Question

O.9471 g sample dimethylphthalate C6H4(COOCH3)2 (M.M.=194.19 g/mole) and unreactive species was reflux with 50.00 ml of 0.1215 M NaOH to hydrolyze the ester group.
After the reaction was complete, the excess NaOH was back-titrated with 24.27ml of 0.1644 M HCl.

Calculate the percentage of dimethylphthalate in the sample

Answer 1

molarity NaOH = 0.1215 M

volume NaOH = 50.00 mL = 0.05000 L

Total moles NaOH = (molarity NaOH) * (volume NaOH)

Total moles NaOH = (0.1215 M) * (0.05000 L)

Total moles NaOH = 0.006075 mol

molarity HCl = 0.1644 M

volume HCl = 24.27 mL = 0.02427 L

Total moles HCl = (molarity HCl) * (volume HCl)

Total moles HCl = (0.1644 M) * (0.02427 L)

Total moles HCl = 0.00399 mol

Unreacted moles of NaOH = Total moles of HCl

Unreacted moles of NaOH = 0.00399 mol

Moles NaOH used in hydrolysis = Total moles NaOH - Unreacted moles of NaOH

Moles NaOH used in hydrolysis = 0.006075 mol - 0.00399 mol

Moles NaOH used in hydrolysis = 0.002085 mol

moles dimethylphthalate = (1/2) * (Moles NaOH used in hydrolysis)

moles dimethylphthalate = (1/2) * (0.002085 mol)

moles dimethylphthalate = 0.001043 mol

mass dimethylphthalate = (moles dimethylphthalate) * (molar mass dimethylphthalate)

mass dimethylphthalate = (0.001043 mol) * (194.19 g/mol)

mass dimethylphthalate = 0.2024 g

percentage of dimethylphthalate in the sample = (mass dimethylphthalate / mass of sample) * 100

percentage of dimethylphthalate in the sample = (0.2024 g / 0.9471 g) * 100

percentage of dimethylphthalate in the sample = 21.38 %