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Chemistry 9 Online
OpenStudy (anonymous):

Help please....If 6.50g of the unknown compound contained 0.217mol of C and 0.433mol of H, how many moles of oxygen, O, were in the sample?

OpenStudy (anonymous):

m(CxHyOz) = 6.50 g mole of C n(C) = 0.217 => mass of carbon m(C) = n(C) * Ar(C) = 0.217 * 12 = 2.604 g mole of H n(H) = 0.433 => mass of hydrogen m(H) = n(H) * Ar(H) = 0.433 * 1=0.433 g so mass of C and H is m(C + H) = 2.604 + 0.433 = 3.037 g m(compound) - m(C and H) = m(Oxygen) m(oxygen) = 6.50 - 3.037 = 3.463 g mole of oxygen is n(O) = 3.463/16 ~ 0.216 (or 0.217) here by Ar I'm writing atomic weights of atoms..

OpenStudy (anonymous):

so my answer would be 0.217?

OpenStudy (anonymous):

sorry im confused?

OpenStudy (anonymous):

I got it thank you very much

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