Question

When 108.4 mg of a compound is completely combusted, 305.3 mg of carbon dioxide and 125.0 mg of water are produced. (a) Determine the empirical formula of the compound. (b) If the molar mass of the compound is 468.8, determine the molecular formula of the compound

Answer 1

Solution (a):-

Step 1:-

Number of moles of CO₂ = Mass ÷ Molar mass

Mass of CO₂ = 305.3 mg = 0.3053 g (given)

Molar mass of CO₂ = (12.01 + 2 × 15.999) = 44 g/mol

Thus, the number of moles of CO₂ = (0.3053g) ÷ (44 g/mol) = 6.94 × 10^-3 moles of CO₂

Since one mole of CO₂ is made up of one mole of C and two moles of O, the 6.94 × 10^-3 moles of CO₂ is formed from 6.94 × 10^-3 moles of C in the sample. Converting 6.94 × 10^-3 moles into g , we have

(6.94 × 10^-3 moles) × (12.01 g /mol) = 0.0833 g of C

Step 2:-

Number of moles of H₂O = Mass ÷ Molar mass

Mass of H₂O = 125.0 mg = 0.125 g. (given)

Molar mass of H₂O = (2 × 1.008 + 15.999) = 18 g/mol

Thus, the number of moles of H₂O = (0.125 moles) ÷ (18 g/mol) = 6.94 × 10^-3 moles of H₂O

Since, one mole of H​​​​​​₂O is made up of two moles of H & one mole of O if we have 6.94 × 10^-3 moles of H₂O , then we have 2 × (6.94 × 10^-3 ) moles = 13.88 × 10^-3 moles of H. Converting the 13 × 10^-3 moles of H into g, we have

(13.88 × 10^-3 moles ) × (1.008 g/mol) = 0.0139 g of H

Step 3:-

Adding the amount of C and H

(0.0833 + 0.0139) g = 0.0973 g

Step 4:-

Amount of O in sample = Total amount of sample - (sum of amount of C and H)

Total amount of sample = 108.4 mg = 0.1084 g

Amount of O in sample = 0.1084 - 0.0973 = 0.0111 g

Number of moles of O = 0.0111 ÷ 15.999 = 0.694 × 10^-3 moles of O

Step 5:-

Overall we have,

6.94 × 10^-3 moles of C

​​​​​​13.88 × 10^-3 moles of H

0.694 × 10^-3 moles of O

Step 6:-

Dividing by the smallest molar amount , we have

C = 10

H = 20

O = 1

Thus, empirical formula of sample = C​​​​​​​​​​​​​​​​​​​​_{10​​​​​​​}H​​​​​​​₂₀O

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Solution (b):-

Empirical mass = (10 × 12.01 + 20 × 1.008 + 15.999 × 1) = 156.26

Molar mass of sample compound = 468.8 g/mol

n = Molar mass / empirical mass = 468.8 / 156.26 = 3

Molecular formula = (Empirical formula) _n

Molecular formula = (C₁₀H₂₀O)₃ = C₃₀H₆₀O₃

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