Question

A powder contains benzoic acid mixed with starch. A student finds that a 2.69 g sample of this powder requires 44.31 mL of a 0.101M sodium hydroxide solution for the acid to be neutralized. What is the percent by mass of benzoic acid in the powder?

Answer 1

Number of moles of NaOH , n = Molarity x volume in L

                                                 = 0.101 M x (44.31 x10 ^-3 L )

                                                 = 4.47x10^-3 mol

The balanced chemical reaction is : C₆H₅COOH + NaOH C₆H₅COONa + H₂O

According to the balanced equation ,

1 mole of C₆H₅COOH reacts with 1 mole of NaOH

4.47x10^-3 mol of C₆H₅COOH reacts with 4.47x10^-3 mol of NaOH

Molar mass of C₆H₅COOH is = (6x12)+(5x1)+12+16+16+1 = 122 g/mol

So mass of C₆H₅COOH is , m = Number of moles x molar mass

                                               = 4.47x10^-3 mol x 122 (g/mol)

                                               = 0.545 g

Given the mass of sample = 2.69 g

So mass percent of benzoic acid = ( mass of benzoic acid / mass of sample) x 100

                                                    = ( 0.545 / 2.69 ) x 100

                                                     = 20.3 %

Therefore the percent by mass of benzoic acid in the powder is 20.3 %