Question

11) A 1.357 gram sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. The combustion produced 1.989 grams of carbon dioxide and 0.8143g of water .

A. Find the empirical formula of the compound
B.if the molar mass of the is found to be 60.06 g/mol what is the molecular formula

Answer 1

The answer to your question is below

Step-by-step explanation:

Data

mass of the sample = 1.357 g

mass of carbon dioxide = 1.989 g

mass of water = 0..8143 g

Process

-Moles of Carbon

Molar mass of CO₂ = 44 g

44 g of CO₂ -------------- 12 g of C

1.989 g of CO₂ --------- x

x = (1.989 x 12) / 44

x = 0.542 g of C

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

0.542 g of C ------------- x

x = (0.542 x 1) / 12

x = 0.0452 moles

-Moles of Hydrogen

Molar mass of water = 18 g

18 g of H₂O --------------- 2 g of H

0.8143 g of H₂O --------- x

x = (0.8143 x 2) / 18

x = 0.0904 g of H

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

0.0904 g of H --------- x

x = (0.0904 x 1) / 1

x = 0.0904 moles of H

-Moles of Oxygen

Mass of Oxygen = 1.357 - 0.542 - 0.0904

= 0.7246 g

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

0.7246 g of O --------- x

x = (0.7246 x 1) / 16

x = 0.04528

2.- Divide by the lowest number of moles

Carbon = 0.0452/0.0452 = 1

Hydrogen = 0.0904/0.0452 = 2

Oxygen = 0.0452 /0.0452 = 1

3.- Write the empirical formula

CH₂O

4.- Find the molecular formula

-Molar mass CH₂O = 12 + 2 + 16

= 30 g

-Divide the molecular mass by the molar mass 60/30 = 2

-Write the molecular formula

2(CH₂O) = C₂H₄O₂