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.

Question

Quizplus · Accepted Answer

1) Calculate the empirical formula:
- Convert the mass percentages to moles: 
C: 65.45g / 12.011 g/mol = 5.45 mol
H: 5.45g / 1.008 g/mol = 5.41 mol
O: 29.09g / 15.999 g/mol = 1.82 mol
- Divide by the smallest number of moles to get the simplest ratio:
C: 5.45 mol / 1.82 mol = 3
H: 5.41 mol / 1.82 mol = 3
O: 1.82 mol / 1.82 mol = 1
- The empirical formula is therefore CH3O.

2) Calculate the molecular formula:
- Calculate the molecular mass of the empirical formula: 
(1 x 12.011 g/mol) + (3 x 1.008 g/mol) + (1 x 15.999 g/mol) = 31.024 g/mol
- Divide the given molar mass by the empirical formula mass to determine the molecular formula ratio: 
100 g/mol / 31.024 g/mol = 3.22
- Multiply the empirical formula by this ratio to obtain the molecular formula: 
(1 x 12.011 g/mol) + (3 x 1.008 g/mol) + (1 x 15.999 g/mol) x 3.22 = 97.996 g/mol
The molecular formula is therefore C3H9O3, which simplifies to C3H9O3.

A Compound Has a Molar Mass of 100 G/mol and the Percent