – (aq) → SO42– (aq) + H2O(l) – – HSO4 (aq) + HCl(aq) → H2SO4(aq) + Cl (aq)

6

Different acid strengths (e.g. one strong, one weak) or different acid types (e.g. one monoprotic, one diprotic)

7

a

b

8

Methanoic acid is a weak acid that only partially ionises in water, producing few ions and so having low conductivity. Hydrochloric acid is a strong acid that ionises completely in water, producing many ions and so having high conductivity. Only the O–H bond has sufficient polarity to make the hydrogen atom acidic. The three C–H bonds are only slightly polar, so these hydrogen atoms are not acidic.

The solutions listed in order of decreasing pH are: NaOH is a strong base, ∴ high pH NH3 is a weak base, ∴ pH > 7 but not too high H2O has pH = 7 CH3COOH is a weak acid, ∴ pH < 7 HNO3 is a strong acid, ∴ low pH H2SO4 is a strong, diprotic acid, ∴ pH < pH of HNO 3



Revision of acids and bases No.

Answer

9

a

HNO3 is a strong, monoprotic acid. It completely ionises in water. + 2.0 ∴ [ H 3 O ] = [HNO 3 ] = 0.050 = 5.0 × 10 M −

(

)

pH = −log10 [H 3 O + ] = −log 10 5.0 × 10 2.0 = 1.3 – b Ba(OH)2 completely dissociates in water to produce 2 OH ions per unit of Ba(OH) 2. ∴

−

[OH ] = 2 × 0.50 = 1.0 = 10 0 −

[ H 3 O + ] × [OH ] = 10

14

−

∴

10 0 × [ H 3 O + ] = 10

∴

[ H 3 O + ] = 10

pH

∴

10

= −

14

−

−

14

−

M

log10 [H 3 O + ] = − log10 10

14

−

=

14

n(H+) = c × V = 10 –3.0 × 0.0200 = 2.00 × 10 –5.0 n 2.00 × 10 5.0 + 4.0 = 10 [H 3 O ] = = 0.200 V −

−

pH = −log 10 [ H 3 O + ] = −log10 10 4.0 = 4.0 A 10-fold dilution produces a change of one pH unit. −

11

c1V 1 = c2V 2 + –pH –2.0 + –pH –2.5 c1 = [H3O ] = 10 = 10 and c2 = [H3O ] = 10 = 10 –2.0 × 50.0 = 10 –2.5 × V 2 ∴ 10 V 2 = 158 mL, therefore 108 mL must be added.

12

n(H ) = n(HCl) = c × V = 0.00100 × 20.00 × 10 = 2.00 × 10 mol – –3 –5 n(OH ) = 2 × n(Ba(OH)2) = 2 × c × V = 2 × 0.00100 × 20.00 × 10 = 4.00 × 10 mol – –5 ∴ OH is in excess by 2.00 × 10 mol n 2.00 × 10 5 4 [OH ] = = mol = 5.00 × 10 3 V 40.00 × 10

+

–3

–5

−

−

−

−

[H 3 O + ] × [OH ] = 10 −

−

4

[ H 3 O + ] = 10

∴

5.00 × 10

∴

[ H 3 O + ] = 2.00 × 10

pH

∴

= −

×

14

−

11

−

14

−

M

log10 [ H 3 O + ] = −log 10 (2.00 × 10

11

−

) = 10.7



Acid-base titrations No.

Answer

1

n(Na2CO3) = c × V = 0.0500 × 0.250 = 0.0125 mol m(Na2CO3) = n × M = 0.0125 × 106.0 = 1.325 g 1 Accurately weigh approximately 1.3 g sample of the primary standard. 2 Transfer the sample to a 250.0 mL volumetric flask. 3 Ensure complete transfer by washing with water. 4 Dissolve the primary standard in water by shaking. 5 Add water to make up the solution to the calibration mark, then mix again.

2

The sodium hydroxide is very concentrated. An exact concentration cannot be calculated using the data available.

3

The ratio of volumes is 20:5 or 1:4. This means the sodium hydroxide is four times more concentrated than the nitric acid. Therefore the concentration is 0.4 M. This ratio only works if the molar ratio in the balanced equation is 1:1.

4

a b

5

a b

2HNO3(aq) + Na2CO3(aq) → 2NaNO3(aq) + H2O(l) + CO2(g) n(HNO3) = c × V = 0.10 × 0.040 = 0.0040 mol 1 n(Na2CO3) = × n(HNO3) = 0.0020 mol 2 n 0.0020 = 10 mL V (Na2CO3) = = 0.20 c HCl(aq) + NH3(aq) → NH4+(aq) + Cl – (aq) + i At the equivalence point, some NH4 exists. This is a weak acid, hence the pH will be about 5. ii After excess HCl is added, the pH is slightly greater than 1. (pH of the acid is 1. The acid will be diluted by the increased volume due to the neutralised ammonia solution, giving a concentration less than 0.1 M and so a pH greater than 1.) iii

6

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) n(NaOH) = c × V = 0.114 × 0.01545 = 0.001761 mol n(CH3COOH) = n(NaOH) n(CH3COOH) in original 25.0 mL = n(CH3COOH) in aliquot × c(CH3COOH) =

n V

=

0.00881 0.0250

100 20.0

= 0.00881 mol

= 0.352 M



Acid-base titrations No.

