Internal sample questions and answers

Achieved
1 Calculate the amount (in moles) of carbon dioxide, CO2, in 23.0 g of the gas?
M(CO2) = 44.0 g mol-1
n(CO2) = = 0.523 mol

Achieved
2 Calculate the mass of 2.65 mol of sodium hydroxide, NaOH.
M(NaOH) = 40.0 g mol-1

m = 2.65 mol x 40.0 g mol–1 = 106 g

Achieved
3 24.5 mL of a solution contains 0.00257 mol of HCl. Calculate the concentration, in mol L–1, of the HCl solution.
c = = 0.105 mol L–1

Achieved
4 Calculate the amount (in mole) of NaOH present in 15.6 mL of 0.152 mol L–1 solution.
n = 0.152 mol L–1 x 15.6 x 10–3 L = 2.37 x 10–3 mol

Merit
5(a) Caffeine is a stimulant which has a mass composition of 49.5% carbon, 5.20% hydrogen, 28.9% nitrogen and 16.5% oxygen.
Calculate the empirical formula of caffeine.

(a) n(C) = = 4.13 mol n(H) = = 5.20 mol
n(N) = = 2.06 mol n(O) = = 1.03 mol
empirical formula C4H5N2O

Amount in moles of each element correct.
Correct process, but incorrect empirical formula used. (achieved)

(b) If the molar mass of caffeine is 194 g mol-1, use your answer to part (a) above to determine the molecular formula of caffeine.

molecular formula C8H10N4O2
Empirical and molecular formulae correct. (merit)

Merit
6. Calculate the percentage of carbon and oxygen in sucrose, C12H22O11.
M(C12H22O11) = 342 g mol–1

% C= x 100 = 42.1 %
% O = x 100 = 51.5 %
Process correct but incorrect molar mass of sucrose used. (Achieved)
% of C and O correct(Merit)

Merit
7. 500 mL of 0.253 mol L-1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L-1 NaHCO3 solution.
What is the concentration of the final solution?

n(NaHCO3)
= (0.253 mol L-1 x 0.500 L) + (0.824 mol L-1 x 0.800 L)
= 0.786 mol
Total volume = 1.30 L
concentration of final solution = = 0.604 mol L-1
Correct total amount of NaHCO3
OR
correct process for calculating concentration using a volume of 1.3 L. (Achieved)
Correct determination of concentration of final solution. (Merit)

Excellence
8. A chemist found that 4.69 g of sulfur combined with fluorine t
o produce 15.81 g of gas. Determine the empirical formula of the compound.

n(S) = = 0.146 mol
m(F) = 15.81 - 4.69 = 11.1 g
n(F) = = 0.584 mol
Hence S : F = 1 : 4 and empirical formula is SF4.

Correct amount of S OR F. (achieved)
Correct process with one minor error. eg Incorrect mass of F (Merit)
Correct determination empirical formula. (excellence)

Excellence
9. What mass of CO2 is produced in the complete combustion of 36.5 g of ethanol according to the following equation?
C2H5OH + 3 O2 2 CO2 + 3 H2O

M(C2H5OH) = 46.0 g mol-1
(i) n(C2H5OH) = 36.5 g / 46.0 g mol-1 = 0.793 mol
(ii) n(CO2) = 2 x n(C2H5OH) = 1.59 mol
(iii) m(CO2) = 1.59 mol x 44.0 g mol-1 = 69.8 g

Correct process used in either step (i) or (iii). (achieved)
Correct process with one minor error. eg An incorrect molar mass. (Merit)
Correct answer for mass of CO2 produced. (excellence)

Excellence
10 .What mass of iron can be produced if 50.0 g of carbon monoxide react with iron(III) oxide according to the following equation?
Fe2O3 + 3 CO 2 Fe + 3 CO2

(i) n(CO) = = 1.79 mol
(ii) n(Fe) = x n(CO) = 1.19 mol
(iii) m(Fe) = 1.19 mol x 55.9 g mol-1 = 66.5 g

Correct process used in either step (i) or (iii).(achieved)
Correct process with one minor error.eg An incorrect molar mass.(Merit)
Correct answer for mass of Fe produced. (Excellence)

Excellence
11. Hydrated magnesium sulfate is heated in a crucible. The following data is collected:
mass of crucible and lid = 26.49 g
mass of hydrated magnesium sulfate = 2.13 g
mass crucible, lid and magnesium sulfate after first heating = 27.55 g
mass of crucible, lid and magnesium sulfate after second heating = 27.53 g
Use these results to determine the formula of hydrated magnesium sulfate.
M(MgSO4) = 120.3 g mol-1 M(H2O) = 18.0 g mol-1

mass of anhydrous MgSO4 = 1.04 g
mass of H2O = 1.09 g
amount of MgSO4 = = 8.65 x 10-3 mol
amount H2O = = 6.06 x 10-2 mol
ratio n(H2O) / n(MgSO4) = 7
formula is MgSO4.7H2O

Correct determination of masses of MgSO4 and H2O.(Achieved)
Correct determination of either amount of MgSO4 or H2O. (Merit)
Correct formula for hydrated salt.(Excellence)