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Question

A compound contains only C, H, and N. Combustion of 28.0 mg of the compound produces 26.8 mg CO2 and 32.9 mg H2O. What is the empirical formula of the compound? (Type your answer using the format CO2 for CO2.)

Answer 1

Atomic weight:
C: 12.011
H: 1.0079
N: 14.007
Molecular weight:
CO2: 44.0095
H2O: 18.015
Therefore, in every 28.0 mg of the compound , there is (12.011/44.0095)x26.8 mg = 7.31 mg of C, and (2x1.0079/18.015)x32.9 mg = 3.68 mg of H, leaving:
28.0 - 7.31 - 3.68 = 17.01 mg of N.
7.31 mg of C is 0.6086mMol C = A Mol C
3.68 mg of H is 3.651mMol H = 6A Mol H, and
17.01 mg of N is 1.2144mMol of N = 2A Mol N
Thus the empirical formula of the compound is: CH6N2