Question

A compound contains 40.0% c, 6.71% h, and 53.29% o by mass. the molecular weight of the compound is 60.05 amu. the molecular formula of this compound is

Answer 1

To determine the molecular formula of the compound given, we first need to determine the empirical formula. To do that we assume to have 100 grams sample of the compound with the given composition. Then, we calculate for the number of moles of each element. We do as follows:
         mass       moles
C       40.0        3.33
H       6.71        6.64
O      53.29       3.33

Dividing the number of moles of each element with the smallest value, we will have the empirical formula:

CH2O

To determine the molecular formula, we multiply a value to the empirical formula. Then, calculate the molar mass and see whether it is equal to the one given (60.05 g/mol).

              molar mass 
CH2O      30.03
C2H4O2  60.06

Most likely, the molecular formula of the compound would be C2H4O2.

Answer 2

The molecular formula of the compound would be C2H4O2

Molecular formula

Molecular formula = [empirical formula]n

Where n = molar mass/empirical formula mass

Empirical formula

C = 40/12 = 3.33

H = 6.71/1 = 6.71

O = 53.29/16 = 3.33

Dividing through by 3.33, we have:

Empirical formula = CH2O

Mass of CH2O = 12 + 2 + 16 = 30

n = 60.05/30 = 2

Thus, the molecular formula would be [CH2O]2 or C2H4O2

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