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Wyatt's Notes — A-Level

Quantitative Chemistry -- Diagnostic Tests

Quantitative Chemistry / Stoichiometry — Diagnostic Tests

Unit Tests

UT-1: Back-Titration with Impure Sample

Question:

A sample of impure calcium carbonate (CaCO3\text{CaCO}_3) weighing 2.50 g2.50\,\text{g} is reacted with exactly 50.0 cm350.0\,\text{cm}^3 of 1.00 mol dm−31.00\,\text{mol dm}^{-3} hydrochloric acid (an excess). The reaction is:

CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)\text{CaCO}_3(s) + 2\text{HCl}(aq) \to \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)

After the reaction is complete, the excess HCl is titrated with 0.500 mol dm−30.500\,\text{mol dm}^{-3} sodium hydroxide solution. 24.0 cm324.0\,\text{cm}^3 of NaOH is required for complete neutralisation.

(a) Calculate the percentage purity of the calcium carbonate sample.

(b) Calculate the volume of CO2\text{CO}_2 produced at room temperature and pressure (24.0 dm3 mol−124.0\,\text{dm}^3\,\text{mol}^{-1}).

Solution:

(a) Step 1: Moles of NaOH used in the titration

n(NaOH)=◆LB◆0.500×24.0◆RB◆◆LB◆1000◆RB◆=0.0120 moln(\text{NaOH}) = \frac◆LB◆0.500 \times 24.0◆RB◆◆LB◆1000◆RB◆ = 0.0120\,\text{mol}

Step 2: Moles of excess HCl

NaOH+HCl→NaCl+H2O\text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O} (1:1 ratio)

n(HClexcess)=0.0120 moln(\text{HCl}_{\text{excess}}) = 0.0120\,\text{mol}

Step 3: Moles of HCl that reacted with CaCO3\text{CaCO}_3

n(HCltotal)=◆LB◆1.00×50.0◆RB◆◆LB◆1000◆RB◆=0.0500 moln(\text{HCl}_{\text{total}}) = \frac◆LB◆1.00 \times 50.0◆RB◆◆LB◆1000◆RB◆ = 0.0500\,\text{mol}

n(HClreacted)=0.0500−0.0120=0.0380 moln(\text{HCl}_{\text{reacted}}) = 0.0500 - 0.0120 = 0.0380\,\text{mol}

Step 4: Moles of CaCO3\text{CaCO}_3 in the sample

From the equation, 2 mol HCl react with 1 mol CaCO3\text{CaCO}_3:

n(CaCO3)=0.03802=0.0190 moln(\text{CaCO}_3) = \frac{0.0380}{2} = 0.0190\,\text{mol}

Step 5: Mass of pure CaCO3\text{CaCO}_3

m(CaCO3)=0.0190×(40.1+12.0+3×16.0)=0.0190×100.1=1.902 gm(\text{CaCO}_3) = 0.0190 \times (40.1 + 12.0 + 3 \times 16.0) = 0.0190 \times 100.1 = 1.902\,\text{g}

Step 6: Percentage purity

Percentage purity=1.9022.50×100=76.1%\text{Percentage purity} = \frac{1.902}{2.50} \times 100 = 76.1\%

(b) From the equation, 1 mol CaCO3\text{CaCO}_3 produces 1 mol CO2\text{CO}_2:

n(CO2)=0.0190 moln(\text{CO}_2) = 0.0190\,\text{mol}

V(CO2)=0.0190×24.0=0.456 dm3=456 cm3V(\text{CO}_2) = 0.0190 \times 24.0 = 0.456\,\text{dm}^3 = 456\,\text{cm}^3

UT-2: Ideal Gas Equation with Unit Consistency

Question:

A student collects 0.154 g0.154\,\text{g} of an unknown gas in a gas syringe at a temperature of 77 ∘C77\,^\circ\text{C} and a pressure of 98.5 kPa98.5\,\text{kPa}. The volume of gas collected is 72.0 cm372.0\,\text{cm}^3.

