What is mass percent benzoic acid in powder?

The equation for the reaction of benzoic acid (C₆H₅COOH) with sodium hydroxide:
C₆H₅COOH + NaOH → C₆H₅COONa + H₂O
Mole ratio C₆H₅COOH : NaOH = 1 : 1

No. of moles of NaOH = (0.107 mol/L) × (35.3/1000 L) = 0.003777 mol
No. of moles of C₆H₅COOH = (0.003777 mol) × (1/1) = 0.003777 mol

Molar mass of C₆H₅COOH = (12.0×7 + 1.0×6 + 16.0×2) g/mol = 122.0 g/mol
Mass of C₆H₅COOH = (0.003777 mol) × (122.0 g/mol) = 0.461 g
Percent by mass of benzoic acid in the powder = (0.461/2.505) × 100% = 18.4%

To determine the number of moles of sodium hydroxide, multiply the volume in liters by the molarity.

n = 0.0353 * 0.107 = 0.0037771

This is the formula of benzoic acid, C6H5COOH, It is organic acid. When it reacts with sodium hydroxide, the products are sodium benzoate and water.

C6H5COOH + NaOH → NaC6H5COO + H2O


The number of moles of benzoic acid is the same as the number of moles of sodium hydroxide. Let’s determine the mass of one mole of benzoic acid.

Mass = 12 * 6 + 5 + 12 + 2 * 12 + 1 = 122 grams
Let’s determine the mass of 0.0037771 mole.

Mass = 0.0037771 * 122 = 0.460062 grams

Use the following equation to determine the percent by mass.

% = 100 * 0.460062 ÷ 2.505

This is approximately 18.4 %. I hope this is helpful for you.

35.3 x 0.107 = 3.77mm benzoic acid x 122mg/mm = 461mg = 0.461g
0.461/2.505 x 100 = 18.4%