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

The % hydrogen in the sample = 4.94%

Step-by-step explanation

 Calculate the number of moles of NaOH required in back titration:

 Moles of HCl  =  0.1644 M * 24.27 mL *1L/1000 mL  =3.89 X 10^-3 moles

1 mole of HCl  reacts with 1 mol of NaOH

Thus, 3.89 X 10^-3 moles of HCl  react with 3.89 X 10^-3 moles  of NaOH

 Calculate the number of mole of unreacted NaOH:

Moles of NaOH =0.1031 M*50.00 mL*1L/1000 mL  = 5.12 X 10^-3 moles

The number of mole of unreacted NaOH =5.12 X 10^-3 mol -3.89 X 10^-3 moles =1.23 X 10^-3 mol

1.23 X 10^-3 mol  of NaOH react with C₆H₄ (COOCH₃)₂.

C₆H₄ (COOCH₃)₂ + 2OH- ---->C₆H₄ (COO)₂^2- + 2CH₃OH

 According to the balanced equation, 2 moles of NaOH react with1mol of C₆H₄ (COOCH₃)₂

Therefore, 1.23 X 10^-3 mol of NaOH  will react with =1.23 X 10^-3 mol of NaOH  *1mol C₆H₄ (COOCH₃)₂/2 mol NaOH =6.15 X 10^-4 mol C₆H₄ (COOCH₃)₂

 Calculate the mass of hydrogen in the sample:

1 mol C₆H₄ (COOCH₃)₂ has 10 moles of hydrogen

Thus, 6.15 X 10^-4 mol C₆H₄ (COOCH₃)₂ will have =6.15 X 10^-4 mol C₆H₄ (COOCH₃)₂ *10 moles of hydrogen /1molmol C₆H₄ (COOCH₃)₂ =6.15 X 10^-3  moles of hydrogen

Mass of hydrogen = 6.15 X 10^-3 moles * 1.008 g/mol =0.0062 g hydrogen

Calculate the % hydrogen in the sample:

The % hydrogen in the sample = (0.0062 g /0.1255 g)*100

The % hydrogen in the sample = 4.94%