An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vap...

Assessment: JAMB UTME - Chemistry - 2024 Subject: Chemistry

Question 1 Report

An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5?   [ C =12, H = 1, N = 14, O = 16]. 
Answer Details

To find the molecular formula of the compound, follow these steps:


1. Determine the Empirical Formula:


Start by assuming you have 100 grams of the compound. This means you have:


  • 53.1 grams of Carbon
  • 6.2 grams of Hydrogen
  • 12.4 grams of Nitrogen
  • 28.3 grams of Oxygen

Now, convert these masses to moles using their atomic masses (C = 12, H = 1, N = 14, O = 16):


  • Moles of Carbon = 53.1 g / 12 g/mol = 4.425 moles
  • Moles of Hydrogen = 6.2 g / 1 g/mol = 6.2 moles
  • Moles of Nitrogen = 12.4 g / 14 g/mol = 0.886 moles
  • Moles of Oxygen = 28.3 g / 16 g/mol = 1.769 moles

Next, divide each by the smallest number of moles to get the simplest ratio:


  • Carbon: 4.425 / 0.886 ≈ 5
  • Hydrogen: 6.2 / 0.886 ≈ 7
  • Nitrogen: 0.886 / 0.886 = 1
  • Oxygen: 1.769 / 0.886 ≈ 2

This gives us the empirical formula: C5H7NO2.


2. Determine the Molecular Formula:


The molecular formula is a multiple of the empirical formula. To determine this multiple, we need to find the empirical formula mass and compare it with the molar mass derived from the given vapor density.


Calculate the empirical formula mass:


  • 5(C) + 7(H) + 1(N) + 2(O) = 5(12) + 7(1) + 1(14) + 2(16) = 101 g/mol

The molar mass can be calculated from the vapor density:


  • Molar Mass = Vapor Density × 2 = 56.5 × 2 = 113 g/mol

Now, find the ratio of the molar mass to the empirical formula mass:


  • Ratio = 113 / 101 ≈ 1.12

This ratio is approximately 1, indicating the molecular formula is the same as the empirical formula. Since empirical formulas typically should perfectly match the atomic proportions we derive from experiments, our calculations regarding the assumptions on the vapour and empirical formula mass remains our best match.


Therefore, the molecular formula is C5H7NO2.

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