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

1 Answer
May 24, 2016

Answer:

#(CFBrO)##xx# #3# = ??

Explanation:

Molar mass of compound (#381.01# #g#/#mol#) has been provided for you.

Now you need to calculate the molar mass of #CFBrO#.
The molar mass of:
#C = 12.0# #g#/#mol#
#F = 19.0# #g#/#mol#
#Br = 79.9# #g#/#mol#
#O = 16.0# #g#/#mol#

The molar mass of empirical formula, therefore (#CFBrO#) is:
#12.0 + 19.0 + 79.9 + 16.0# = #126.9# #g#/#mol#.

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

i.e. #381.01/126.9# = #3.00#
Rounded to nearest whole number

Now, multiply the whole number by the Empirical Formula.
Like this #(CFBrO)##xx# #3# = #C_3F_3Br_3O_3#.

Therefore your molecular formula is #C_3F_3Br_3O_3#.