# Chemistry

Please help me solve this empirical formula.Im so confused

C=58.5%
H=4.1%
N=11.4%
and O=26.0%

Thank you so much

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asked by Roger
1. Take 100 g sample for
58.5g C
4.1g H
11.4 g N
26.0 g O

Convert to mols. I'll use approximations.
58.5/12 = 4.875
4.1/1 = 4.1
11.4/14 = 0.814
26/16 = 1.625
Now find the ratio in small whole numbers with the smallest number being 1.00. The easy way to do that is to divide all of the numbers by the smallest number. Round to whole numbers. That gives us
C = 4.875/0.814 = about 5.99 = about 6
H = 4.1/0.814 = 5.04 = about 5
N = 0.814/0.814 = 1.00 = 1
O = 1.625/0.814 = 1.996 = about 2
So the empirical formula is
C6H5NO2

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posted by DrBob222

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