Analysis of a sample of an organic compound showed it contains 39.9%carbon,6.9%hydrogen and 53.2%oxygen.

Question

a)calculate the empirical formula
b)if the relative molecular mass is 60,what is the molecular formula of the compound.(C=12,H=1,O=16)

Answer 1

Take a 100 g sample which gives you
39.9 g C
6.9 g H
53.2 g O

Convert to mols.
mols C = 39.9/12 = approx 3.3
mols H = 6.9/1 = approx 6.9
mols O = 53.2/16 = 3.3

Now find the ratio of the three to each other with the lowest number being 1. The easy way to do this is to divide the smallest by itself (which gives 1.0), then divide the other numbers by that same small number.
3.3/3.3 = 1
6.9/3.3 = 2
3.3/3.3 = 1
So the ratio is CH2O and that is the empirical formula.
Empirical mass is 12+2+16 = 30
Since the molar mass is 60, there must be how many units in the molecule; i.e., the molecular formula is (CH2O)2 or C2H4O2.

Answer 2

analysis of a sample of an organic compound showed it to contain 39%,carbon 6.9% hydrogen and 53.2% oxygen.caculate the emperical formular

Answer 3

thanks elot

Answer 4

Analysis of a simpe organic compound showed it contain 39.9% corbon 6.9% hydrogen and 53.2% oxygen calculate the empirical formula. solution

Answer 5

Thank alot

Answer 6

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