A 0.9679-g sample containing dimethylphthalate, (194.19 g/mol), and unreactive species was refluxed with 50.00 mL of 0.1215 M to hydrolyze the ester groups (this process is called saponification).

C6H4(COOCH3)2 + 2OH>> C6H4(COO)-2 + H2O

After the reaction was complete, the excess NaOH was back titrated with 32.25mL of 0.1251M HCl. Calculate the percentage of dimethylphthalate in the sample.

20.44% is the percentage of dimethylphthalate in the sample.

Explanation:

Molarity of NaOH =

Volume of NaOH consumed in back titration =

Molarity of HCl =

Volume of HCl =

Volume of NaOH used in hydrolysis of ester = 50.00 mL - 33.21 mL = 16.79 mL[/tex]

Moles of NaOH in 16.79 ml of 0.1215 M solution : n

Volume of solution = 16.79 mL = 0.01679 L ( 1mL=0.001 L)

1 mole of NaOH has 1 mole of hydroxide ions than 0.002040 moles of NaOH will have :

1 × 0.002040 mol = 0.002040 mol of hydroxide ions

Moles of hydroxide ions = 0.002040 mol

According to reaction, 2 moles of hydroxide ion reacts with 1 mole of  dimethylphthalate , then 0.002040 moles of hydroxide ion swill react with ;

of dimethylphthalate

Mass of 0.001020 moles of dimethylphthalate :

194 g/mol × 0.001020 mol = 0.1979 g

Mass of sample = 0.9679 g

Mass of dimethylphthalate = 0.1979 g

Percentage of dimethylphthalate  in sample;

20.44% is the percentage of dimethylphthalate in the sample.

1.solid

2.carbon dioxide

3.helium