A compound contains 12.8% C, 2.1% H, and 85.1% Br (by mass). Calculate the empirical formula and the molecular formula of this compound, given that the molar mass is 188 g/mol.

The answer to your question is Molecular formula C₂H₄Br₂

                                                    Empirical formula  CH₂Br

Explanation:

Data

Carbon  12.8%

Hydrogen 2.1%

Bromine 85.1%

Empirical formula = ?

Molecular formula = 188 g/mol

Process

1.- Convert the percents to grams

Carbon  12.8 g

Hydrogen 2.1 g

Bromine 85.1 g

2.- Convert the grams to moles

Carbon

                            12 g 1 mol

                             12.8 g  x

                              x = (12.8 x 1) / 12

                              x = 1.07 mol

Hydrogen

                             1 g 1 mol

                             2.1 g  x

                             x = (2.1 x 1)/1

                             x = 2.1 mol

Bromine

                             80 g 1 mol

                             85.1 g x

                             x = (85.1 x 1)/80

                             x = 1.06

3.- Divide by the lowest number of moles

Carbon         1,07/1.06  = 1

Hydrogen      2.1/1.06 = 1.99 = 2

Bromine        1.06/1.06 = 1

4.- Write the empirical formula

                       CH₂Br

5.- Write the molecular formula

Calculate the molar mass of the empirical formula

CH₂Br = 12 + 2 + 80 = 94

Divide the molar mass by the molar mass of the empirical formula

                188/94 = 2

Multiply the empirical formula by two

               C₂H₄Br₂

i'm pretty sure its b i might be wrong though!

particle   p is an electron and particle q is a neutron