# How to calculate the number of atoms of each element in a urea?

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?

• Something is off with the answer from the text: the number of moles of nitrogen given above is on the order of 100 trillion. – Todd Minehardt Sep 8 '16 at 23:00
• 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. – MaxW Sep 10 '16 at 15:37
• 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. – KeatonB Sep 11 '16 at 0:14
• 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, – porphyrin Oct 1 '19 at 7:44