I have a stoichiometry question which I answered incorrectly and I am not sure where I went wrong. The question is as follows:

Urea, $\ce{(NH2)2CO}$, is used for fertilizer and many other things. Calculate the number of $\ce{N}$, $\ce{C}$, $\ce{O}$, and $\ce{H}$ atoms in $1.68 \times 10^4~\mathrm{g}$ of urea.

First, I found the total atomic weights of each element in this compound to find the percent composition:

  • N: 28.02 amu
  • H: 4.032 amu
  • C: 12.01 amu
  • O: 16 amu
  • Total atomic weight of urea: 60.062

Then, I found the percent composition of urea:

  • N: 46.65%
  • H: 6.71%
  • C: 20.00%
  • O: 26.64%

Then, I converted $1.68 \times 10^4~\mathrm{g}$ into moles by dividing this amount in grams by 60.062, and found that it is approximately 279.71 moles.

After that, I took percentages of this to calculate the number of moles of each element in the sample of urea (for example, 46.65% of 279.71 to find the number of moles of nitrogen).

Finally, I multiplied each of these numbers in moles by Avogadro's number ($6.022\times10^{23}$) to obtain the number of atoms. However, when I checked my answers in my textbook, I was wrong by a significant amount.

\begin{array}{lrr} & \text{My answer} & \text{Correct answer}\\\hline \ce{N}:& 7.83 \times 10^{25} & 3.37 \times 10^{37}\\ \ce{H}:& 1.13 \times 10^{25} & 6.74 \times 10^{26}\\ \ce{C}:& 3.37 \times 10^{25} & 1.69 \times 10^{26}\\ \ce{O}:& 4.49 \times 10^{25} & 6.74 \times 10^{26}\\\hline \end{array}

Do you know where I am going wrong?

  • 3
    $\begingroup$ Something is off with the answer from the text: the number of moles of nitrogen given above is on the order of 100 trillion. $\endgroup$ Commented Sep 8, 2016 at 23:00
  • 1
    $\begingroup$ the answers given are just wrong. From the molecular formula there should be as many oxygen atoms as carbon atoms. There should be 4 times as many hydrogen atoms as carbon atoms. There should be twice as many nitrogen atoms as carbon atoms. $\endgroup$
    – MaxW
    Commented Sep 10, 2016 at 15:37
  • $\begingroup$ I think that whoever added all the answers into the chart may have accidentally messed up the books answer for atoms of oxygen, because I am positive that when I answered the question the number of atoms of oxygen was equal to the number of atoms of carbon that is listed (1.69×10^26) which is correct. The only incorrect answer was for Nitrogen and only the exponent was wrong. $\endgroup$
    – KeatonB
    Commented Sep 11, 2016 at 0:14
  • $\begingroup$ There are $\approx 280 \times 6.0\cdot10^{23}\times 8 $ atoms. N atoms are present as 2/8 of these H as 4/8, and C=O=1/8, $\endgroup$
    – porphyrin
    Commented Oct 1, 2019 at 7:44
  • $\begingroup$ You don't need complicated calculations once you know the number of moles of urea. everything else is just that number time the number of atoms in a single molecule of each element in a single urea. $\endgroup$
    – matt_black
    Commented Mar 2, 2021 at 0:20

3 Answers 3


You were correct in converting the grams of urea to mol of urea, but the next step would be to convert the mol of urea to molecules of urea using avogadro's number (which converts mol to molecules or atoms). Then you want to multiply the molecules of urea by the number of atoms of the element in question that are present in Urea. So for example there are 2 atoms of N per molecule of Urea, therefore multiplying the molecules of urea you calculated earlier by 2 will yield the total number on nitrogen atoms present. As was mentioned in the comments, the value for N atoms present given by your book was incorrect and should have been 3.37 × 10^26 atoms of Nitrogen. See if you can calculate the correct number of each of the remaining elements, as the answers given were correct.


do not stuck into percentage,this method is incorrect.here by formula you can say that one mole of urea contain 2 mole of N, 4 mole of H,1 mole of C and! mole of O. Also the no. of moles you calculated(for urea) are 279.71,thus no. OF moles of N will be 2x279.71 that of H will be 4x279.71 and so on, since now you know the number of moles of each atom you can now calculate no. of atom of each element.

  • $\begingroup$ why downvote??? $\endgroup$ Commented Sep 11, 2016 at 17:46

I see a couple subtle errors in your calculations, both of them involving units.

First, when you get the number of moles (279.71) you have the number of moles of molecules. To count individual atoms you have to multiply the percentage composition (which I will say more about, below) by the total number of moles of atoms. So, convert your moles of molecules to moles of atoms by multiplying by the eight (coubt them) atoms per molecule. So you have, if I follow significant figures, 2237.7 moles of atoms in your urea sample.

Next is the matter of percentages you use. You used mass percentages. You need molar percentages. So instead of multiplying 46.25 mass% nitrogen by those 2237.7 moles, you should multiply 25 mole% nitrogen, this being two nitrogen atoms divided by eight total atoms in the molecule. You could also have multiplied the two nitrogen atoms per molecules times 279.71 moles of molecules directly, saving a conversion step (a good idea is to try both ways and see that they are equivalent). Do the same for the other elements.

Finally, you or the book misrecorded one of the answers. The exponent in the nitrogen atom count should be 26, not 37.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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