Answer

7

Indirect titrations are performed when the substance to be analysed is: insoluble • volatile (so the titration method is not fast enough to carry out the analysis before the • concentration of the solution changes as evaporation occurs ) a weak acid or a weak base (it is sometimes difficult to obtain sharp endpoints in these • titrations ).

8

a

b c

9

CH3COO – is present at the equivalence point. This weak base will mean that the pH will be above 7 at the equivalence point. An indicator such as methyl red would be a poor choice because it does not change colour until pH 3, long after the equivalence point has been reached.

a i

Water ii Solution going into it—hydrochloric acid iii Solution going into it—sodium carbonate iv Water One method to try with this style of question is to picture a very exaggerated amount of water in the piece of glassware and to ask whether this will affect the experiment in any way. b 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) n(Na2CO3) = c × V = 0.100 × 0.0183 = 0.00183 mol n(HCl) = 2 × n(Na2CO3) = 2 × 0.00183 = 0.00366 mol n(HCl) in original 10.0 mL = n(HCl) in aliquot × c(HCl) =

10

n V

=

0.0458 0.0100

250.0 20.00

= 0.00366 ×

250.0 20.00

= 0.0458 mol

= 4.58 M

2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2H2O(l) n(Mg(OH)2) = c × V = 0.10 × 0.0250 = 0.00250 mol n(HNO3) = 2 × n(Mg(OH)2) = 2 × 0.00250 = 0.00500 mol n 0.00500 = 0.370 M c(HNO3) = = 0.0135 V



A back titration No.

Answer

1

a CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) b HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

2

a b c d

–3

–2

n(HCl) initially = c × V = 1.020 × 40.00 × 10 = 4.080 × 10 mol –3 –3 n(NaOH) = c × V = 0.275 × 25.56 × 10 = 7.029 × 10 mol –3 –3 n(HCl) unreacted = n(NaOH) = 0.275 × 25.56 × 10 = 7.029 × 10 mol n(HCl)reacting = n(HCl) initially – n(HCl)unreacted

= 4.080 × 10 –2 – 7.029 × 10 –3 = 3.377 × 10 –2 mol

e n(CaCO3) =

1 2

× n(HCl) reacting =

f m(CaCO3) = n × M = g % CaCO3 in marble =

1 2

1 2

–2

× 3.377 × 10 = 1.689 mol

× 3.377 × 10 –2 × 100.1 = 1.690 g

m(CaCO3 ) m( marble)

×

100 1

1.690

=

1.740

×

100 1

= 97.13%

3

Solid sodium hydroxide absorbs water from the atmosphere. Solid samples may therefore be damp and therefore impure. Sodium hydroxide solutions react with carbon dioxide in the atmosphere, decreasing the concentration of the solution. 2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l)

4

CO2 is an acidic oxide. It reacts with NaOH. If the CO 2 was not removed, more NaOH would be required for the titration, leading to an increased value for the unreacted HCl, and hence a decreased value for the CaCO3 percentage.

5

Calcium carbonate is not soluble. The carbonate ion is a weak base that gives an indistinct endpoint in a direct titration.

6

a This would dilute the acid. More would be required to react with the marble, giving an increased percentage for the carbonate. b This would dilute the NaOH. More would be required to react with the HCl, giving an increased value for the unreacted HCl. This, in turn, would give a decreased value for the reacting HCl, and so a decreased percentage for the carbonate. c This has no effect. Flasks should be rinsed with water.

7

n(CO2) =

pV

=

765 × 101.3 × 95.0 × 10

3

−

= 0.00394 mol

RT 760 × 8.31 × 296 n(CaCO3) = n(CO2) = 0.00394 mol m(CaCO3) = n × M = 0.00394 × 100.1 = 0.394 g 100 0.394 m(CaCO3 )

% CaCO3 in marble = 8

m(marble)

×

1

=

0.411

×

100 1

= 95.9%

Small amounts of gas may have been lost in the collection process. Some carbon dioxide may remain dissolved in the reaction solution.



Oxidation numbers and redox equations No.

Answer

1

Examples:

2

a +6

3

a b c d e f

a NF3

b +6

c +4

b N2O5

Not redox Oxidant = I2O5, reductant = CO Not redox 2+ Oxidant = Hg , reductant = N2H4 – Oxidant = NO3 , reductant = H 2S Oxidant = NO2, reductant = NO 2

