Calculating masses of unknown quantities of reactants.

[thatguythere]

Homework Statement

Calculate the number of moles and mass of BaCl₂ and NaCl in the original mixture.

Homework 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)

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.

 

[Borek]

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.

Moles BaSO₄= 1.769gBaSO₄ x (1 mol BaSO₄/233.39gBaSO₄) = .00768 moles BaSO₄

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

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

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