Drawing and naming the carboxylic acids

3 marks · Normal

Vinegar contains ethanoic acid as its main component. A student finds that a 75 g bottle of vinegar contains 3.0 g of pure ethanoic acid.

Calculate the percentage by mass of ethanoic acid in this sample of vinegar. Give your answer to 2 significant figures.

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.

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4.0% (±0.2)

Also accept: 4%; 4.00%

Common mistakes: Dividing 75 by 3; Multiplying instead of dividing

Show mark scheme and worked solution

Marking points

M1 · 1 mark · Writes or uses the correct formula: (mass of solute / mass of solution) x 100

M2 · 1 mark · Substitutes values correctly: (3.0 / 75) x 100

A1 · 1 mark · Correct final answer to 2 significant figures: 4.0% (follow-through)

Worked solution

Percentage by mass = (mass of ethanoic acid / total mass of vinegar) 100

Therefore, the percentage by mass of ethanoic acid in the vinegar is 4.0% (2 significant figures).