A hydrocarbon has a weight of 78 gives carbon dioxide and water on combustion with 3.385 of carbon and 0.682 of water find the formula
Calculate the moles of C and of H formed in the combustion:
"\\dfrac{3.385}{44}=0.0769 moles" CO2 * ("\\dfrac{1 mole}{1moles_{CO_{2}}})" "=0.0769moles" of C
"\\dfrac{0.682}{18}=0.0379moles" H2O * ("\\dfrac{2 mole}{1moles_{H_{2}O}})" = "0.0758moles" of H
Divide each by the smaller:
C = "\\dfrac{0.0769}{0.0758}=1.0"
H = "\\dfrac{0.0758}{0.0758} = 1.0"
Since the empirical formula mass is 13 "( \\tfrac{g}{mole})" dividing the molar mass of the compound by the formula mass gives "\\tfrac{78}{16}\t=6" .
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