Solution: (1) calculate the empirical formula, (2) compare "EFW" to molecular weight, (3) multiply empirical formula by proper scaling factor.
1) Calculate the empirical formula:
carbon: 49.98 g ÷ 12.011 g/mol = 4.16
hydrogen: 5.19 g ÷ 1.008 g/mol = 5.15
nitrogen: 28.85 g ÷ 14.007 g/mol = 2.06
oxygen: 16.48 g ÷ 15.999 g/mol = 1.03
carbon: 4.16 ÷ 1.03 = 4.04 = 4
hydrogen: 5.15 ÷ 1.03 = 5
nitrogen: 2.06 ÷ 1.03 = 2
oxygen: 1.03 ÷ 1.03 = 1
2) Empirical formula is C4H5N2O. The "empirical formula weight" is about 97.1, which gives a scaling factor of two.
3) The molecular formula is:
C4H5N2O times 2 = C8H10N4O2 <--- that's the molecular formula
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