Question

A gas sample is found to contain 39.10% carbon,7.y7% hydrogen, 26.11%oxygen, 16.82% Phosphorus and 10.30% flourine. If the molecular mass is 184.1g/mol. What is the molecular formula?

Answer 1

Explanation:

Assuming 100 g of unknown, you have...

39.10 g C x 1 mol/12 g = 3.26 mole C

7.67 g H x 1 mol/1 g = 7.67 mol H

26.11 g O x 1 mol/16 g = 1.63 mol O

16.82 g P x 1 mol/31 g = 0.543 mol P

10.30 g F x 1 mol/19 g = 0.542

Dividing all by 0.542 we get

3.26/0.542 = 6 mol C

7.67/0.542 = 14 mol H

1.63/0.542 = 3 mol O

0.543/0.542 = 1 mol P

0.542/0.542 = 1 mol F

Empirical formula = C6H14PO3F. This has a molar mass of 72 + 14 + 31 + 48 + 19 = 184

Since the unknown has a molar mass of 184, the molecular formula is the same as the empirical formula