a +6

5

C = 0 H = +1 O = –2

6

a

c

d

e

d NH3

d –1

4

b

c N2

b

+2

c

+7

+

–

Oxidation: CH3CH2OH(aq) + H2O(l) → CH3COOH(aq) + 4H (aq) + 4e 2– + – 3+ Reduction: Cr 2O7 (aq) + 14H (aq) + 6e → 2Cr (aq) + 7H2O(l) 2– + Redox: 3CH3CH2OH(aq) + 2Cr 2O7 (aq) + 16H (aq) 3+ → 3CH3COOH(aq) + 4Cr (aq) + 11H 2O(l) Oxidation: 2I – (aq) → I2(aq) + 2e – – + – – Reduction: BrO3 (aq) + 6H (aq) + 6e → Br (aq) + 3H2O(l) Redox: 6I – (aq) + BrO3 – (aq) + 6H+(aq) → 3I2(aq) + Br – (aq) + 3H2O(l) 2+ 3+ – Oxidation: Fe (aq) → Fe (aq) + e – + – 2+ Reduction: MnO4 (aq) + 8H (aq) + 5e → Mn (aq) + 4H2O(l) Redox: 5Fe 2+(aq) + MnO4 – (aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) + – Oxidation: H2O2(aq) → O2(g) + 2H (aq) + 2e – + – 2+ Reduction: MnO4 (aq) + 8H (aq) + 5e → Mn (aq) + 4H2O(l) – + 2+ Redox: 5H2O2(aq) + 2MnO4 (aq) + 6H (aq) → 5O2(g) + 2Mn (aq) + 8H2O(l) + – Oxidation: H2S(g) → S(s) + 2H (aq) + 2e Reduction: Cr 2O72– (aq) + 14H+(aq) + 6e – → 2Cr 3+(aq) + 7H2O(l) 2– + 3+ Redox: 3H2S(g) + Cr 2O7 (aq) + 8H (aq) → 3S(s) + 2Cr (aq) + 7H2O(l)



Recovering silver from solution No.

Answer

1

a b c

2

a b

AgNO3 solution NaCl solution deionised water +

–

Ag (aq) + Cl (aq) → AgCl(s) + – Ag (aq) + CrO4 (aq) → AgCrO4(s)

3

n(NaCl)

4

There is uncertainty in the pipette and burette volumes, and in the judging of the endpoint.

5

Dry and weigh the sample until the mass was constant.

6

a

= c × V = 0.100 × 0.02000 = 0.002000 mol n(AgNO3) = n(NaCl) = 0.002000 mol 0.002000 n = 0.1035 M = c(AgNO3) = V 0.01932

b

+

2+

2Ag (aq) + Cu(s) → 2Ag(s) + Cu (aq) Silver ion

7

n(AgNO3)

= c × V = 0.1035 × 0.02000 = 0.002070 mol n(Ag) = n(AgNO3) = 0.002070 mol m(Ag) present = n × M = 0.002070 × 107.9 = 0.2234 g mass recovered 0.206 % recovery = × 100 = × 100 = 92.2% mass present 0.2234

8

• Reaction between the Ag ions in solution and the Cu spiral was incomplete. • Some silver remained adhering to the copper spiral. • Small amounts of silver were lost in the transfer and filtering process.

9

S2O42– (aq) + 2H2O(l) → 2SO32– (aq) + 4H+(aq) + 2e –

10

2H2O(l) → O2(g) + 4H+(aq) + 4e –

+



Redox titrations No.

Answer

1

a b

2

a b

3

a

b c

I2(aq) + 2e – → 2I – (aq) If iodine is present a blue colour will be present. When the iodine has all reacted (i.e. the endpoint has been reached), the blue colour will disappear. –

+

c(CH3CH2OH)

a

b c d

–

CH3CH2OH(aq) + H2O(l) → CH3COOH(aq) + 4H (aq) + 4e Cr 2O72– (aq) + 14H+(aq) + 6e → 2Cr 3+(aq) + 7H2O(l) 2– + 3+ 2Cr 2O7 (aq) + 16H (aq) + 3CH3CH2OH(aq) → 4Cr (aq) + 11H2O(l) + 3CH3COOH(aq) 2– n(Cr 2O7 ) = c × V = 0.200 × 0.01945 = 0.00389 mol 3 2– n(CH3CH2OH) = × n(Cr 2O7 ) = 1.5 × 0.00389 = 0.00584 mol 2 n(CH3CH2OH)in original = n(CH3CH2OH)in aliquot ×

4

+

C6H4O2(OH)4(aq) + I2(aq) → C6H4O4(OH)2(aq) + 2I (aq) + 2H (aq) n(I2) = c × V = 0.10 × 0.0250 = 0.0025 mol n(C6H4O2(OH)4) = n(I2) = 0.0025 mol 0.0025 n = 0.140 M = c(C6H4O2(OH)4) = V 0.0178

2+

=

n =

V

0.0730 0.025

250.0 20.00

= 0.00584 ×

250.0 20.00

= 0.0730 mol

= 2.92 M

3+

Fe (aq) → Fe (aq) + e MnO4 – (aq) + 8H+(aq) + 5e – → Mn2+(aq) + 4H2O(l) – + 2+ 2+ 3+ MnO4 (aq) + 8H (aq) + 5Fe (aq) → Mn + 4H2O(l) + 5Fe (aq) The pink colour of the MnO4 – persists when the endpoint has been reached. – n(MnO4 ) = c × V = 0.110 × 0.0128 = 0.00141 mol 2+ – n(Fe ) = 5 × n(MnO4 ) = 5 × 0.00141 = 0.00704 mol 0.00704 n 2+ = 0.352 M = c(Fe ) = V 0.020



Redox titrations No.

