empirical formula?

Question

Gas X is found to be 24% carbon and 76% flourine by mass.
(a) determine the empirical formula of gas X.
(b) if the molar mass of gas X is 200.04 g/mol, determine its molecular formula.

Answer 1

A)
Assume you have a 100g sample. That means 24% of the 100g is carbon and 76% is flourine.

0.24 x 100 = 24g C
0.76 x 100 = 76g F

24g C = 2 mol C
76g F = 4 mol F
2 C : 4 F = 1C:2F

As a chemical formula, 1:2 ratio is written as CF2

B)Molar mass of CF2 = 50 g/mol
200.04/50 = 4

That means there are 4 times as many elements in the molecular formula as there are in the empirical formula.
1C x 4 = 4C
2F x 4 = 8F

Molecular Formula = C4F8

Answer 2

assume you have 100 g sample. So you have 24 g carbon and 76 g fluorine.

Divide the grams by the respective mol. mass of each atom to get moles. Now convert this to a ratio of whole numers, for example. 1 carbon to 4 fluorine. This will be the EF

Calculate the mass of the EF and compare to the 200.04 g/mol. If the EF is a multiple of this, adjust to get the MF.

Answer 3

Empirical formula: CF2
Molecular formula: C4F8

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