What is the percent compisition of C9 H11 NO4?

The task is simple, calculate the percent yield of each element in C9 H11 NO4. This is what I have done:

1. Calculated molecular weight of the mixture:

108.1(AW of C [9]) + 11.09 (AW of H [11]) + 78 (MW of NO4) = 197.2 g/mol

2. Divided the aw of each element by the 197.2 g/mol

108.1/197.2= 5.48 C g/mol
11.1/197.2= .056 H g/mol
78/197.2= .4 NO4 g/mol
Combined total is 5.936, not equal to 197.2

I stopped here since I know I must be doing something wrong if my answer to #2 does not total to my answer to #3. What have I done wrong?

Answer:
NO4 is off. Nitrous Oxide is (N20)4. If it is Nitrous Oxide, that is your problem.
Also, your confused on step two. Once you divide the individual by the total, the end resualt shouldn't equal the original number when all three are added.
If the NO4 is 1 nitrogen, 4 oxygen, step one is right. On step two, 108.1 divided by 197.2 is .548. Once you turn those numbers into percents, you get just about 100 percent making your work correct.
54.8 %C, 56 %H, 40% NO4. The reason for the percentages being a little over 100% is you rounded some of your numbers.

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