Combustion analysis of a hydrocarbon produced 33.01 g CO2 and 4.84 g H2O.

I was asked to find the empirical formula of the hydrocarbon. I used these molar mass values for my calculations:

C - 12.0107

H - 1.00794

O - 15.9994

Here is my work:

Molar mass of CO_2 = 12.0107 + 2*15.9994 = 44.0095
Molar mass of H_2O = 2*1.00794 + 15.9994 = 18.01528

We now have 33.01g of CO2 and 4.84g of H2O, which means:

33.01/44.0095 = 0.750 mol CO_2
4.84/18.01528 = 0.26866 mol H_2O
0.26866*2 = 0.53732 mol H

From the calculation we know that the ratio of C to H is 0.750:0.537, which is approximately 3:2. However, the empirical formula C3H2 is not the right answer. Can anyone tell me why I am wrong? Thank you!

  • 1
    $\begingroup$ @CaptainToad For future reference: When computing with quantities like $\pu{mol}$, $\pu{kg}$, etc. an equation sign indicates both 1) equality of the numbers on the right and left hand side, and 2) equality of the dimensions on the left and on the right hand side. Thus, units (like in $\pu{0.750 mol}$ $\ce{CO2}$ above) do not suddenly appear out of nothing (on occasion, they may yield a dimension less quantity, e.g., Reynold's number, though). Keep track of them (e.g., for dimensional analysis). $\endgroup$
    – Buttonwood
    Commented Oct 7, 2021 at 11:25

2 Answers 2


Let Hydrocarbon be of form $\ce{C_xH_y}$

Our combustion reaction becomes $\ce{C_xH_y}+(2x+\frac{y}{2})\ce{O_2}\implies x\ce{CO_2}+\frac{y}{2}\ce{H_2O}$

$1$ mole of hydrocarbon produce $x*44g$ of $\ce{CO_2}$

If $\ce{CO_2}$ produced is $33.01g$, then $x=\frac{3}{4}$

Similarly $1$ mole of hydrocarbon produces $\frac{y}{2}$ moles of $\ce{H_2O}$

If $\ce{H_2O}$ produced = $4.84g$, then $\frac{y}{2}=4.84/18\implies y=\frac{1.61}{3}$

Therefore $\frac{x}{y}\approx1.4$ Therefore empirical formula becomes $\ce{C_7H_5}$


0.750//0.53732 = 1.396/1;Hydrocarbon, no heteroatoms, needs even number of H so ratio becomes 2.792/2. x4 = 11.18/8 X5= 13.96/10 X6= 16.72/12 best fit is 5times so becomes C14H10. Notice if one rounds off immediately the answer is nice and neat. Don't ever fall into that trap. Next step is MW determination; Rast method or freezing point in benzene or infrared spectrum and mass spec.

  • $\begingroup$ Why was this answer downvoted? It is the only answer without guessing [ or a MW that fits] or approximating and does show that empirical measurements and/or made up problems do have errors. $\endgroup$
    – jimchmst
    Commented Jan 3, 2022 at 1:35

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