percent by mass of acetic acid

Question

Calculate the percent by mass of acetic acid in a vinegar solution which is 0.794 M acetic acid. Show all calculations.

Answer 1

≈ 4.77% acetic acid by mass

Work with a 1.00 L basis

Molarity is per litre, so taking exactly 1 L makes the numbers fall out directly.

Step 1 — mass of solute in that litre

n = (0.794 mol/L)(1.00 L) = 0.794 mol

Molar mass of CH₃COOH (C₂H₄O₂): 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol

m = (0.794 mol)(60.05 g/mol) = 47.7 g acetic acid

Step 2 — mass of the whole solution

This is the step the problem depends on, and it is worth being explicit about: you need the density of the solution, which was not given. Vinegar is dilute aqueous, so its density is very close to water's — take 1.00 g/mL:

m_solution = (1000 mL)(1.00 g/mL) = 1000 g

Step 3 — percent by mass

% = (mass solute / mass solution) × 100 = (47.7 / 1000) × 100 = 4.77%

Note the denominator is the mass of the whole solution, not of the water. Dividing by the solvent mass gives percent by mass of solute-to-solvent, a different quantity.

Say the assumption out loud in your write-up

Since density was not supplied, state that you assumed 1.00 g/mL. The real value for 5% vinegar is about 1.005 g/mL, which would give 4.74% — a difference of 0.03 percentage points, entirely negligible here. But an unstated assumption is the kind of thing that costs marks, and in a more concentrated solution the same shortcut would introduce real error.

Reasonableness check — this one is satisfying

Commercial vinegar is labelled 5% acetic acid, and the FDA requires at least 4% for a product sold as vinegar. Getting 4.77% from a 0.794 M solution says the number is right and that 0.794 M is a realistic concentration for actual vinegar rather than an invented one.

If your answer had come out near 0.05% or 477%, the likely culprit is a factor of 1000 — mixing grams with milligrams, or millilitres with litres.