Need helping finding empirical formula

The answer to your question is    C₃H₆O

Explanation:

Process

1.- Express the percents to grams

Carbon = 62 g

Hydrogen = 10.4 g

Oxygen = 27.4 g

2.- Convert the grams to moles

                    12 g of C 1 mol

                    62 g of C  x

                       x = (62 x 1)/12

                       x = 5.17 moles of Carbon

                     1 g of H 1 mol

                    10.4 g       x

                       x = (10.4 x 1) / 1

                       x = 10.4 moles of H

                     16 g of O 1 mol

                      27.4 g of O  x

                      x = (27.4 x 1)/16

                      x = 1.71 moles of O

3.- Divide by the lowest number of moles (1.71)

Carbon   = 5.17/1.71 = 3

Hydrogen = 10.4/1.71 = 6

Oxygen = 1.71/1.71 = 1

4.- Write the empirical formula

                     C₃H₆O

The empirical formula is C3H6O

Explanation:

Step 1: Data given

Suppose mass of comound X = 100.0 grams

Compound X contains:

62.0 % carbon = 62.0 grams

10.4 % hydrogen = 10.4 grams

27.5 % oxygen = 27.5 grams

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2: Calculate moles

Mol = mass / molar mass

Mol C = 62.0 grams / 12.01 g/mol

Mol C = 5.16 moles

Mol H = 10.4 grams / 1.01 g/mol

Mol H = 10.3 moles

Mol O = 27.5 grams / 16.00 g/mol

Mol O = 1.72 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 5.16 moles / 1.72 moles = 3

H = 10.3 moles / 1.72 moles =  6

O = 1.72 moles / 1.72 moles = 1

The empirical formula is C3H6O

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