A powder contains benzoic acid mixed with starch. A student finds that a 2.505 g sample of this powder requires 35.3 mL of a 0.107M sodium hydroxide solution for the acid to be neutralized. What is the percent by mass of benzoic acid in the powder?

Question

Adrien Taylor

Chemistry

14 June, 02:09

A powder contains benzoic acid mixed with starch. A student finds that a 2.505 g sample of this powder requires 35.3 mL of a 0.107M sodium hydroxide solution for the acid to be neutralized. What is the percent by mass of benzoic acid in the powder?

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Answers (1)

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Carl · Answer 1

Answer is: the percent by mass of benzoic acid in the powder is 18.41%.

1) Balanced chemical reaction:

C₆H₅COOH (aq) + NaOH (aq) → C₆H₅COONa (aq) + H₂O (l).

V (NaOH) = 35.3 mL : 1000 mL/L.

V (NaOH) = 0.0353 L; volume of the sodium hydroxide.

c (NaOH) = 0.107 mol/L; molarity of the sodium hydroxide.

n (NaOH) = c (NaOH) · V (NaOH).

n (NaOH) = 0.0353 L · 0.107 mol/L.

n (NaOH) = 0.00378 mol; amount of the sodium hydroxide.

2) From balanced chemical reaction: n (NaOH) : n (C₆H₅COOH).

n (NaOH) = n (C₆H₅COOH).

n (C₆H₅COOH) = 0.0038 mol; amount of the benzoic acid.

M (C₆H₅COOH) = 122.12 g/mol; molar mass of the benzoic acid.

m (C₆H₅COOH) = n (C₆H₅COOH) · M (C₆H₅COOH).

m (C₆H₅COOH) = 0.0038 mol · 122.12 g/mol.

m (C₆H₅COOH) = 0.461 g; mass of the benzoic acid.

m (powder) = 2.505 g; mass of a powder.

ω (C₆H₅COOH) = m (C₆H₅COOH) : m (powder) · 100%.

ω (C₆H₅COOH) = 0.461 g : 2.505 g · 100%.

ω (C₆H₅COOH) = 18.41%; mass percentage of the benzoic acid in a powder.