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?

  • 2
    $\begingroup$ Something is off with the answer from the text: the number of moles of nitrogen given above is on the order of 100 trillion. $\endgroup$ – Todd Minehardt Sep 8 '16 at 23:00
  • $\begingroup$ 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 Sep 10 '16 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 Sep 11 '16 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 Oct 1 '19 at 7:44

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.

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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.

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  • $\begingroup$ why downvote??? $\endgroup$ – Vidyanshu Mishra Sep 11 '16 at 17:46

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