Question

A compound is 37.48% C, 12.58% H, and 49.93% O by mass.

What is the empirical formula of the compound? CH4O, C3HO4, C2H8O2, C4HO4

Answer 1

The answer to your question is CH₄O

Step-by-step explanation:

Data

Carbon = 37.48 %

Hydrogen = 12.58 %

Oxygen = 49.93%

Process

1.- Express the percent as grams

Carbon = 37.48 g

Hydrogen = 12.58 g

Oxygen = 49.93 g

2.- Convert the grams to moles

12 g of C ---------------- 1 mol

37.48 g ------------------ x

x = (37.48 x 1) / 12

x = 3.12 moles

1 g of H ------------------- 1 mol

12.58 g ------------------- x

x = 12.58 moles

16 g of O ----------------- 1 mol

49.93 g ------------------- x

x = (49.93 x 1) / 16

x = 3.12 moles

3.- Divide by the lowest number of moles

Carbon 3.12/3.12 = 1

Hydrogen 12.58/ 3.12 = 4.0

Oxygen 3.12 / 3.12 = 1

4.- Write the empirical formula

CH₄O