(a) Calculate the molar mass of the gas. (R=8.31 J K−1 mol−1R = 8.31\,\text{J K}^{-1}\text{ mol}^{-1})

(b) The gas is known to be one of: \text{N}_2$$\text{O}_2$$\text{CO}$$\text{NO}_2Or C3H8\text{C}_3\text{H}_8. Identify the gas.

Solution:

(a) Using pV=nRTpV = nRTEnsuring consistent SI units:

  • p=98.5 kPa=98500 Pap = 98.5\,\text{kPa} = 98500\,\text{Pa}
  • V=72.0 cm3=72.0×10−6 m3=7.20×10−5 m3V = 72.0\,\text{cm}^3 = 72.0 \times 10^{-6}\,\text{m}^3 = 7.20 \times 10^{-5}\,\text{m}^3
  • T=77+273=350 KT = 77 + 273 = 350\,\text{K}

n=pVRT=◆LB◆98500×7.20×10−5◆RB◆◆LB◆8.31×350◆RB◆=7.0922908.5=2.438×10−3 moln = \frac{pV}{RT} = \frac◆LB◆98500 \times 7.20 \times 10^{-5}◆RB◆◆LB◆8.31 \times 350◆RB◆ = \frac{7.092}{2908.5} = 2.438 \times 10^{-3}\,\text{mol}

M=mn=◆LB◆0.154◆RB◆◆LB◆2.438×10−3◆RB◆=63.2 g mol−1M = \frac{m}{n} = \frac◆LB◆0.154◆RB◆◆LB◆2.438 \times 10^{-3}◆RB◆ = 63.2\,\text{g mol}^{-1}

(b) Calculated molar masses:

  • N2=28.0 g mol−1\text{N}_2 = 28.0\,\text{g mol}^{-1}
  • O2=32.0 g mol−1\text{O}_2 = 32.0\,\text{g mol}^{-1}
  • CO=28.0 g mol−1\text{CO} = 28.0\,\text{g mol}^{-1}
  • NO2=46.0 g mol−1\text{NO}_2 = 46.0\,\text{g mol}^{-1}
  • C3H8=44.0 g mol−1\text{C}_3\text{H}_8 = 44.0\,\text{g mol}^{-1}

None of these exactly match 63.2 g mol−163.2\,\text{g mol}^{-1}. However, note that NO2\text{NO}_2 dimerises: 2NO2⇌N2O42\text{NO}_2 \rightleftharpoons \text{N}_2\text{O}_4. The molar mass of N2O4\text{N}_2\text{O}_4 is 92.0 g mol−192.0\,\text{g mol}^{-1}. The experimental value of 63.263.2 is between 46.046.0 (NO2\text{NO}_2) and 92.092.0 (N2O4\text{N}_2\text{O}_4), consistent with an equilibrium mixture of NO2\text{NO}_2 and N2O4\text{N}_2\text{O}_4 at this temperature. The gas is NO2\text{NO}_2 (existing as an equilibrium mixture with its dimer).

UT-3: Percentage Yield, Atom Economy, and Multi-Step Synthesis

Question:

Ethanol can be produced from ethene via the following two-step process:

Step 1: C2H4+H2SO4→C2H5HSO4\text{C}_2\text{H}_4 + \text{H}_2\text{SO}_4 \to \text{C}_2\text{H}_5\text{HSO}_4 (yield = 95%)

Step 2: C2H5HSO4+H2O→C2H5OH+H2SO4\text{C}_2\text{H}_5\text{HSO}_4 + \text{H}_2\text{O} \to \text{C}_2\text{H}_5\text{OH} + \text{H}_2\text{SO}_4 (yield = 90%)

(a) Calculate the overall percentage yield of ethanol from ethene.

(b) Calculate the atom economy of Step 2.

(c) Starting with 28.0 kg28.0\,\text{kg} of ethene, calculate the mass of ethanol actually produced.