Answer

5

a

Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) 2+ – b Fe(s) → Fe (aq) + 2e 2H+(aq) + 2e – → H2(g) c HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) n(HCl) initially added = c × V = 1.08 × 0.100 = 0.108 mol n(NaOH) = c × V = 1.00 × 0.0459 = 0.0459 mol n(HCl) remaining after reaction with iron = n(NaOH) = 0.0459 mol n(HCl) reacting with iron = 0.108 – 0.0459 = 0.0621 mol 1 0.0621 × n(HCl) = = 0.0311 mol n(Fe) = 2 2 m(Fe) = n × M = 0.0311 × 55.8 = 1.732 g 1.732 m( Fe) % Fe = × 100 = × 100 = 60.4% 2.87 m( nail)



Analysis of iron in iron ore No.

Answer

1

An absorbance spectrum would be created by testing the absorbance of the complex ion solution at various wavelengths. The wavelength of maximum absorbance would be chosen (provided this wavelength was not also absorbed by other components in the ore solution).

2

To relate the absorbance of the solution to its concentration.

3

Absorbance of 0.400 corresponds to a concentration of 0.0125 M. 0.0125 mol in 1.0 L means 0.00125 mol in 100.0 mL. m(Fe) = n × M = 0.00125 × 55.8 = 0.0698 g 0.0698 m(Fe) % Fe = × 100 = × 100 = 69.8% 0.100 m(ore)

4

MnO4 (aq) + 8H (aq) + 5e → Mn (aq) + 4H2O(l) Fe2+(aq) → Fe3+(aq) + e – – + 2+ 2+ 3+ MnO4 (aq) + 8H (aq) + 5Fe (aq) → Mn (aq) + 4H2O(l) + 5Fe (aq)

5

n(MnO4

6

The determined value would be lower, because less permanganate solution would be required for titration (since it will only react with the Fe2+, not the Fe 3+).

7

a

8

A pure sample of known composition could not be obtained because: the precipitate may contain a mixture of iron(II) and iron(III) hydroxides • the iron hydroxide precipitates are not stable when heated. •

9

2Fe → 2Fe(OH)2 → Fe2O3

–

+

–

2+

–

) = c × V = 0.0335 × 19.75 × 10 –3 mol 2+ – n(Fe) = n(Fe ) = 5 × n(MnO4 ) –3 m(Fe) = n × M = 5 × 0.0335 × 19.75 × 10 × 55.8 = 0.1846 g 0.1846 m(Fe) % Fe = × 100 = × 100 = 68.9% m(ore) 0.268

2+

–

Fe (aq) + 2OH (aq) → Fe(OH)2(s)

n(Fe2O3)

=

m M

=

1.08 159.6

b

3+

–

Fe (aq) + 3OH (aq) → Fe(OH)3(s)

= 0.00677 mol

n(Fe)

= 2 × n(Fe2O3) m(Fe) = n × M = 2 × 0.00677 × 55.8 = 0.755 g 0.755 m(Fe) % Fe = × 100 = × 100 = 74.8% 1.01 m(ore) 10

Precipitates other than iron hydroxide may have formed, leading to a larger precipitate mass.

11

a

12

Atomic absorption spectroscopy of a solution prepared by dissolving ore in acid can be used.

Spectroscopic

b

Gravimetric

c

Gravimetric

d

Spectroscopic



Colorimetric determination of manganese in steel No

Answer

1

a

2+

–

b c 2

Straight line graph of y = 0.076 x.

3

2.56 ppm = 2.56 mg L –1

4

+

–

Mn (aq) + 4H2O(l) → MnO4 (aq) + 8H (aq) + 5e IO4 – (aq) + 2H+(aq) + 2e – → IO3 – (aq) + H2O(l) 2+ – – + – 2Mn (aq) + 3H2O(l)+ 5IO4 (aq) → 2MnO4 (aq) + 6H (aq) + 5IO3 (aq)

– n(MnO4 )

=

m M

=

2.56 × 10

3

−

mol

118.9

–

n(Mn) = n(MnO4 ) m(Mn) = n × M =

% Mn m/m =

2.56 × 10 118.9

m(Mn )

×

m(sample)

−

3 ×

54.9 = 1.18 × 10

100 =

1.18 × 10

–3

g

3

−

2.890

×

100 = 0.0409%

5

Steel is mostly composed of iron. Step 1 oxidised the iron to iron(II) then to iron(III).

6

8.0 ppm MnO 4 = 8.0 mg per 1.0 L (1000 mL) – ∴ 0.8 mg of MnO 4 in 100 mL

–

−3

0.8 × 10 ∴

118.9

mol of MnO4 – in 100 mL

−3

0.8 × 10 ∴

118.9

mol of KMnO4 in 100 mL

−3

0.8 × 10 ∴

118.9

× 158.0

g of KMnO4 in 100 mL

1.1 × 10 –3 g of KMnO 4

∴

7

Everything added to the sample (nitric acid, potassium persulfate, periodate and phosphoric acid) should be added in the same proportions, in case any of these substances absorb light in the region being analysed.

8

400 nm is in the blue–violet region, which is close to the colour of MnO 4 and therefore will not be absorbed strongly. 520 nm is closer to green, and will be more strongly absorbed.

9

If all absorbances were shifted by the same quantity, it would not have made a difference.

10

Atomic absorption spectrometry could be used.

–



Analysing mass spectra No.

