A compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% O. Determine the empirical formula and the molecular formula.

Empirical formula: C3H30

Molecular formula: C6H6O2

Explanation:

The empirical formula can be calculated thus:

65.45% C - 65.45g

5.45% H - 5.45g

29.09% O - 29.09g

We convert each mass to mole by dividing by their respective atomic mass:

C = 65.45/12 = 5.45mol

H = 5.45/1 = 5.45mol

O =29.09/16 = 1.82mol

Next, we divide each mole value by the smallest value;

C = 5.45mol/1.82 = 2.99

H = 5.45mol/1.82 = 2.99

O = 1.82mol/1.82 = 1

Approximately, the empirical ratio of C, H and O is 3:3:1 i.e. C3H3O

Molecular formula is calculated thus:

(C3H3O)n = 100g/mol

{12(3) + 1(3) + 16}n = 100

(36+3+16)n = 100

(55)n = 100

n = 100/55

n = 1.82

n = 2

Molecular formula= (C3H3O)2

C6H6O2

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it is a solid

explanation: