1.4 Exam Practice

Complete combustion of 2.50 g of an organic compound X produced 3.67 g of CO_2 and 1.50 g of H_2O. The compound contains only carbon, hydrogen and oxygen.

(a) Calculate the mass of carbon and hydrogen in the sample. [2]

(b) Determine the mass of oxygen in the compound and hence the empirical formula of X. [3]

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(a) Mass of C: 12.01/44.01 × 3.67 = 1.001 g [1]

Mass of H: 2 × 1.008/18.02 × 1.50 = 0.168 g [1]

(b) Mass of O: 2.50 - 1.001 - 0.168 = 1.331 g [1]

Moles: C = 1.001/12.01 = 0.0834, H = 0.168/1.008 = 0.167, O = 1.331/16.00 = 0.0832 [1]

Ratio C : H : O = 1 : 2 : 1, so empirical formula = CH_2O [1]

Examiner tip: When calculating empirical formulae, divide each mole value by the smallest mole value. If the ratio is not whole numbers (e.g. 1 : 1.5 : 1), multiply all values by 2. Show all working clearly for full marks.

A student prepares a stock solution by dissolving 5.85 g of NaCl (M = 58.44 g mol⁻¹) in water to make 250 cm³ of solution.

(a) Calculate the concentration of the stock solution in mol dm⁻³. [2]

(b) The student takes 25.0 cm³ of the stock solution and dilutes it to 100 cm³. Calculate the concentration of the diluted solution. [2]

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(a) Moles of NaCl = 5.85/58.44 = 0.1001 mol [1]

Concentration = 0.1001/0.250 = 0.400 mol dm⁻³ [1]

(b) Using c_1V_1 = c_2V_2: 0.400 × 25.0 = c_2 × 100 [1]

c_2 = 0.100 mol dm⁻³ [1]

Examiner tip: Always convert cm³ to dm³ (divide by 1000) before using the concentration formula. The dilution formula c_1V_1 = c_2V_2 works with any volume unit as long as both sides use the same unit.

Calcium carbonate decomposes on heating: CaCO_3(s) → CaO(s) + CO_2(g)

A student heated 10.0 g of CaCO_3 and obtained 4.82 g of CaO.

(a) Calculate the theoretical yield of CaO from 10.0 g of CaCO_3. [2]

(b) Calculate the percentage yield. [1]

(c) Calculate the atom economy for the production of CaO in this reaction. [1]

(d) Give one reason why the actual yield is less than the theoretical yield. [1]

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(a) Moles of CaCO_3 = 10.0/100.09 = 0.09991 mol [1]

1:1 ratio, so theoretical mass of CaO = 0.09991 × 56.08 = 5.604 g [1]

(b) Percentage yield = 4.82/5.604 × 100 = 86.0% [1]

(c) Atom economy = 56.08/100.09 × 100 = 56.0% [1]

(d) Any one of: incomplete decomposition / some CaCO_3 did not fully decompose / transfer losses / some CaO absorbed moisture from the air [1]

Examiner tip: Questions on yield and atom economy frequently appear on Paper 2. Remember that percentage yield compares actual vs theoretical, while atom economy compares desired product mass vs total product mass from the balanced equation.

Links to: R2.1 Amount of Change (stoichiometry and limiting reagents)

25.0 cm³ of a solution of sodium hydroxide of unknown concentration was titrated against 0.150 mol dm⁻³ sulfuric acid. The mean titre was 20.0 cm³.

The equation for the reaction is: 2NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + 2H_2O(l)

(a) Calculate the number of moles of H_2SO_4 used. [1]

(b) Determine the number of moles of NaOH in the 25.0 cm³ sample. [1]

(c) Calculate the concentration of the NaOH solution. [1]

(d) State the colour change observed if methyl orange is used as the indicator. [1]

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(a) n = 0.150 × 0.0200 = 3.00 × 10⁻³ mol [1]

(b) Mole ratio 2:1, so moles NaOH = 2 × 3.00 × 10⁻³ = 6.00 × 10⁻³ mol [1]

(c) c = 6.00 × 10⁻³/0.0250 = 0.240 mol dm⁻³ [1]

(d) Yellow to orange / red (at the end point, in the flask) [1]

Examiner tip: Always check the stoichiometric ratio from the balanced equation before using mole ratios. A 2:1 ratio means you need twice the moles of NaOH compared to H_2SO_4. This is one of the most common errors in titration calculations.

Links to: R3.1 Proton Transfer (acid-base reactions and indicators)