Revision paper 2 questions quantitative chemistry
1. A 0.496g sample of an unknown hydrocarbon was completely burned in oxygen. The sample produced 1.56g
of carbon dioxide and 0.638g of water.
(a) How many moles of carbon dioxide were formed?
Moles of CO2 = 1.56/44.01 = 0.0354 moles = 0.0354 moles of C
(b) How many moles of water were formed? 0.638/18.01 = 0.0354
(c) What is the empirical formula? CH2
C
0.0354
0.0354/0,0354 = 1
H
0.0354 x 2 = 0.0708 (2 moles of H in 1 mole of H20)
0.0708/0.0354 = 2
2. A 25.00 cm3 sample of vinegar was titrated with a 0.614 mol dm-3 NaOH solution. An appropriate indicator
was added to the sample and the sample was titrated to the end point. The initial volume of NaOH was
0.50 cm3 and the final volume of NaOH was 31.35 cm3.
(a) How many cm3 of NaOH were required for the titration? 31.35 – 0.50 = 30.85 cm3
(b) What is the concentration of ethanoic acid in the vinegar sample?
Moles of NaOH = 0.03085 x 0.614 = 0.0189 moles
Ratio NaOH : CH3COOH = 1:1
Moles of CH3COOH = 0.0189 moles
Concentration CH3COOH = 0.0189 moles/0.025 dm3 = 0.76 mol dm-3
(c) How many grams of ethanoic acid are there in the 25.00 cm3 sample? 0.0189 moles x 60.06 = 1.14g
(d) Calculate the percentage of vinegar (acid) in the sample. 1.14g/25.00g x100 = 4.56%.
(assume the density of vinegar is 1.0 g cm-3)
3. An organic compound was found by CHN analysis to contain 40.45% C; 7.86% H and 15.73% N.
A separate experiment determined the relative molecular mass of the compound to be 89.0 g mol-1.
(a) Determine the empirical formula of the compound.
C
40.45/12.01
3.37/1.12
3
H
7.86/1.01
7.78/1.12
7
N
15.73/14.01
1.12/1.12
1
O
35.96/16.01
2.25/1.12
2
C3H7NO2
(b) What is the molecular formula of the compound? Molar mass/empirical mass = 89/89 = 1
Molecular formula = C3H7NO2 x 1 = C3H7NO2
4. A certain solid is known to be a mixture of Al2(SO4)3 and (NH4)2SO4. A sample of this mixture weighing
2.8506 g is dissolved in water. A solution of ammonia is added, until in excess and the solution is heated to
boiling. White Al(OH)3 precipitates out and is filtered and dried in the air. It is then heated in a crucible for 20
minutes and is completely converted into Al2O3. The final product weighs 0.4163 g.
(a) How many moles of Al2O3 were obtained in this section? 0.4163/101.87 = 4.08 x 10-3
(b) How many moles of Al were in the original sample? 4.08 x 10-3 x 2= 8.16 x 10-3
(c) Calculate the percentage by mass of Al in the original sample?
8.16 x 10-3 x 26.92 = 0.219g;
% = 0.22/2.8506 x 100 = 7.70%
(d) Calculate the percentage by mass of Al2(SO4)3 in the original sample.
Molar ratio Al3+ : SO42- = 2 : 3 = 8.16 x 10-3 : 1.22 x 10-2 moles
Topic 1 paper 2 revision 1
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Mass of Al2(SO4)3 = 0.219g (Al) + (1.22 x 10-2 moles x 96.06) = 0.219 + 1.182 = 1.401g
% = 1.401/2.8506 x 100 = 49.18%
(e) Write a balanced equation for the reaction that occurs when ammonia is added to a solution of Al2(SO4)3.
Al2(SO4)3 + 6NH4OH  3(NH4)2SO4 + 2Al (OH)3
5.
(a) Anhydrous sodium carbonate was used in an experiment to determine the concentration of an unknown
hydrochloric acid solution. A sample of the carbonate was heated, cooled and weighed until a constant
weight was obtained. Explain why this procedure was followed and comment on its importance.
Heating to remove any water in the anhydrous sodium carbonate to ensure all water of crystallization
was removed. Water would dilute acid.
(b) After being treated as described in (a) above, 1.377g of anhydrous sodium carbonate was dissolved in
water and made up to 250 cm3 in a volumetric flask. A 25.00 cm3portion of this solution was titrated with
the hydrochloric acid, using methyl orange as an indicator. The titration required 22.65 cm3 of the acid
for complete neutralization.
(i)
Write a balanced equation describing the neutralization reaction.
