What is the empirical formula for 54.53% C, 9.15% H, and 36.32%O by mass.
CHO (please add subscripts or superscripts)
The molecular mass of the compound is 132 amu. What is the molecular formula?
2 Answers
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C = 54.53/12.011 = 4.5400
H = 9.15/1.008 = 9.0774
O = 36.32/15.999 = 2.2701
Divide by smallest:
C = 2.0
H = 4.0
O = 1
Empirical formula = C2H4O
Molar mass C2H4O = 12.011*2 +1.008*4 +15.999 = 44.053
132/44 = 3
Molecular formula = (C2H4O)3 = C6H12O3
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0.5453 = 12*C/132 gives C = 6
0.0915 = 1*H/132 gives H = 12
0.3632 = 16*O/132 gives O = 3
as subscripts
C6H12O3