Practical: making and testing the smell of some esters in the lab
3 marks · NormalA student prepares an ester in the lab by reacting 1.0 cm³ of propanoic acid (density = 1.01 g/cm³, molar mass = 74 g/mol) with an excess of methanol in the presence of a few drops of concentrated sulfuric acid. After the reaction, they transfer the contents to water to isolate the ester layer.
Calculate the maximum mass of methyl propanoate that could be formed from the propanoic acid used.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
1.01 g of methyl propanoate (to 2 significant figures) g (±0.05)
Also accept: 1.0 g; 1.014 g
Common mistakes: Forgetting to calculate moles of acid; Using wrong molar mass; Assuming 1 cm³ = 1 g for propanoic acid
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Calculate the mass of propanoic acid using density × volume: 1.0 cm³ × 1.01 g/cm³ = 1.01 g.
M2 · 1 mark · Find moles of propanoic acid: mass/molar mass = 1.01 g / 74 g/mol = 0.0136 mol (2 or 3 s.f. accepted). (follow-through)
A1 · 1 mark · Use correct stoichiometry (1:1) and calculate the maximum mass of methyl propanoate formed: moles × molar mass (88 g/mol) = 0.0136 mol × 88 g/mol = 1.20 g. (Accept answers in range 1.0–1.2 g with correct working shown.) (follow-through)
Worked solution
Step 1: Calculate the mass of propanoic acid used:
Step 2: Calculate the number of moles of propanoic acid:
Step 3: Use the 1:1 mole ratio (for every 1 mole of acid, 1 mole of ester is formed):
Step 4: Calculate the maximum mass of methyl propanoate formed (molar mass = 88 g/mol):
Final answer: The maximum mass of methyl propanoate that could be formed is about 1.01 g (to 2 significant figures).