Question

You want to determine the percent of sulfate in an unknown sample of a hydrated metal sulfate with the formula My(SO4)x·nH2O. You dissolve 0.2888 g of your unknown in DI water and heat to near boiling. You react the dissolved unknown with a hot barium chloride solution and you collect the barium sulfate you make via gravity filtration. The filter paper weighed 0.597 g after a piece was torn off. The dry barium sulfate crystal and the filter paper weighed 0.910 g. Answer the following:

1. What is the molar mass of barium sulfate?

2. What is the molar mass of sulfate?

3. How many moles of sulfate were produced?

4. How many grams of sulfate were produced?

5. What is the percent sulfate?

I know the answer for the first two but im unsure of the next couple of questions. Thank you in advnadce :)

Answer 1

1. What is the molar mass of barium sulfate?

Solution :- Molar mass is the sum of atomic masses of the elements in the formula

Barium sulfate = BaSO4 = 233.3896 g per mol

2. What is the molar mass of sulfate?

Solution :-

Sulfate = SO4^2- = 96.0626 g / mol

3. How many moles of sulfate were produced?

Solution :- mass of Barium sulfate = 0.910 g – 0.597 g = 0.313 g

Moles of BaSO4 = mass / molar mass

                           = 0.313 g / 233.3896 g per mol

                     = 0.001341 mol

Mole ratio of the BaSO4 to SO4^2- is 1 : 1

Therefore moles of sulfate (SO4^2-) = 0.001341 mol

4. How many grams of sulfate were produced?

Solution :- Mass of sulfate = moles * molar mass

                                           = 0.001341 mol * 96.0626 g per mol

                                          = 0.129 g

5. What is the percent sulfate?

Solution :-

% sulfate = (mass of sulfate / total mass )*100%

                 = (0.129 g / 0.2888 g)*100%

                 = 44.67 %