A 1 g sample of an organic compound was burned in a combustion analyzer and yielded 1.42 g carbon dioxide and 0.872 g of water. Calculate the percent carbon and hydrogen in the sample. From the percentages, calculate the empirical formula for the compound.


Your initial compound weights 1g, and it is only made of C and H. The products are CO2 and H2O. In other words (the reaction is not balanced, but for our purpose we don't care):

CxHy + O2 --> CO2 + H2O

From the grams of CO2 you can get back to the moles (and therefore grams) of C in the initial compound, while from the grams of water you can get to the moles (and grams) of H.

moles of C = 1.42 g CO2 x $\frac{\text{1 mol CO2}}{\text{44.0 g CO2}}$ x $\frac{\text{1 mol C}}{\text{1 mol CO2}}$ = 0.0323 mol C You then multiply by the molar mass of carbon, which is 12 g/mol, and you get 0.388 g.

We do the same thing for hydrogen:

moles of C = 0.872 g H2O x $\frac{\text{1 mol H2O}}{\text{18.0 g H2O}}$ x $\frac{\text{2 mol H}}{\text{1 mol H2O}}$ = 0.0969 mol H = 0.0989 g

Note that we did the same thing twice, but since we have two moles of H in water we multiplied by 2 in the second equation. Now we just have to find the percentage: You get

$\frac{\text{0.388 g}}{\text{1 g sample}}$ *100 % = 38.8 % C

$\frac{\text{0.0989 g}}{\text{1 g sample}}$ *100 % = 9.89 % H

I hope this helps!

  • $\begingroup$ Since the percentages don’t add up to 1, is there oxygen in the compound as well? $\endgroup$ – Katie Jan 25 '19 at 3:39
  • $\begingroup$ Yes, there are some other elements for sure (oxygen, or maybe nitrogen) $\endgroup$ – Pier Jan 25 '19 at 3:43
  • $\begingroup$ Would the empirical formula be CH3? $\endgroup$ – Katie Jan 25 '19 at 3:49
  • $\begingroup$ No, CH3 would be part of the empirical formula, but we don't know the complete one since we don't have all the elements :) $\endgroup$ – Pier Jan 26 '19 at 14:34

Mass of carbon =(12/44)X1.42=0.39g So percent mass of carbon= (0.39/1)X100 = 39 Mass of hydrogen = (2/18)X0.872 =0.097g So percent mass of hydrogen = (0.097/1)X100 = 9.7

P.S.: Please always attempt such simple questions at home; you are sure to get the answer, no?

  • $\begingroup$ Here we cannot conclude about the empirical formula since there are possibly one or more other element(s). $\endgroup$ – Nandakumar U K Jan 25 '19 at 4:56
  • $\begingroup$ Thank you! That’s what I thought as well. This was a question on my homework. Since the percentages I calculated didn’t add to 1 I thought I was calculating the percentages incorrectly. $\endgroup$ – Katie Jan 25 '19 at 5:27
  • 1
    $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$ – andselisk Jan 25 '19 at 13:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.