CxHy+[x+4y]O2⟶xCO2+2yH2O
Moles of CO2 =443.08=0.07 moles=x
Moles of H2O =180.72=0.04 moles=2y
x=0.07g atoms of carbon
y=0.08g atoms of hydrogen
x:y0.070.0878
Empirical formula becomes C7H8.
JEE Main 2013 — Chemistry Physical Chemistry
A gaseous hydrocarbon on combustion gives 0.72g of water and 3.08g CO2. What is the empirical formula of the hydrocarbon?
Held on 7 Apr 2013 · Verified 6 Jul 2026.
C6H5
C7H8
C2H4
C3H4
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