S6 Chemistry Paper: Organic Combustion Analysis, Empirical and Molecular Formulae, Esterification Yield, Addition Polymerisation and Fermentation (100 marks)

2026-11-13

S6 Chemistry – organic calculations

S6 Chemistry organic paper: 6 questions, 24 parts on combustion analysis, formulae, esterification yield, alkene addition, polymerisation and fermentation, with full answers.

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Paper details　S6 Chemistry practice paper on organic combustion analysis, empirical formula, esterification percentage yield, addition polymerisation and fermentation, with answers and marking points.

Section A: Combustion Analysis and Empirical Formula (18 marks)

Organic compound X contains only carbon, hydrogen and oxygen. 2.30 g of X burns completely to give 4.40 g of carbon dioxide and 2.70 g of water. (Relative atomic masses: C = 12.0, H = 1.0, O = 16.0; CO2 = 44.0, H2O = 18.0)

#	Question	Answer and marking notes
1	Calculate the moles of carbon dioxide and hence the mass of carbon in X.	Answer: 4.40 / 44.0 = 0.100 mol; mass of C = 0.100 x 12.0 = 1.20 g. (Marking: moles 2, mass 2)
2	Find the mass of hydrogen in X.	Answer: moles of water = 2.70 / 18.0 = 0.150 mol; moles of H = 2 x 0.150 = 0.300 mol; mass of H = 0.300 g. (Marking: water moles 2, H moles 2, mass 1)
3	Find the mass and the number of moles of oxygen in X.	Answer: 2.30 – 1.20 – 0.300 = 0.800 g; 0.800 / 16.0 = 0.0500 mol. (Marking: mass 2, moles 2)
4	Deduce the empirical formula of X.	Answer: C : H : O = 0.100 : 0.300 : 0.0500 = 2 : 6 : 1, so the empirical formula is C2H6O. (Marking: ratio 3, formula 2)

Section B: Molecular Formula, Isomerism and Functional Group Test (16 marks)

Another sample Y has the same empirical formula as X and a molar mass of 46.0 g/mol. Y has two isomers: ethanol (C2H5OH) and dimethyl ether (CH3OCH3).

#	Question	Answer and marking notes
5	Find the molecular formula of Y.	Answer: molar mass of C2H6O = 2×12.0 + 6×1.0 + 16.0 = 46.0; 46.0 / 46.0 = 1, so the molecular formula is C2H6O. (Marking: formula mass 2, multiplier 1, formula 1)
6	Name the two isomers and state the functional group of each.	Answer: ethanol CH3CH2OH with a hydroxyl group (-OH); dimethyl ether CH3OCH3 with an ether linkage (-O-). (Marking: ethanol 2, dimethyl ether 2)
7	Describe one chemical test to distinguish the two isomers and the observations.	Answer: add sodium; ethanol gives off hydrogen gas (bubbles) while dimethyl ether shows no obvious reaction. (Marking: reagent 2, observation 2)
8	0.100 mol of ethanol reacts with excess sodium. Find the moles of hydrogen released.	Answer: 2C2H5OH + 2Na -> 2C2H5ONa + H2; mole ratio 2 : 1, so H2 = 0.100 / 2 = 0.0500 mol. (Marking: equation 2, ratio 1, answer 1)

Section C: Esterification and Percentage Yield (18 marks)

0.200 mol of ethanol is mixed with 0.150 mol of ethanoic acid and warmed with concentrated sulphuric acid to make ethyl ethanoate (molar mass 88.0 g/mol). The experiment gives 9.68 g of ethyl ethanoate.

