Question

Need helping finding empirical formula

Answer 1

The empirical formula is C3H6O

Step-by-step 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

Answer 2

The answer to your question is C₃H₆O

Step-by-step 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