Question

1

Vaporisation, ionisation and deflection

2

a b

3

Answer

The vaporised sample is bombarded with high-energy electrons to produce positive ions by knocking electrons off the sample atoms or molecules. Positive ions moving in a magnetic field are deflected to varying degrees depending on their charge to mass ratio. The greater the charge, the greater the deflection; the greater the mass, the smaller the deflection. Thus ions are separated by mass.

Peak at mass 20 is Peak at mass 36 is

4

Ar = Σ(RIM

5

a M =

20

+

, 21 is 36 + Ar , 38 is 20 10

Ne

21 10 38 20

Ne Ar

+

and 22 is and 40 is

22 10 40 20

Ne Ar

+

× abundance fraction) Let a be the percentage abundance of the lighter isotope. ∴ 121.75 × 100 = (120.90 × a) + (122.90 × (100 – a)) ∴ a = 57.5% b c d

+

CH3CH2CH2CH(OH)CH3 + M – 1 = CH 3CH2CH2C(OH)CH3 + M – 15 = CH 3CH2CH2CH(OH) + M – 43 = CH(OH)CH 3

6

Compound I. The spectrum shows prominent peaks at mass-to-charge ratios of 57 and 29. + + These would correspond to the masses of fragments CH 3CH2CO and CH3CH2 . These fragments are likely to form from compound I. Compound II would be expected to form fragments at mass-to-charge ratios of 71 (CH3CH2CH2CO+), 43 (CH3CH2CH2+) and 15 + (CH3 ).

7

Comparison of the mass spectrum of an unknown substance with computer files of mass spectra of many compounds allows the unknown substance to be identified. The fragmentation pattern can be used for ‘fingerprinting’ a substance (in much the same as the fingerprint region of the infrared spectrum is used).



Spectroscopic analysis of organic compounds I No.

Answer

1

a i b c d

2

5 ii 2 i 3:2:2:2:1 ii 9:1 2,2-dimethylpropanal (peaks in the ratio 9:1) The fingerprint regions of the infrared spectra of the unknown could be compared with that of 2,2-dimethylpropanal.

a i

ii

b i ii

3

4

5

Infrared radiation is absorbed by molecules that move from ground state to excited vibrational energy states. The energy is absorbed as bonds within the molecules bend and stretch. –1 The peak at 1710 cm suggests a C=O bond, present in both structures. The –OH group in compound II would be expected to produce a peak in the 2500–3000 cm –1 range. The absence of this peak suggests that the unknown is compound I. Radio waves Compound I would produce a spectrum with two peaks with the area ratio 1:1. Compound II would produce a spectrum with three peaks with the area ratio 3:2:1. The unknown is therefore likely to be compound I (methyl ethanoate).

Compound I would produce a spectrum with four peaks with the area ratio 3:2:2:3. Compound II would produce a spectrum with two peaks with the area ratio 6:4 or 3:2. The unknown is therefore compound II, 3-pentanone. Notice that the peak at chemical shift 1.2 ppm, caused by the CH3 nuclei, is split into three peaks, called a triplet. The peak at chemical shift 2.5, caused by the CH2 nuclei, is split into four peaks, called a quartet. The splitting follows an n + 1 rule. A peak will be split n + 1 times by n adjacent nuclei. In the case of the peak at 1.2 ppm, the CH 3 nuclei have two neighbouring protons (CH 2), so the peak is split 3 (i.e. 2 + 1) times (a triplet). For the peak at 2.5 ppm, the CH2 nuclei have three neighbouring protons (CH 3), so the peak is split 4 (i.e. 3 + 1) times (a quartet). a i b c d

3 ii 3 i 3:2:3 ii 3:2:3 i One single, one triple and one quartet ii One single, one triple and one quartet The chemical shift values could be used to distinguish between them. For example, the CH3CO group (shift 2.1) and the RCOOCH 2R group (shift 4.1).



Amino acid analysis using TLC

No.

Answer

1

R f (W) = 0.5, R f (X) = 0.35, R f (Y) = 0.2 and R f (Z) = 0.1

2

W could be leucine or isoleucine, X could be proline or tyrosine or valine, Y is alanine or possibly threonine and Z is arginine.

3

R f (W) = 0.7, R f (X) = 0.8, R f (Y) = 0.5 and R f (W) = 0.55

4

W could be leucine or isoleucine, X is proline, Y is alanine and Z is arginine.

5

Some components may not separate in a given solvent. Using a second solvent makes it more likely that separation can be achieved. The double development also allows for a longer development time, increasing the chance of resolution of component spots.

6

If the spot is below the solvent level, the amino acids will dissolve into the solvent, rather than travel up the filter paper.

7

The amino acids are colourless, and the spots developed using ninhydrin fade quickly in light. It is therefore necessary to mark their positions quickly.

8

Perspiration on the skin contains amino acids. These could interfere with the analysis if they were transferred to the plate.

9

Separation of the components occurs because the components adsorb (stick) to the stationary phase with different strengths. The stronger the adsorption, the more slowly the component moves as the mobile phase (the solvent) sweeps over the stationary phase (the plate). Components undergo continuous adsorption and desorption (unsticking) to differing degrees. The components move at different rates depending on their strength of adsorption and the tendency to desorb.



GLC (Gas liquid chromatography)

No.

Answer

1

A primary alkanol is one with the –OH functional group on carbon 1.

