A 0.1255 g sample containing dimethylphthalate, C6H4(COOCH3)2 and unreactive species was refluxed with 50.00 mL of 0.1031 M NaOH to hydrolyze the ester groups. C6H4 (COOCH3 )2 + 2OH- C6H4 (COO)2 2- + 2CH3OH After the reaction was complete, the excess NaOH was back titrated with 24.27 mL of 0.1644 M HCl. Calculate the % hydrogen in the sample

sample contains 4.64 % H.

Step-by-step explanation

The reaction is C₆H₄(COOCH₃)₂ + 2OH^-?→? C₆H₄(COO)₂^2- + 2CH₃OH

First let us calculate the amount of NaOH consumed.

moles of NaOH consumed = moles of NaOH - moles of HCl from back titration

To easily calculate the moles, let us convert the volume in mL to liters. For NaOH, 50 mL is equal to 0.05 L while for HCl, 24.27 mL is equal to 0.02427 L.

moles of NaOH = (0.1031 M NaOH) (0.05 L) - (0.1644 M HCl) (0.02427 L)

moles of NaOH = 0.001165012 moles

Second, calculate the amount of hydrogen.

Using moles of NaOH, use stoichiometric ratio to solve for moles C₆H₄(COOCH₃)₂, then moles of H in C₆H₄(COOCH₃)₂, the to mass of hydrogen by molar mass

mass H = 0.001165012 moles NaOH x (1 mol C₆H₄(COOCH₃)₂ / 2 mol NaOH) x (10 mol H/ 1 mol C₆H₄(COOCH₃)₂ ) x 1 g/mol H

mass H = 0.00582506 grams H

Third, obtain % H by dividing the mass of hydrogen and mass of sample multiplied by 100%.

% H = mass of H/mass of sample x 100%

% H = 0.00582506 grams H/0.1255 g sample x 100%

% H = 4.64 %