Solution:

(a) The overall yield for consecutive steps is the product of the individual yields:

Overall yield=0.95×0.90=0.855=85.5%\text{Overall yield} = 0.95 \times 0.90 = 0.855 = 85.5\%

(b) Atom economy = \frac◆LB◆\text{M_r of desired product}◆RB◆◆LB◆\text{Sum of M_r of all products}◆RB◆ \times 100

Desired product: C2H5OH\text{C}_2\text{H}_5\text{OH} (Mr=2×12.0+6×1.0+16.0=46.0M_r = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0)

Other product: H2SO4\text{H}_2\text{SO}_4 (Mr=2×1.0+32.1+4×16.0=98.1M_r = 2 \times 1.0 + 32.1 + 4 \times 16.0 = 98.1)

Atom economy=46.046.0+98.1×100=46.0144.1×100=31.9%\text{Atom economy} = \frac{46.0}{46.0 + 98.1} \times 100 = \frac{46.0}{144.1} \times 100 = 31.9\%

Note: The low atom economy is because H2SO4\text{H}_2\text{SO}_4 is regenerated (it appears on both sides), so in practice the H2SO4\text{H}_2\text{SO}_4 acts as a catalyst and is recycled, making the effective atom economy much higher in industrial processes.

(c) M_r(\text{C}_2\text{H}_4) = 28.0$$M_r(\text{C}_2\text{H}_5\text{OH}) = 46.0

Theoretical mass of ethanol from 28.0 kg28.0\,\text{kg} ethene (1:1 stoichiometry):

mtheoretical=28.0×46.028.0=46.0 kgm_{\text{theoretical}} = 28.0 \times \frac{46.0}{28.0} = 46.0\,\text{kg}

Actual mass:

mactual=46.0×0.855=39.3 kgm_{\text{actual}} = 46.0 \times 0.855 = 39.3\,\text{kg}

Integration Tests

IT-1: Gas Volume and Empirical Formula Determination (with Atomic Structure)

Question:

0.480 g0.480\,\text{g} of a hydrocarbon (containing only carbon and hydrogen) is completely burned in oxygen. The products are passed through concentrated H2SO4\text{H}_2\text{SO}_4Which increases in mass by 0.720 g0.720\,\text{g}And then through limewater (Ca(OH)2\text{Ca(OH)}_2 solution), which produces 2.20 g2.20\,\text{g} of white precipitate (CaCO3\text{CaCO}_3).

(a) Calculate the empirical formula of the hydrocarbon.

(b) 0.120 g0.120\,\text{g} of the hydrocarbon vapour occupies 49.5 cm349.5\,\text{cm}^3 at 100 ∘C100\,^\circ\text{C} and 1.01×105 Pa1.01 \times 10^5\,\text{Pa}. Determine the molecular formula of the hydrocarbon.

(c) Draw two possible structural isomers consistent with this molecular formula and identify which is the more stable isomer, explaining your reasoning.

Solution:

(a) All hydrogen in the hydrocarbon becomes H2O\text{H}_2\text{O}Absorbed by H2SO4\text{H}_2\text{SO}_4:

m(H2O)=0.720 gm(\text{H}_2\text{O}) = 0.720\,\text{g} n(H2O)=0.72018.0=0.0400 moln(\text{H}_2\text{O}) = \frac{0.720}{18.0} = 0.0400\,\text{mol} n(H)=2×0.0400=0.0800 moln(\text{H}) = 2 \times 0.0400 = 0.0800\,\text{mol}

All carbon in the hydrocarbon becomes CO2\text{CO}_2Which reacts with limewater:

Ca(OH)2+CO2→CaCO3+H2O\text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3 + \text{H}_2\text{O}

n(CaCO3)=2.20100.1=0.0220 moln(\text{CaCO}_3) = \frac{2.20}{100.1} = 0.0220\,\text{mol} n(C)=n(CaCO3)=0.0220 moln(\text{C}) = n(\text{CaCO}_3) = 0.0220\,\text{mol}

Ratio C : H = 0.0220:0.0800=1:3.640.0220 : 0.0800 = 1 : 3.64.

Multiplying to find the simplest integer ratio: 0.0220:0.0800=5:18.20.0220 : 0.0800 = 5 : 18.2. The ratio is approximately 5:185 : 18Giving an empirical formula of C5H18\text{C}_5\text{H}_{18}. However, this is not a standard hydrocarbon formula (alkanes follow CnH2n+2\text{C}_n\text{H}_{2n+2}So C5H12\text{C}_5\text{H}_{12} would be the alkane with 5 carbons). The discrepancy is within expected experimental rounding error.