2

It must constantly be heated so that the mixture being analysed remains gaseous at all times.

3

It means that the sample is heated very quickly so that it enters the column as vapour.

4

This shows whether or not the machine has exactly the same parameters for each analysis. If the 1-hexanol peak occurs at the same retention time and the same height each time, the machine is properly calibrated.

5

The 1-hexanol peak was constant in terms of height and retention time.

6

Oxygen is too reactive and with the heat could have oxidised the alkanol mixture.

7

They are not easily vaporised and are more susceptible to decomposing with heat.

8

Dispersion forces

9

Ethanol is a smaller molecule. Hence, the dispersion forces formed between ethanol and the stationary phase are weaker than the bonds formed between 1-butanol and the stationary phase. Ethanol will spend less time adsorbed to the stationary phase and will pass through the column more quickly.

10

Retention time is qualitative; peak height is quantitative.

11

m(C4H9OH) n(C4H9OH)

= density × volume =

m M

=

0.806 × 3 ×10 74.1

3

−

–5

= 3 × 10 mol

12

Ethanol, 1-butanol and 1-pentanol

13

A peak height of 45 was obtained for 1-butanol in 3 mL of mixture. A peak height of 21 was obtained for a 3 µL (3 × 10 –5 mol) sample of 1-butanol. 45 –5 –5 mol ∴ n(C4H9OH) = × 3 × 10 = 6 × 10 21

14

CH3CH2CH2CH2CH2OH

15

Due to its branched structure 2-propanol would have slightly weaker dispersion forces than 1-propanol and so would have been marginally more weakly adsorbed. It would therefore have had a retention time slightly less than 2.43 min.



A summary of analytical techniques

No. 1

Question

Answer Procedure

Procedure is part of the technique of GA

Preparation of a standard solution or solutions Selection of wavelength to be used for analysis Production of a precipitate

VA

SP

CH







 

Determination of R f or Rt values



In some forms the sample to be tested must first be vaporised A calibration curve is plotted using peak areas of samples of known concentration

2

Analysis required









Analysis could be conducted using GA

To detect the presence of a C=O group in a molecule To test a water sample for the presence of chromium To check the amount of ammonia in a window cleaner To determine the ratio of 16O to 18O in an ice sample To determine the level of iron in an iron ore sample To identify a performance-enhancing drug in urine samples To separate the pigments in plant leaf extracts

VA

SP

CH

    



 


 


A summary of analytical techniques No. 3

Question

Answer Procedure

Procedure is part of the technique of AAS

Radiation is absorbed by electrons. Spectral information is given in chemical shift values. The sample for analysis is in a vaporised state. Radiation is absorbed by nuclei. Transitions occur between atomic electronic energy levels. Spectral information is given in wave numbers. Transitions occur between nuclear spin energy levels. The sample for analysis is in solution. It does not involve the absorption of radiation by the sample.

4

Mass

Infrared

UV-vis



NMR

 



 

   







Procedure

Procedure is part of the technique of TLC

HPLC

The stationary phase is a solid.





The mobile phase is a liquid.





Used for volatile organic compounds.

GLC



Results are presented as a chromatogram for which the retardation factor ( R f ) of each component is found. It relies on the principle of separation of components by selective adsorption to a stationary phase. R f values of known samples are compared with those of unknown samples in order to identify the unknown. Results are presented as a chromatogram showing a series of peaks at times taken for the components to travel a set distance.



















Naming organic compounds

No.

Answer

1

a

2-fluoropropene b propane c 3-methylbutanoic acid d 2-methyl-1-propanol e trichlorofluoromethane

2

3

a b c d

4

n

n

n

n

n

n

n –

n

a

b

5

C H2 + 2 C H2 C H2 + 1OH C 1H2 – 1COOH

Butane has the higher boiling point, because the more linear molecule means that molecules pack well together, giving greater contact between adjacent molecules and hence stronger dispersion forces. Stronger forces mean a higher boiling point. The branched molecules of 2-methylpropane do not pack together very well.

1-pentene, 2-pentene, 2-methyl-1-butene, 2-methyl-2-butene, 3-methyl-1-butene, dimethylpropene



Naming organic compounds No.

Answer

6



Organic reaction pathways

No.

Answer

1

Compound A: ethene CH 2CH2 Compound B: ethanol CH 3CH2OH Compound C: ethanoic acid CH 3COOH Compound D: propane CH 3CH2CH3 Compound E: 1-chloropropane CH 3CH2CH2Cl Compound F: 1-propanol CH 3CH2CH2OH Compound G: 1-propyl ethanoate CH 3COOCH2CH2CH3

2

Test with bromine solution. Compound A (unsaturated) will decolourise bromine, while D (saturated) will not.

3

C is acidic and will produce a solution with pH less than 7. G is neither acidic nor basic; its solution will be neutral.

4

B and F belong to the primary alkanol homologous series. F is larger than B and so will have larger dispersion forces between molecules, leading to a higher boiling point.

5

Isomers

6

Compound L is an acid (i.e. contains an acidic functional group).

