# Chemistry

How do I do the empirical formula for 37.01% carbon, 2.22% hydrogen, 18.50% nitrogen, and 42.27% oxygen? My teacher gave me the answer (C7H5N3O6), but I can never get that answer.

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1. Take a 100 g sample which gives you
37.01g C]
2.22 g H
18.50 g N
42.27 g O

Convert grams to mols mols = grams/atomic mass.

37.01/12 = approx 3.08
2.22/1 = approx 2.22
18.50/14 = approx 1.32
42.27/16 = approx 2.64

Now find the ratio of these elements to each other with the smallest number no smaller than 1.00. The easy way to do that is to divide the smallest number by itself, then divide all of the other numbers by the same small number.
3.08/1.32 = approx 2.33
2.22/1.32 = 1.68
1.32/1.32 = 1.00
2.64/1.32 = 2.00

You want whole numbers but 0.33 and 0.68 are too large to be thrown away (and if you do this more accurately than my approximations above it may be a little better but you still must try and find whole numbers.) What I do is multiply all of the numbers by 2,3,4,5 etc until you get whole numbers (or at least very close to whole numbers). I can see that the way to get rid of that .33 is to multiply by 3 so let's try that. (Try multiplying by 2 and see what you get.)
3.08/1.32 = approx 2.33*3 = 6.99 C
2.22/1.32 = 1.68*3 = 5.04 H
1.32/1.32 = 1.00*3 = 3.00 N
2.64/1.32 = 2.00*3 = 6.00 O
6.99 will round to 7.0 easily for C.
5.04 rounds to 5.0 for H.
and 3.00 and 6.00 are ok so the empirical formula is C7H5N3O6

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