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What is the empirical formula if you have 134g Fe, 76.9g S and 115g O?

The answer to your question is  FeSO₃

Explanation:

Data

Fe = 134 g

S = 76.9 g

O = 115 g

Process

1.- Convert the grams to moles for each element

For Fe

                        56 g of Fe 1 mol

                       134 g of Fe  x

                        x = (134 x 1) / 56

                        x =  134/56

                        x = 2.39 moles

For S

                        32 g of S 1 mol

                        76.9 g of S  x

                         x = (76.8 x  1) / 32

                         x = 2.4 moles

For O

                        16 g of O 1 mol

                       115 g of O  x

                         x = (115 x 1)/ 16

                         x = 7.18 moles

2.- Divide by the lowest number of moles

Fe    = 2.39/2.39 = 1

S      = 2.4/2.39 = 1

O     = 7.18/2.39  = 3

3.- Write the empirical formula

                            FeSO₃

he changes his ways

Explanation: