What is the molecular formula of a compound if its empirical formula is CFBrOCFBrO and its molar mass is 381.01 g/mol?

1 Answer
May 24, 2016

(CFBrO)(CFBrO)xx× 33 = ??

Explanation:

Molar mass of compound (381.01381.01 gg/molmol) has been provided for you.

Now you need to calculate the molar mass of CFBrOCFBrO.
The molar mass of:
C = 12.0C=12.0 gg/molmol
F = 19.0F=19.0 gg/molmol
Br = 79.9Br=79.9 gg/molmol
O = 16.0O=16.0 gg/molmol

The molar mass of empirical formula, therefore (CFBrOCFBrO) is:
12.0 + 19.0 + 79.9 + 16.012.0+19.0+79.9+16.0 = 126.9126.9 gg/molmol.

In order to get the molecular formula of CFBrOCFBrO,
divide molar mass of compound by molar mass of empirical formula, like this: "molar mass of compound"/" molar mass of empirical formula"molar mass of compound molar mass of empirical formula = "whole number"whole number

i.e. 381.01/126.9381.01126.9 = 3.003.00
Rounded to nearest whole number

Now, multiply the whole number by the Empirical Formula.
Like this (CFBrO)(CFBrO)xx× 33 = C_3F_3Br_3O_3C3F3Br3O3.

Therefore your molecular formula is C_3F_3Br_3O_3C3F3Br3O3.