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?
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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
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