Na2CO3 + 2HCl  NaCl + CO2 + H2O
(ii)
Determine the molar concentration of the sodium carbonate solution.
1.377/105.99 = 0.013 moles /0.250 dm-3 = 0.052 mol dm-3
Determine the molar concentration of the unknown hydrochloric acid solution.
Moles of Na2CO3 in 25.00 cm3 = 0.052 x 0.025 = 0.0013;
Molar ratio Na2CO3 : HCl = 1:2 = 0.0013: 0.0026;
Concentration = 2.6 x10-3 /0.02265 = 0.115 mol dm-3
(iii)
(c) A second solution was prepared by dissolving 0.50 g of crystalline sodium carbonate in water. This
solution was then titrated with another portion of the hydrochloric acid solution. It was found that 30.45
cm3 of acid was required for complete neutralization.
(i)
(ii)
Calculate the apparent molar mass of the sodium carbonate used in the procedure.
Moles of HCl used: 0.115 x 0.03045 = 3.495 x 10-3
Moles of Na2CO3 = 3.495 x 10-3 /2 = 1.747 x 10-3
Molar mass = mass/ moles = 0.50/ 1.747 x 10-3 = 286.20 g
Determine the molecular formula of the crystals used in this procedure.
Mass of water = apparent molar mass of crystalline or hydrated sodium carbonate – molar
mass of anhydrous sodium carbonate = 286.20 g - 105.99 = 180.21g
Moles of Na2CO3 = 1
Moles of H2O= 180.21/18 = 10
Ratio = 1: 10
Na2CO3.10H2O
6. The mineral dolomite consists of mainly calcium carbonate and magnesium carbonate (CaCO 3 and
MgCO3) together with several other substances.
(a) 2.356 g of dolomite was reacted with 50.0 cm 3 of 1.0 mol dm-3 HNO3, which was an excess of acid.
When the excess acid was titrated with 0.050 mol dm-3 NaOH, 60.0 cm3 was required for neutralization.
i.
Write a balanced equation, including state symbols, for the reaction of either CaCO3 or
MgCO3 with nitric acid.
CaCO3 + 2HNO3 Ca(NO3)2 + CO2 + H2O
Topic 1 paper 2 revision 1
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ii.
iii.
iv.
Calculate the number of moles of nitric acid neutralized by the NaOH.
Moles of NaOH = 0.050 x 0.06 = 0.0030 = moles of HNO3
Determine the number of moles of nitric acid that reacted with dolomite.
Moles HNO3 at start = 0.05 x 1.0 = 0.05
Moles left = 0.003
Moles reacted = 0.05 – 0.003 = 0.047
Find the number of moles of carbonate in the dolomite sample.
Molar ratio HNO3: CO32- = 2 : 1 = 0.047 : 0.0235 moles of carbonate
(b) The calcium in the ore was determined by precipitating calcium oxalate, CaC2O4, from the
solution above and heating the CaC2O4 to form calcium oxide. 0.672 g of CaO were obtained
from the original 2.356g of dolomite.
i. Calculate the number of moles of calcium oxide produced. 0.672/56.08 = 0.012 moles
ii. Determine the number of moles of calcium carbonate in the dolomite sample.
Molar ratio CaCO3 : CaO = 1: 1 = 0.012 moles of CaCO3
iii. Find the number of moles magnesium carbonate in the dolomite sample.
Moles of MgCO3 = moles of carbonate – moles of calcium carbonate
= 0.0235 – 0.012 = 0.0115
(c) Calculate the percentage by mass of
i. Calcium carbonate in the dolomite sample. 0.012 x 100.08 = 1.20/2.356 x100 = 50.9%
ii. Magnesium carbonate in the dolomite sample. 0.0115 x 84.31 = 0.97/2.356 x 100 =
41. 2 %
7. Two different samples of lead oxides were analysed.
(a) The first sample contained 90.8% lead by mass. Calculate the simple empirical formula of this oxide.
Pb
90.8/207.19
0.438/0.438
1
3
O
9.2/16.00
0.575/0.438
1.31
4
Pb3O4
(b) The second oxide sample was completely reduced (oxygen removed) to lead using hydrogen as the
reducing agent. From this sample 0.928 g of lead and 0.0807g of water were produced.
(i)
(ii)
(iii)
Calculate the number of moles of lead in the second sample. Moles of Pb = 0.928/207.19 =
0.0045
Calculate number of moles of oxygen in the second sample. Moles of H2O = 0.0807/18.01 =
0.0045 = moles of O; mass of O = 0.0045 x 16 = 0.0717g
Write a balanced equation that represents the reduction of the oxide sample with hydrogen.