7

Compound H: 1-butene CH 2CHCH2CH3 Compound I: 2-butene CH 3CHCHCH3 Compound J: 1-chlorobutane CH 3CH2CH2CH2Cl Compound K: 2-chlorobutane CH 3CH(Cl)CH2CH3 Compound L: 1-butanol CH 3CH2CH2CH2OH Compound M: butanoic acid CH 3CH2CH2COOH Compound N: polybut-2-ene …-CHCH 3CHCH3CHCH3CHCH3-…



Spectroscopic analysis of organic compounds II No. 1

Answer m(C)

in sample = m(C) in the CO 2 produced =

12.0

× 1.32 = 0.36 g 44.0 2.00 × 0.72 = 0.080 g m(H) in sample = m(H) in H2O produced = 18.0 m(O) in sample = m(sample) – m(C) – m(H) = 0.60 – 0.36 – 0.080 = 0.16 g C:H:O =

0.36 0.080 0.16 12.0

:

1.0

:

16.0

= 0.03 : 0.08 : 0.01 = 3 : 8 : 1

The empirical formula (EF) of the compound is C 3H8O. 2

From the mass spectrum, RMM is 60. EFM = 60. The molecular formula of the compound is also C3H8O.

3

A peak at M – 15 is often due to the loss of a methyl group. The molecule probably contains a methyl group or groups.

4

I CH3CH(OH)CH3 II CH3CH2CH2OH III CH3CH2OCH3

5

Compound I would have 2 peaks only (the CH 3 nuclei and the CH(OH) nucleus). The other two compounds would each produce three peaks.

6

a

b

7

8

Compound I would have three peaks in the ratio 6:1:1. Compound II would have four peaks in the ratio 3:2:2:1. Compound III would have three peaks in the ratio 3:2:3. The peak at chemical shift 1.2 ppm is due to the CH3 nuclei. The neighbouring carbon 1 has one H nucleus, so the peak is split in two ( n + 1). The peak at chemical shift 3.9 ppm is due to the CH nucleus. The neighbouring carbons have six 1H nuclei, so the peak is split in seven ( n + 1).

Compare the fingerprint regions of the infrared spectra of the unknown and 2-propanol, or compare the fragmentation patterns of the mass spectra of the unknown and 2-propanol. m(C)

in sample = m(C) in the CO 2 produced =

12.0

× 2.78 = 0.758 g 44.0 2.00 m(H) in sample = m(H) in H2O produced = × 1.14 = 0.127 g 18.0 m(O) in sample = m(sample) – m(C) – m(H) = 1.560 – 0.758 – 0.127 = 0.675 g C:H:O =

0.758 0.127 0.675 12.0

:

1.0

:

16.0

= 0.063:0.127:0.042 = 1.5:3.0:1.0 = 3:6:2

The empirical formula of the compound is C 3H6O2.



Spectroscopic analysis of organic compounds II No.

Answer

9

From the mass spectrum, RMM is 74. EFM = 74, so the molecular formula of the compound is also C3H6O2.

10

Isomer I

Isomer II

Isomer III

CH3COOCH3

HCOOCH2CH3

CH3CH2COOH

2

3

3

One single peak, one quartet and one triplet

One quartet, one triplet and one single peak

Structural formula

Number of peaks on the 1H NMR spectrum Pattern of peak 1 splitting on the H NMR spectrum 11

Two single peaks

No. Both isomers II and III fit the pattern. 1

12

The chemical shift values would be needed to identify the types of H nuclei present. For example, if the single peak has a shift of 11–12 (due to the 1H of the OH), isomer III 1 would be correct. If the single peak has a shift of 8–9 (due to the H of the –COOH), then isomer II would be correct.

13

The peak in the 1700–1750 cm region is typical of the C=O group, present in both isomers II and III. The peak in 2500–3000 cm –1 region is probably due to the OH of the carboxylic acid group, suggesting that isomer III is the correct structure.

14

Isomer III is acidic, and so would be expected to produce a solution with pH less than 7, and react with a solution of Na 2CO3 to produce carbon dioxide gas. Isomers I and II are esters and are not acidic. They may have a distinctive fruity smell.

–1



Protein structure and function

No.

Answer

1

a b

Refer to figure 8.2.1 on page 171 of the coursebook. Note that the COOH and NH 2 groups are attached to the same carbon atom. Refer to the diagrams of the twenty amino acids shown on the examination data booklet.

2

3

4

a b c d

5

1 2 4 999 (in general n – 1)

a i

b

c

–NHCH2CONHCH2CONHCH2CO– ii –NH(CH2)3CONH(CH2)3CONH(CH2)3CO– iii –CH2CH2CH2 – i Condensation ii Condensation iii Addition A protein has a complex sequence of many different amino acids. It does not use a single monomer like this.

6

The oppositely charged dipole ends along the chain align, as the chain forms a helix.



Protein structure and function No.

Answer

7

a b c

8

9

An enzyme is a biological catalyst. It is a protein with a specific 3D structure. It is this structure/shape that allows it to catalyse a particular reaction. For example: lactase is required for the digestion of milk; sucrase for the breakdown of sucrose and catalase to break down peroxide molecules. See the lock and key diagram of enzyme functioning (figure 8.2.8) on page 174 of the coursebook.

‘Inhibitors’ or blockers can act by either filling the active site on the enzyme, thus making it no longer a catalyst, or by changing the shape of the active site so that it no longer performs its usual function. This is an example of denaturation of the protein (casein) in the milk. Denaturation is a breakdown of the complex tertiary structure of the protein.

