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Chemistry 20 Online
OpenStudy (anonymous):

gen chem 1: a 2.5600 gram sample of a sulfur-containing compound is analyzed by precipitating the sulfur as BaSO4. If 1.1756 g of BaSO4 are obtained, what is the percentage of sulfur in the sample?

OpenStudy (anonymous):

my work: 1.1756 g BaSO4 x 1mol BaSO4/233.39 g BaSO4 x 1 mol S/1mol BaSO4 x 32.065 g s/1mol s x 1/2.5600 g sample x 100% = 6.3091% Is this correct?

OpenStudy (anonymous):

n(BaSO4) = 1.1756/233 = 0.005 mol m(S in BaSO4) = 0.005*32= 0.16 g (this sulfur is the sulfur in the sample as well) w(S in the sample) = 0.16/5600 * 100% = 0.0029%

OpenStudy (anonymous):

We it the same way except the sample is 2.5600g instead of 5600.

OpenStudy (anonymous):

Thanks

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