Answer to Question #294935 in General Chemistry for dli

Question

A 516.7-mg sample containing a mixture of K2SO4 and (NH4)2SO4 was dissolved in water and treated with BaCl2, precipitating the SO4 2– as BaSO4. The resulting precipitate was isolated by filtration, rinsed free of impurities, and dried to constant weight, yielding 863.5 mg of BaSO4. What is the %w/w K2SO4 in the sample?

Answer 1

Mass of K2SO4 = 516.7mg/1000mg = 0.5167g

Mass of product (BaSO4) = 863.5mg/1000mg =0.8635g

Let Xg be the mass of K2SO4 in the mixture;

Thus mass of the other salt (NH4)SO4 = (0.5167 -X )g

Moles of BaSO4 produced =mass/MM =0.8635g/233.4g/mol

=0.003699mol

Moles of Ka2SO4 = mass/MM = Xg/174.3g/mol

= 0.00573X mol

Moles of (NH4)2SO4 = (0.5267-X)g/132.1g/mol

= (0.00391-0.00757X) mol

Total number of moles SO42- ions obtained = 0.00573mol + (0.00394-0.00757X)mol

= (0.00391-0.001833X) mol

Mole ratios of SO42- ions to Ba2+ ions = 1:1

Thus moles of Ba2+ ions = (1/1)x0.00391-0.001833X =(0.00391-0.001833X)mol

But 0.003699mol of BaSO4 precipitate is formed

Hence

(0.00391-0.001833X)mol = 0.003699mol

0.001833X = 0.00391- 0.003699mol

= 0.000212mol

Therefore,

X = 0.000212/0.001833 = 0.1156g = 115.6mg

% by mass = (115.6mg/516.7mg)x100 = 22.36% K2SO4