An impure sample of benzoic acid (C₆H₅COOH,122.12 g/mol) is titrated with 0.8067 M NaOH.A 5.109-g sample requires 36.97 mL of titrant to reach the endpoint.What is the percent by mass of benzoic acid in the sample?

C₆H₅COOH(aq) + NaOH(aq) $\to$ NaC₆H₅COO(aq) + H₂O(l)

Question

An impure sample of benzoic acid (C₆H₅COOH,122.12 g/mol) is titrated with 0.8067 M NaOH.A 5.109-g sample requires 36.97 mL of titrant to reach the endpoint.What is the percent by mass of benzoic acid in the sample?

C₆H₅COOH(aq) + NaOH(aq)

\to

NaC₆H₅COO(aq) + H₂O(l)

Quizplus · Accepted Answer

To find the percent by mass of benzoic acid, calculate the moles of NaOH used: 0.8067 M * 0.03697 L = 0.02983 mol. Since the reaction is 1:1, moles of benzoic acid = moles of NaOH = 0.02983 mol. Mass of benzoic acid = 0.02983 mol * 122.12 g/mol = 3.642 g. Percent by mass = (3.642 g / 5.109 g) * 100% = 71.29%.

An Impure Sample of Benzoic Acid (C6H5COOH,122 →\to→ NaC6H5COO(aq)+ H2O(l) A) 0

An Impure Sample of Benzoic Acid (C₆H₅COOH,122 $\to$ NaC₆H₅COO(aq)+ H₂O(l)
A) 0