Question

A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of carbon dioxide and 21.6 g of water. Determine the empirical formula of the compound.

Answer 1

The answer to your question is C₃H₆O

Step-by-step explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

44 g of CO₂ --------------- 12 g of C

52.8 g --------------- x

x = (52.8 x 12)/44

x = 633.6/44

x = 14.4 g of C

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

14.4 g of C --------------- x

x = (14.4 x 1)/(12)

x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

18 g of H₂O --------------- 2 g of H

21.6 g of H₂O ------------- x

x = (21.6 x 2) / 18

x = 2.4 g of H

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

2.4 g of H ----------------- x

x = (2.4 x 1)/1

x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

= 6.4 g

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

6.4 g of O -------------- x

x = (6.4 x 1)/16

x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

C₃H₆O