10

Strong base, ultraviolet light and other organic molecules can all cause denaturation.

11

Several examples are outlined on page 193 of the coursebook.



DNA structure and function

No.

Answer

1

a b c d

2

a b c d

Deoxyribonucleic acid Ribonucleic acid The nitrogenous bases, adenine, thymine, cytosine, guanine and uracil Polymerase chain reaction—the laboratory process in which a DNA fragment is replicated many times (amplified) A sequence of three nitrogenous bases on the DNA strand. This sequence usually codes for a particular amino acid. The specific bonding that occurs between certain base pairs on opposing strands of DNA in the double-stranded molecule. These pairs are A and T, and C and G. The copying of a double-stranded DNA molecule to produce another identical, doublestranded DNA molecule. The combination of a deoxyribose sugar (in DNA) or ribose sugar (in RNA) molecule with a nitrogenous base and a phosphate molecule. A nucleotide is the basic ‘building’ block of nucleic acid polymers.

3

Refer to figure 8.3.4 on page 177 of the coursebook. Base + sugar → nucleoside

4

Refer to figure 8.3.4 on page 177 of the coursebook. Phosphate + nucleoside → nucleotide

5

Refer to figure 8.3.4 on page 177 of the coursebook. … + nucleotide + nucleotide → polynucleotide

6

The section contains 400 bases. 40 adenine molecules must base pair with 40 thymine molecules, leaving 320 molecules. Since each cytosine must pair with a guanine molecule, there must be 160 of each molecule present.

7

a b

8

a

9

a

Section A: (from top to bottom) GGGCTATTAG; Section B: TTACGAAGGT The C–G link involves 3 hydrogen bonds and so is stronger than the A–T link which has 2 hydrogen bonds. Section A has more C–G links than section B, so section B would be easier to separate (i.e. require less energy).

Translation b Transcription c Replication b c

18 (9 per strand) 4 (note that 5 sugar-phosphate bonds already exist within the nucleotides) 66 (3 per amino acid. In practice, more are needed for the ‘stop’ and ‘start’ of the translation process.)



Biochemical fuels No. 1

Some search results and comments Biofuel

Ethanol

Biogas

Generation/source

It is produced by fermentation, the action of yeast enzymes on glucose under anaerobic conditions.

It is formed when organic matter decays in an oxygen-free (anaerobic) environment. The biogas is generated by the action of microorganisms on organic molecules. Biodiesel It is formed when triglycerides are reacted with concentrated sodium hydroxide and methanol. The sodium hydroxide breaks the ester bond in the triglyceride, allowing the methanol to form an ester with the long fatty acid molecule.

2

3

4

Chemical structure

CH3CH2OH

CH4

Refer to figure 8.4.5 on page 187 of the coursebook.

Energy production

Burnt in air, or mixed with petrol for combustion: CH3CH2OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) Reacted with oxygen in a fuel cell CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

biodiesel + O2(g) → CO2(g) + H2O(g)

Brazil has used ethanol blends for many years. Sugarcane is often their source of ethanol. The energy from the ethanol needs to outweigh the cost of producing the crops. Corn is a common starting point in the US. There are several steps to purifying the ethanol. Purdue yeast offers the possibility of using cellulose (cheap forest waste) to get glucose to get ethanol. The Sunshine tip and Windermere piggery are examples of biogas production. The main product is methane, which can be burnt to produce energy. There are several Australian ‘recipes’ for biodiesel. Biodiesel can be used in normal diesel engines. Unrefined vegetable oil can only be used in blends with normal diesel. Rapeseed (also known as canola) is a common oil chosen. Waste oil from fish and chip shops can be used also. European countries actually sell blends of biodiesel and diesel. Page 1 © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008. This page from the Chemistry Dimensions 2, Teacher’s Resource may be reproduced for classroom use.


Preparation and analysis of aspirin

No.

Answer

1

Analgesic: pain reliever Antipyretic: reduces fever Anti-inflammatory: reduces production of prostaglandins that lead to inflammation

2

Used by the ancient Greeks and by the American Indians, who chewed the leaves.

3

a

4

See figure 8.5.2 on page 189 of the coursebook.

Carboxylic acid b Hydroxy c Ester

5

6

a b

7

a b

8

It is a catalyst. It is used in many ester-forming reactions. The binder holds the tablet together so that it is a single, solid tablet. The binder must be harmless, stable and edible—starch is suitable for this.

a m(acetylsalicylic

b

6

3

acid) = 325 × 10 mg = 325 × 10 g 325 × 10 3 m n(acetylsalicylic acid) = = 180 M n(salicylic acid) = n(acetylsalicylic acid) 325 × 10 3 3 × 138 = 249 × 10 g = 249 kg = n × M = 180 The process is far from 100% efficient. Reaction is incomplete and isolation of the product is difficult because the solubility of the aspirin makes it hard to crystallise out from the reaction solution.

9

Stomach bleeding may occur. In addition, the blood is thinned too much, making it unlikely to congeal and so interfering with the clotting process.

10

Methyl salicylate (oil of wintergreen) is produced. See figure 8.5.2 on page 189 of the coursebook.


Chapter 1-8 Worksheet Solutions

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