Question

A sodium benzoate sample was received for analysis. 24.921g of the solid was heated at 110°C for 24 hours, reducing the mass to 24.773g, A 623.1 mg portion of the dried sample required 42.98 mL of 0.1000N perchloric acid to titrate it.

Report the moisture content of the sample.

Report the % sodium benzoate as determined by titration.

Does this sample meet release specifications according to these two criteria?

Answer 1

1. Report the moisture content of the sample.

24.921g of the solid was heated at 110°C for 24 hours
110°C

Therefore,

= ((24.921 - 24.773 ))/(24.921 )

= 0.148/24.921 X100 = 0.5938 %

(2) Report the % sodium benzoate as determined by titration.
The neutralization reaction performed by titration is,

C₆H₅COONa_(aq) + HClO_4(aq) → C₆H₅COOH_(aq) + NaClO_4(aq)
Firstly, Calculate the number moles of HClO₄ are used
For HClO₄, 1 M = 1 N, hence,

0.1000 M HClO₄ means 0.1000 N HClO₄
Now,
No. of moles = Molarity X Volume
                        = (0.1000) X (42.98)

= 4.298 mmol
Secondly, We will calculate the mmoles of C₆H₅COONa in titrated sample:
= (4.298) X (1 mole C₆H₅COONa / 1 mole HClO₄)

= 4.298 mmol C₆H₅COONa.
Now, Calculate the mass of C₆H₅COONa in 4.298 mmol C₆H₅COONa.
= 4.298 X (molar mass of sodium benzoate)

= 4.298 X 144.11

= 0.0298 mg of C₆H₅COONa is present in dried aliquot.
Lastly, Calculate the % of C₆H₅COONa in the dried sample,
% of sodium benzoate = (sodium benzoate present in dried aliquot)/(623.1 mg)

= 0.0298 / 623.1 X 100

= 0.00478 % sodium benzoate

3) Does this sample meet release specifications according to these two criteria?
As the criterion has been not provided, please answer by your own way.