When 4.748 grams of a hydrocarbon,
CxHy, were burned in a combustion analysis
apparatus, 16.05 grams of CO2 and
3.286grams of H2O were produced.
In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
|empirical formula =||?|
|molecular formula =||?|
Empirical formula is the least coefficient formula, that is
we must find x,y via gravimetry.
Typically we do this relating to moles of C,H
mol of CO2 = mass of CO2/MW of CO2 = (16.05)/(44) = 0.3647 mol of CO2
1 mol of CO2 = 1 mol of C ---> 0.3647 mol of CO2 = 0.3647 mol of C
mol of H2O = mass of H2O/MW of H2O= (3.286)/(18) = 0.1825 mol of H2O
1 mol of H2O= 2 mol of H ---> 0.1825 mol of H2O= 0.1825*2 = 0.365 mol of H
H:C =0.365 / 0.3647 = 1:1
CH is the empirical formula
if MW 26.04
MW of empirical = 12 + 1 = 13
ratio = 2x
C2H2 is the sample
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