#	Question	Answer and marking notes
9	Identify the limiting reagent and explain your answer.	Answer: ethanol and ethanoic acid react in a 1 : 1 ratio; 0.150 mol of acid is less than 0.200 mol of ethanol, so the acid is limiting. (Marking: ratio 2, comparison 2, conclusion 1)
10	Find the theoretical yield of ethyl ethanoate in moles and in grams.	Answer: 0.150 mol; 0.150 x 88.0 = 13.20 g. (Marking: moles 2, mass 2)
11	Find the moles of ethyl ethanoate actually obtained.	Answer: 9.68 / 88.0 = 0.110 mol. (Marking: substitution 2, answer 2)
12	Calculate the percentage yield and suggest one way to raise it.	Answer: 0.110 / 0.150 x 100% = 73.3%; remove the water formed with a Dean and Stark trap (or use excess ethanol) to shift the equilibrium to the right. (Marking: yield 3, method 2)

Section D: Addition Reactions of Alkenes (16 marks)

0.0800 mol of propene reacts completely with hydrogen bromide to give 1-bromopropane (molar mass 122.9 g/mol). The actual yield is 85.0% of the theoretical yield.

#	Question	Answer and marking notes
13	Find the moles of hydrogen bromide required.	Answer: the addition ratio is 1 : 1, so 0.0800 mol is needed. (Marking: ratio 2, answer 2)
14	Find the theoretical mass of 1-bromopropane.	Answer: 0.0800 x 122.9 = 9.83 g. (Marking: moles 2, mass 2)
15	Find the actual mass obtained.	Answer: 9.83 x 0.85 = 8.36 g. (Marking: use of percentage 2, answer 2)
16	Explain why propene undergoes addition rather than substitution.	Answer: propene has a C=C double bond whose pi bond is weaker and breaks easily, so H and Br add to the two carbon atoms; substitution normally needs UV light and occurs in alkanes. (Marking: double bond 2, mechanism 2)

Section E: Addition Polymerisation and Degree of Polymerisation (16 marks)

A sample of poly(ethene) has a molar mass of 1.40 x 10^5 g/mol. The monomer is ethene (C2H4, molar mass 28.0 g/mol).

#	Question	Answer and marking notes
17	Calculate the molar mass of the ethene monomer.	Answer: 2 x 12.0 + 4 x 1.0 = 28.0 g/mol. (Marking: calculation 2, unit 2)
18	Find the average degree of polymerisation of the sample.	Answer: 1.40 x 10^5 / 28.0 = 5.00 x 10^3, that is 5000. (Marking: formula 2, answer 2)
19	Distinguish addition polymerisation from condensation polymerisation with one example each.	Answer: addition polymerisation releases no small molecule, for example poly(ethene); condensation polymerisation releases a small molecule such as water at each step, for example polyester or nylon. (Marking: difference 2, examples 2)
20	0.500 mol of ethene polymerises completely. Find the mass of polymer formed and the moles of polymer chains it contains.	Answer: 0.500 x 28.0 = 14.0 g; 14.0 / 1.40 x 10^5 = 1.00 x 10^-4 mol. (Marking: mass 2, chain moles 2)

Section F: Calculations on Fermentation of Glucose (16 marks)

Glucose (C6H12O6, molar mass 180.0 g/mol) ferments with yeast: C6H12O6 -> 2C2H5OH + 2CO2. 3.60 g of glucose ferments completely. The molar volume of a gas at room conditions is 24.0 dm3/mol.

#	Question	Answer and marking notes
21	Find the moles of glucose.	Answer: 3.60 / 180.0 = 0.0200 mol. (Marking: substitution 2, answer 2)
22	Find the moles of ethanol formed.	Answer: mole ratio 1 : 2, so 0.0200 x 2 = 0.0400 mol. (Marking: ratio 2, answer 2)
23	Find the mass of ethanol formed.	Answer: 0.0400 x 46.0 = 1.84 g. (Marking: molar mass 2, answer 2)
24	Find the volume of carbon dioxide at room conditions and explain why the actual volume is smaller.	Answer: 0.0400 x 24.0 = 0.960 dm3; some carbon dioxide dissolves in the aqueous mixture and some escapes or is absorbed in the apparatus, so the volume collected is smaller. (Marking: moles 1, volume 2, explanation 1)

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