Question

3. A 6.0 g sample that contains a mixture of CaCl2 and NaCl was dissolved in water, and the solution treated with sodium oxalate (Na2C204) to precipitate the calcium as calcium oxalate (CaC204). Suppose 3.50 g of calcium oxalate was isolated and recovered, what is the weight percent of NaCl in the original 6.0 g sample? + 2 NaCl (aq) CaCl2 (aq) + NaCl (aq) + Na2C2O4 (aq) NazC204 (aq) CaC204 (5) No reaction -

Answer 1

Molar mass of CaCl ₂ = 40.08 + ( 2 $\times$ 35.45 ) = 110.98 g / mol

Molar mass of CaC₂O₄ = 40.08 + ( 2 $\times$ 12.01 ) + ( 4 $\times$ 16.00) = 128.10 g / mol

Consider a reaction, CaCl ₂ (aq) + Na₂C₂O₄ $\rightarrow$ CaC₂O₄ (s) + 2 NaCl (aq)

From above reaction, we can write 1 mol CaC₂O₄$\equiv$ 1 mol  CaCl ₂  

$\therefore$  128.10 g CaC₂O₄$\equiv$ 110.98 g CaCl ₂  

$\therefore$ 3.50 g CaC₂O₄$\equiv$ 110.98 $\times$ 3.50 / 128.10 g CaCl ₂  

3.50 g CaC₂O₄$\equiv$ 3.03 g CaCl ₂  

i e 6.0 g sample contain 3.03 g CaCl ₂ .

Hence, Mass of NaCl = 6.0 g - 3.03 g = 2.97 g

% weight NaCl = [ Mass of NaCl / Mass of sample ] 100

= 2.97 g / 6.0 g ) 100

= 49.5 %

ANSWER : % weight NaCl in the sample = 49.5 %