Suppose I am trying to determine a compound that has 15.87% carbon (C), 2.15 % hydrogen (H), 18.5% nitrogen (N), & 63.41% oxygen (O) (n.b., this does not equal exactly 100%): what is the empirical formula of the compound?

1 Answer
Stefan V. · Media Owl
Jan 17, 2015

For my piece of mind, I'll add something to each percentage to make the total 100% - 0.02% to the first three, and 0.01% to oxygen.

So, you know that your compound has "15.89% C"15.89% C, "2.17% H"2.17% H, "18.52% N"18.52% N, and "63.42% O"63.42% O. The first step in determining it empirical formula is to divide each element's percentage by its molar mass

"For C": (15.89%)/(12.0) = 1.34For C:15.89%12.0=1.34

"For H": (2.17%)/(1.00) = 2.17For H:2.17%1.00=2.17

"For N": (18.52%)/(14.0) = 1.32For N:18.52%14.0=1.32

"For O": (63.42%)/(16.0) = 3.96For O:63.42%16.0=3.96

The next step is to divide all these numbers by the smallest one; this is done in order to get the ratios of the four elements in the molecule

"For C": 1.34/1.32 ~= 1.00For C:1.341.32≅1.00

"For H": 2.17/1.32 = 1.64For H:2.171.32=1.64

"For N": 1.32/1.32 = 1.00For N:1.321.32=1.00

"For O": 3.96/1.32 = 3For O:3.961.32=3

The empirical formula must only contain subscripts that are integers, which means that your actual empirical formula is

C_3H_5N_3O_9C3H5N3O9

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