Question

A gas sample is found to contain 39.10% carbon, 7.67% hydrogen, 26.11% oxygen, 16.82% phosphorus, and 10.30% fluorine. If the molecular mass is 184.1 g/mol, what is the molecular formula?

Answer 1

Answer:

C6H14O3F

Explanation:

The first step is to divide each compound by its molecular weight

Carbon

= 39.10/12

= 3.258

Hydrogen

= 7.67/1

= 7.67

Oxygen

= 26.11/16

= 1.63

Phosphorous

= 16.82/31

= 0.542

Flourine

= 10.30/19

= 0.542

The next step is to divide by the lowes value

3.258/0.542

= 6 mol of C

7.67/0.542

= 14 mol of H

1.63/0.542

= 3 mol of O

0.542/0.542

= 1 mol of P

0.542/0.542

= 1 mol of F

Hence the molecular formula is C6H14O3F