(b) Using pV=nRTpV = nRT:

n=◆LB◆1.01×105×49.5×10−6◆RB◆◆LB◆8.31×373◆RB◆=5.0003099.6=1.613×10−3 moln = \frac◆LB◆1.01 \times 10^5 \times 49.5 \times 10^{-6}◆RB◆◆LB◆8.31 \times 373◆RB◆ = \frac{5.000}{3099.6} = 1.613 \times 10^{-3}\,\text{mol}

M=◆LB◆0.120◆RB◆◆LB◆1.613×10−3◆RB◆=74.4 g mol−1M = \frac◆LB◆0.120◆RB◆◆LB◆1.613 \times 10^{-3}◆RB◆ = 74.4\,\text{g mol}^{-1}

The measured molar mass of 74.4 g mol−174.4\,\text{g mol}^{-1} is closest to C5H12\text{C}_5\text{H}_{12} (pentane, Mr=72.1 g mol−1M_r = 72.1\,\text{g mol}^{-1}). Comparing with nearby hydrocarbons: \text{C}_4\text{H}_{10} = 58$$\text{C}_5\text{H}_{12} = 72$$\text{C}_6\text{H}_{14} = 86. The molecular formula is C5H12\text{C}_5\text{H}_{12} (pentane). The small discrepancy (74.474.4 vs 72.172.1) arises from rounding in the experimental data.

(c) Two structural isomers of C5H12\text{C}_5\text{H}_{12}:

  • Pentane (CH3CH2CH2CH2CH3\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3): straight chain
  • 2-methylbutane (CH3CH(CH3)CH2CH3\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3): branched

Pentane is the straight-chain isomer and has a higher boiling point due to greater surface area for van der Waals interactions. 2-methylbutane is more branched and has a slightly lower boiling point. Both are stable alkanes; pentane is the least sterically hindered.

IT-2: Hydrated Salt Formula Determination (with Thermodynamics)

Question:

5.00 g5.00\,\text{g} of hydrated barium chloride (BaCl2⋅xH2O\text{BaCl}_2 \cdot x\text{H}_2\text{O}) is heated strongly until all the water of crystallisation is removed, leaving 4.26 g4.26\,\text{g} of anhydrous BaCl2\text{BaCl}_2.

(a) Determine the value of xx in the formula BaCl2⋅xH2O\text{BaCl}_2 \cdot x\text{H}_2\text{O}.

(b) The enthalpy of solution of anhydrous BaCl2\text{BaCl}_2 is −13.2 kJ mol−1-13.2\,\text{kJ mol}^{-1}While the enthalpy of solution of BaCl2⋅2H2O\text{BaCl}_2 \cdot 2\text{H}_2\text{O} is +8.8 kJ mol−1+8.8\,\text{kJ mol}^{-1}. Use a Hess’s law cycle to calculate the enthalpy change for the reaction:

BaCl2(s)+2H2O(l)→BaCl2⋅2H2O(s)\text{BaCl}_2(s) + 2\text{H}_2\text{O}(l) \to \text{BaCl}_2 \cdot 2\text{H}_2\text{O}(s)

Solution:

(a) Mass of water lost: 5.00−4.26=0.74 g5.00 - 4.26 = 0.74\,\text{g}

n(H2O)=0.7418.0=0.0411 moln(\text{H}_2\text{O}) = \frac{0.74}{18.0} = 0.0411\,\text{mol}

n(BaCl2)=◆LB◆4.26◆RB◆◆LB◆137.3+2×35.5◆RB◆=4.26208.3=0.02045 moln(\text{BaCl}_2) = \frac◆LB◆4.26◆RB◆◆LB◆137.3 + 2 \times 35.5◆RB◆ = \frac{4.26}{208.3} = 0.02045\,\text{mol}

x=◆LB◆n(H2O)◆RB◆◆LB◆n(BaCl2)◆RB◆=0.04110.02045=2.01≈2x = \frac◆LB◆n(\text{H}_2\text{O})◆RB◆◆LB◆n(\text{BaCl}_2)◆RB◆ = \frac{0.0411}{0.02045} = 2.01 \approx 2

The formula is BaCl2⋅2H2O\text{BaCl}_2 \cdot 2\text{H}_2\text{O}.

