From the following percentage compositions and true molecular (formula) mass det
ID: 492447 • Letter: F
Question
From the following percentage compositions and true molecular (formula) mass determine the molecular formula of the compounds shown A compound was analyzed and found to be 40.00% C, 8.71% H and 53.29% O. What is the empirical formula for this compound? The molecular weight was found to be about 120. What is the true formula? A compound whose molecular weight is 112 was found to contain 42.8% C, 7.20% H and 50.0% N. What is the true formula? A substance containing 27.27% C, 4.58% H, 36.33% O and 31.8% N was found to have a molecular weight of 88.1. What is the formula of this substance?Explanation / Answer
Element % At.Wt Relative number simple ratio
C 40 12 40/12 = 3.33 3.33/3.33 = 1
H 6.71 1 6.71/1 = 6.71 6.71/3.33 = 2
O 53.29 16 53.29/16 = 3.33 3.33/3.33 =1
Empirical formula = CH2O
molecular formula = (empirical formula)n
n = M.Wt/E.F.Wt = 120/30 = 4
molecular formula = (CH2O)4 = C4H8O4
10. Element % At.Wt Relative number simple ratio
C 42.8 12 42.8/12= 3.56 3.56/3.56 = 1
H 7.2 1 7.2/1 = 7.2 7.2/3.56 = 2
N 50 14 50/14 = 3.57 3.57/3.56 =1
Empirical formula = CH2N
molecular formula = (empirical formula)n
n = M.Wt/E.F.Wt
= 112/28 = 4
molecular formula + (CH2N)4 = C4H8N4
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