PbO + H2  H2O + Pb
8. Aspirin, C9H8O4, is made by reacting ethanoic acid anhydride, C4H6O3, with 2-hydrobenzoic acid, C7H6O3,
according to the equation
2 C7H6O3 + C4H6O3 
2C9H8O4
+
H2O
(a) If 15.0 g 2 hydroxybenzoic acid is reacted with 15.0 g of ethanoic anhydride, determine the limiting
reagent.
[2]
Topic 1 paper 2 revision 1
3
Moles of C7H6O3
15.0/138.13
0.108/0.108
1
limiting reagent
Moles of C4H6O3
15.0/102.09
0.147/0.108
1.36
(b) Calculate the maximum mass of aspirin that could be obtained in this reaction.
[2]
Molar ratio C7H6O3 : C9H8O4 = 1 : 1= 0.108 moles of aspirin
Mass of aspirin = 0.108 x 180.1 = 19.4g
(c) If the mass obtained in this experiment was 13.7 g, calculate the percentage yield of aspirin. [1]
13.7/19.4 x 100 = 70.6%
9. The percentages of carbon, hydrogen and oxygen of an unknown monoprotoic acid are
C 70.6 %
H 5.89%
O 23.5%
(a) Calculate the empirical formula.
C
70.6/12
5.88/1.47
4
H
5.89/1
5.89/1.47
4
[3]
O
23.5/16
1.47/1.47
1
C4H4O
(b) A 0.542 g sample of this acid was dissolved in distilled water and titrated with 0.136 M NaOH.
The endpoint of this titration was reached after 29.30 cm3 of NaOH was added. Calculate the
molar mass of the unknown acid.
[2]
Moles of NaOH = 0.136 mol dm-3 x 0.0293 dm3 = 3.98 x 10-3 moles
Molar ratio NaOH : CH3COOH = 1 : 1
Molar mass CH3COOH = 0.542/3.98 x 10-3 = 136g
(c) Find the molecular formula mass of the unknown acid.
Molar mass /empirical mass = 136/68 = 2
Molecular formula = empirical formula x 2 = C4H4O x 2 = C8H8O2
10.
[1]
(d) Write a balanced equation for the complete combustion of a sample of this acid.
C8H8O2 + 9O2  8CO2 + 4H2O
[1]
(e) Calculate the mass of CO2 that would be formed by the combustion of 0.150 g of the acid.
Moles of acid = 0.150/136.1 = 1.10 x 10-3
Moles of CO2 = 1.10 x 10-3 x 8 = 8.80 x 10-3
Mass of CO2 = 8.80 x 10-3 x 44 = 0.388g
[2]
27.82 g of hydrated sodium carbonate crystals, Na2CO3. xH2O, was dissolved in water and made up
3
3
3
to 1.000 dm . 25.00 cm of this solution was neutralized by 48.80 cm of hydrochloric acid of
concentration 0.1000 mol dm
(a)
−3
.
Write an equation for the reaction between sodium carbonate and hydrochloric acid.
Topic 1 paper 2 revision 1
[2]
4
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
(b)
2
Calculate the molar concentration of the sodium carbonate solution neutralized by the hydrochloric
acid.
[3]
n(Na2CO3) =
1
2
n(HCl);
n(HCl) = 48.80 × 0.1000 = 0.00 488 mol;
1000
 0.0024  1000
25 = 0.0976 mol dm−3;
concentration of Na2CO3
Award [3] for correct answer.
Award [3] for correct answer based on equation in (a),
i.e. allow ECF from (a). Note –1(SF) is possible.

(c)

3
Determine the mass of sodium carbonate neutralized by the hydrochloric acid and hence the mass
of sodium carbonate present in the1.000dm3 of solution.
[3]
Mr Na2CO3 = 2(22.99) + 12.01 + 3(16.00) = 105.99;
Accept 106.
mass of Na2CO3 reacting with HCl(aq) = 0.00244 × 105.99 = 0.259 g;
Allow ECF from (b) and M.
mass of Na2CO3 in 1.000 dm3 = 0.259 × 1000 = 10.36 g;
25
(d)
3
Calculate the mass of water in the hydrated crystals and hence find the value of x.
mass of water in crystals = (27.82 −10.36) = 17.46 g;
Allow ECF from (b) and (c).
[4]
number of moles of water = 17.46 = 0.9689;
18.02
Accept 0.97
mole ratio Na2CO3 : H2O = 0.0976 : 0.9689;
x = 10;
Topic 1 paper 2 revision 1
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