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

Question

Luna Chen · Accepted Answer

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

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#.

Noah Aldrich · Answer

undefined The compound's molecular formula is C2F2Br2O2.