(b) Hess’s law cycle:

BaCl2(s)+2H2O(l)→ΔHBaCl2⋅2H2O(s)\text{BaCl}_2(s) + 2\text{H}_2\text{O}(l) \xrightarrow{\Delta H} \text{BaCl}_2 \cdot 2\text{H}_2\text{O}(s)

Route 1 (direct): ΔH=?\Delta H = ?

Route 2 (via aqueous solution):

  • BaCl2(s)→BaCl2(aq)\text{BaCl}_2(s) \to \text{BaCl}_2(aq): ΔH1=−13.2 kJ mol−1\Delta H_1 = -13.2\,\text{kJ mol}^{-1}
  • BaCl2⋅2H2O(s)→BaCl2(aq)+2H2O(l)\text{BaCl}_2 \cdot 2\text{H}_2\text{O}(s) \to \text{BaCl}_2(aq) + 2\text{H}_2\text{O}(l): ΔH2=+8.8 kJ mol−1\Delta H_2 = +8.8\,\text{kJ mol}^{-1}

Route 2 gives: BaCl2(s)+2H2O(l)→BaCl2(aq)\text{BaCl}_2(s) + 2\text{H}_2\text{O}(l) \to \text{BaCl}_2(aq) (reverse of the second step, so −ΔH2-\Delta H_2)

ΔH=ΔH1+(−ΔH2)=−13.2−8.8=−22.0 kJ mol−1\Delta H = \Delta H_1 + (-\Delta H_2) = -13.2 - 8.8 = -22.0\,\text{kJ mol}^{-1}

The hydration of BaCl2\text{BaCl}_2 to form the dihydrate releases 22.0 kJ mol−122.0\,\text{kJ mol}^{-1}.

IT-3: Titrations with Multiple Equivalence Points (with Acids and Bases)

Question:

25.0 cm325.0\,\text{cm}^3 of a solution containing both HCl\text{HCl} and CH3COOH\text{CH}_3\text{COOH} is titrated with 0.100 mol dm−30.100\,\text{mol dm}^{-3} NaOH\text{NaOH} using phenolphthalein indicator. 20.0 cm320.0\,\text{cm}^3 of NaOH\text{NaOH} is required to reach the end point. In a separate experiment, 25.0 cm325.0\,\text{cm}^3 of the same solution is titrated with 0.100 mol dm−30.100\,\text{mol dm}^{-3} NaOH\text{NaOH} but using a pH meter. The pH curve shows the first equivalence point at 10.0 cm310.0\,\text{cm}^3 and the second at 20.0 cm320.0\,\text{cm}^3.

(a) Calculate the concentration of HCl\text{HCl} in the original solution.

(b) Calculate the concentration of CH3COOH\text{CH}_3\text{COOH} in the original solution.

(c) Explain why the pH at the second equivalence point is greater than 7, and explain the significance of the first equivalence point.

Solution:

(a) The first equivalence point at 10.0 cm310.0\,\text{cm}^3 corresponds to neutralisation of the strong acid HCl\text{HCl} (since HCl\text{HCl} is neutralised first as it fully dissociates, and the pH\text{pH} rises steeply first around this point due to the strong acid being consumed before the weak acid):

n(HCl)=n(NaOH)first eq. pt.=0.100×10.01000=1.00×10−3 moln(\text{HCl}) = n(\text{NaOH})_{\text{first eq. pt.}} = 0.100 \times \frac{10.0}{1000} = 1.00 \times 10^{-3}\,\text{mol}

[HCl]=◆LB◆1.00×10−3◆RB◆◆LB◆25.0/1000◆RB◆=0.0400 mol dm−3[\text{HCl}] = \frac◆LB◆1.00 \times 10^{-3}◆RB◆◆LB◆25.0/1000◆RB◆ = 0.0400\,\text{mol dm}^{-3}

