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3 years ago

Please with this problem

Chemistry

1 answer:

olasank [31]3 years ago

7 0

<h3>Answer:</h3>

Empirical Formula = C₃H₈O₃

Molecular Formula = C₃H₈O₃

<h3>Solution:</h3>

Data Given:

Mass of Sample = 9.2 g

Mass of Carbon = 3.6 g

Mass of Hydrogen = 0.8 g

Mass of Oxygen = 9.2 - (3.6 + 0.8) = 4.8 g

Step 1: Calculate Moles of each Element;

Moles of C = Mass of C ÷ At.Mass of C

Moles of C = 3.6 ÷ 12.01

Moles of C = 0.2997 mol

Moles of H = Mass of H ÷ At.Mass of H

Moles of H = 0.8 ÷ 1.01

Moles of H = 0.7920 mol

Moles of O = Mass of O ÷ At.Mass of O

Moles of O = 4.8 ÷ 16.0

Moles of O = 0.3000 mol

Step 2: Find out mole ratio and simplify it;

C H O

0.2997 0.7920 0.3000

0.2997/0.2997 0.7920/0.2997 0.3000/0.2997

1 2.64 1.001

Multiply by 3,

3 7.92 ≈ 8 3

Hence, Empirical Formula = C₃H₈O₃

Step 3: Calculating Molecular Formula:

Molecular formula is calculated by using following formula,

Molecular Formula = n × Empirical Formula ---- (1)

Also, n is given as,

n = Molecular Weight / Empirical Formula Weight

Molecular Weight = 92 g.mol⁻¹

Empirical Formula Weight = 12 (C₃) + 1.01 (H₈) + 16 (O₃) = 92.08 g.mol⁻¹

So,

n = 92 g.mol⁻¹ ÷ 92 g.mol⁻¹

n = 1

Putting Empirical Formula and value of "n" in equation 1,

Molecular Formula = 1 × C₃H₈O₃

Molecular Formula = C₃H₈O₃

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