# When neutralizing a mixture of palmitic and stearic acids weighing 106.6 g, a mixture of sodium palmitate

When neutralizing a mixture of palmitic and stearic acids weighing 106.6 g, a mixture of sodium palmitate and sodium stearate weighing 115.4 g was obtained. Determine the mass fractions of acids in the initial mixture.

Required data: stearic acid – C17H35COOH (M = 284 g / mol), palmitic acid – C15H31COOH (M = 256 g / mol). To simplify the problem, we will assume that the acids were neutralized with sodium hydroxide according to the following scheme: C17H35COOH + NaOH -> C17H35COONa + H2O; C15H31COOH + NaOH -> C15H31COONa + H2O. M (C17H35COONa) = 306 g / mol, M (C15H31COONa) = 278 g / mol. Let us denote the mass fraction of stearic acid in the mixture as x. This means that the proportion of palmitic acid in the initial mixture is (1 – x). The masses of acids in the mixture are 106.6 * x and 106.6 * (1-x) g, respectively. Let us turn to the amount of acid substance in the initial mixture. n (C17H35COOH) = (106.6 * x) / 284 mol, n (C15H31COOH) = (106.6 * [1 -x]) / 256 mol. According to the reaction equations, the amount of the salt substance is formed as much as the acid initially reacts. Therefore, the mass of sodium stearate is [(106.6 * x) / 284] * 306 g, sodium palmitate – [(106.6 * [1 -x]) / 256] * 278 g. By condition, the sum of the masses of salts is 115.4 g. Solving the equation for x, we get x = 0.399.

Answer: 39.9% stearic acid; 60.1% palmitic acid.

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