I got another Chem Question, Stoic, sorry guys, haha, i don't have the textbook answers so I feel really insecure xD... Hope you guys can bear with me

A sample of hydrated copper (II) sulfate (CuSo4.5H2O) is contaminated with silica. A 13.86g sample of the compound was dissolved in water and filtered to remove the insoluble silica. An excess of barium chloride solution was added to the filtrate, and the resulting precipitate collected and dried. If the sample contained 5.05 g of silica, what mass of barium sulfate precipitate would be expected?
So basically I did the m(sample) - m (silica) = m(CuSo4)
Which would be 13.86g - 5.05g = 8.81g
and then I used that m(cuSo4) and did my mol etc..etc..
My final answer was m(BaSO4) = 12.8g
Not sure if its right ^^..Hope I can get it confirmed tih someone