Answer to Question #54947 in General Chemistry for neelai
A 2.5600g sample of a compound that contains sulfur is analyzed by precipitating all the sulfur as BaSO4. If 1.1756g of BaSO4 are obtained in the analysis, what is the percentage of sulfur in the original compound
n(S) = n(BaSO4)=m(BaSO4)/MW(BaSO4) m(S) = n(S)*MW(S) w(S) = m(S)/M = m(BaSO4)*MW(S)/MW(BaSO4)*M MW were taken from Wikipedia w(S) = 1.1756*32.065/233.43*2.5600 = 0.063080 = 6.3080%
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