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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.