an unkown hydrocarbon contain carbon hydrogen and oxygen combustion of 1.6 g of the hydrocarbon produces...

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an unkown hydrocarbon contain carbon hydrogen and oxygencombustion of 1.6 g of the hydrocarbon produces 1.0286 g h2o and3.7681 g co2. what is the empiral formula of this compound?

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4.4 Ratings (869 Votes)

CO2 moles = mass of CO2 / molar mass of CO2

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3.7681 / 44

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0856

moles of C = 0.0856

mass of C = 0.0856 x 12

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.0276 g

moles of H2O = 1.0286 / 18

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0571

moles of H = 2 x 0.0571

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.1143

mass of H = 0.1143 x 1 = 0.1143 g

mass of O = sample mass - (mass of C + mass of H)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.6 - (1.0276 + 0.1143)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.4581 g

moles of O = 0.4581 / 16 = 0.0286

CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â HÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â O

0.0856 Â Â Â Â Â Â Â Â 0.1143 Â Â Â Â Â Â Â Â Â Â Â 0.0286

3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 4 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -----------------> by dividing smallest number

C3H4O

Empirical formula = C3 H4 O

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