(b) The second equivalence point at 20.0 cm320.0\,\text{cm}^3 corresponds to total acid neutralised. The additional NaOH\text{NaOH} between the two equivalence points neutralises the CH3COOH\text{CH}_3\text{COOH}:

n(CH3COOH)=0.100×20.0−10.01000=1.00×10−3 moln(\text{CH}_3\text{COOH}) = 0.100 \times \frac{20.0 - 10.0}{1000} = 1.00 \times 10^{-3}\,\text{mol}

[CH3COOH]=◆LB◆1.00×10−3◆RB◆◆LB◆25.0/1000◆RB◆=0.0400 mol dm−3[\text{CH}_3\text{COOH}] = \frac◆LB◆1.00 \times 10^{-3}◆RB◆◆LB◆25.0/1000◆RB◆ = 0.0400\,\text{mol dm}^{-3}

(c) At the second equivalence point, all HCl\text{HCl} and CH3COOH\text{CH}_3\text{COOH} have been neutralised. The solution contains NaCl\text{NaCl} (neutral from the strong acid-strong base reaction) and CH3COONa\text{CH}_3\text{COONa} (a salt of a weak acid and strong base). The acetate ion (CH3COO−\text{CH}_3\text{COO}^-) is a weak base that hydrolyses in water:

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)\text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq)

This produces OH−\text{OH}^- ions, making the solution alkaline (pH>7\text{pH} \gt 7).

The first equivalence point (at 10.0 cm310.0\,\text{cm}^3) represents the point where all HCl\text{HCl} has been neutralised but the CH3COOH\text{CH}_3\text{COOH} has not yet been titrated. The solution at this point contains CH3COOH\text{CH}_3\text{COOH} and NaCl\text{NaCl}So the pH is determined by the weak acid alone (pH\text{pH} approximately equal to 12pKa−12log⁡[HA]\frac{1}{2}\text{p}K_a - \frac{1}{2}\log[\text{HA}]).

Additional Practice Problems

UT-4: Percentage Yield in Multi-Step Synthesis

Question: In a three-step synthesis, the percentage yields are 85%, 72%, and 90% respectively. If 10.0 g10.0\,\mathrm{g} of starting material is used, calculate the mass of final product obtained.

Solution:

Overall yield =0.85×0.72×0.90=0.551=55.1%= 0.85 \times 0.72 \times 0.90 = 0.551 = 55.1\% (1 mark).

Mass of final product =10.0×0.551=5.51 g= 10.0 \times 0.551 = 5.51\,\mathrm{g} (assuming 1:1 molar ratio in each step; if the molar ratios differ, the calculation must account for the molar mass changes at each step) (1 mark).

This illustrates the importance of high yields in each step of a multi-step synthesis. Even with relatively good individual yields (72—90%), the overall yield drops to 55%, meaning nearly half the starting material is lost.

UT-5: Ideal Gas Equation Unit Consistency

Question: A student calculates the volume of gas produced using V=nRT/pV = nRT/p with n = 0.050\,\mathrm{mol}$$R = 8.314$$T = 298And p=100p = 100. They obtain V=123.7V = 123.7. Identify the error and give the correct answer in cm3\mathrm{cm}^3.

Solution:

The student used p=100p = 100 without units. If they intended 100 kPa100\,\mathrm{kPa}They needed to convert to pascals: p=100000 Pap = 100000\,\mathrm{Pa} (1 mark).

V=nRTp=◆LB◆0.050×8.314×298◆RB◆◆LB◆100000◆RB◆=123.9100000=1.24×10−3 m3=1.24 dm3=1240 cm3V = \frac{nRT}{p} = \frac◆LB◆0.050 \times 8.314 \times 298◆RB◆◆LB◆100000◆RB◆ = \frac{123.9}{100000} = 1.24 \times 10^{-3}\,\mathrm{m}^3 = 1.24\,\mathrm{dm}^3 = 1240\,\mathrm{cm}^3

The student’s answer of 123.7123.7 is actually correct numerically but lacks units. If they meant dm3\mathrm{dm}^3Their answer is close. The key error was likely not tracking units through the calculation (1 mark).