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

Do you know where I am going wrong?
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.