Combustion analysis of an unknown compound containing only carbon and hydrogen p

Question

Combustion analysis of an unknown compound containing only carbon and hydrogen produced 4.554 g of CO2 and 2.322 g of H2O. What is the empirical formula of the compound? this would be so much easier if they gave the grams of the unknown compound. Please help!

Explanation / Answer

1.5 g CxHyOz + 5.376 g O2 -> 4.554 g CO2 + 2.322 g H2O CxHyOz + .168 mol O2 -> .1035 mol CO2 + .129 mol H2O CxHyOz + 13 O2 -> 8 CO2 + 10 H2O Balance the C and H. C8H20Oz + 13 O2 -> 8 CO2 + 10 H2O Balance the O. C8H20 + 13 O2 -> 8 CO2 + 10 H2O Because you can't stick 20 hydrogens on a skeleton of 8 carbons, the molecular formula is likely to be C4H10, and since both 4 and 10 are divisible by 2, the empirical formula is C2H5. 1.5 g of C4H10 (butane) yield those products 1.5 g C4H10 = .0259 mol .0259 mol C4H10 + .168 mol O2 -> .1035 mol CO2 + .129 mol H2O 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O

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