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

Answer:

The answer to your question is    CH₄O

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