# Calculating masses of unknown quantities of reactants.

1. Oct 18, 2013

### thatguythere

1. The problem statement, all variables and given/known data
Calculate the number of moles and mass of BaCl2 and NaCl in the original mixture.

2. Relevant equations
We prepared a solution of 0.35M Na2SO4. We then obtained an unknown mixture of BaCl2•2H2O and NaCl, weighed 1 g and added it to 200 mL of water and 10 mL HCl. Finally we added 10mL of the Na2SO4 solution and heated the mixture for several minutes to form a precipitate which was then dried and weighed. Using the weight of the precipitate, which was 1.769 grams, we are asked to calculate the masses and mass percent of both BaCl2 and NaCl in the original mixture. The chemical equation given is:
BaCl2(aq) + Na2SO4 (aq) → BaSO4(s) + 2Na+(aq) + 2Cl-(aq)

3. The attempt at a solution
Moles BaSO4= 1.769gBaSO4 x (1 mol BaSO4/233.39gBaSO4) = .00768 moles BaSO4
Since the molar ratio between BaSO4 and BaCl2 is 1:1

Moles BaCl2=0.00768 mol BaSO4 x (1 mol BaCl2/1 mol BaSO4)
= 0.00768 moles BaCl2

Grams BaCl2= 0.00768 mol BaCl2 x (208.23gBaCl2/1 mol BaCl2
=1.60g BaCl2

This is what makes me believe I have done something quite incorrect, since the original mass of my mixture was only 1 gram. Please advise. Thank you.

2. Oct 18, 2013

### Staff: Mentor

There is something wrong with your calculations in at least two places, sadly, it doesn't change the problem - mass of the barium chloride dihydrate is higher than the original sample mass. My bet is that you have not dried the BaSO4 correctly, that's the most common error in such cases.

Check your math. Molar mass of BaSO4 is OK, but number of moles isn't.

You said you used dihydrate, not anhydrous barium chloride, so the molar mass is incorrect.