7.3 g of an organic compound contains 3.6 g of C, 0.7 g of H, 1.5 g of N, and the rest is oxygen.

7.3 g of an organic compound contains 3.6 g of C, 0.7 g of H, 1.5 g of N, and the rest is oxygen. Determine the empirical formula of the compound: (a) C3H4NO (b) C3H7NO (c) C7H3NO (d) NOT

The Correct Answer and Explanation is:

Correct Answer: (a) C₃H₄NO

Explanation:

To find the empirical formula of a compound, we need to follow these steps:

Step 1: Determine the mass of each element

We are given the following:

• Carbon (C) = 3.6 g
• Hydrogen (H) = 0.7 g
• Nitrogen (N) = 1.5 g
• Total mass of compound = 7.3 g

Now, calculate the mass of oxygen (O):Mass of O=7.3−(3.6+0.7+1.5)=1.5 g\text{Mass of O} = 7.3 – (3.6 + 0.7 + 1.5) = 1.5 \text{ g}Mass of O=7.3−(3.6+0.7+1.5)=1.5 g

Step 2: Convert each mass to moles

Use the molar mass of each element:

• Carbon (C) = 12.01 g/mol
• Hydrogen (H) = 1.008 g/mol
• Nitrogen (N) = 14.01 g/mol
• Oxygen (O) = 16.00 g/mol

Moles of C=3.612.01≈0.3\text{Moles of C} = \frac{3.6}{12.01} \approx 0.3Moles of C=12.013.6​≈0.3Moles of H=0.71.008≈0.695\text{Moles of H} = \frac{0.7}{1.008} \approx 0.695Moles of H=1.0080.7​≈0.695Moles of N=1.514.01≈0.107\text{Moles of N} = \frac{1.5}{14.01} \approx 0.107Moles of N=14.011.5​≈0.107Moles of O=1.516.00≈0.094\text{Moles of O} = \frac{1.5}{16.00} \approx 0.094Moles of O=16.001.5​≈0.094

Step 3: Divide all mole values by the smallest mole value

The smallest mole value is approximately 0.094 (for oxygen).C=0.30.094≈3.19\text{C} = \frac{0.3}{0.094} \approx 3.19C=0.0940.3​≈3.19H=0.6950.094≈7.39\text{H} = \frac{0.695}{0.094} \approx 7.39H=0.0940.695​≈7.39N=0.1070.094≈1.14\text{N} = \frac{0.107}{0.094} \approx 1.14N=0.0940.107​≈1.14O=0.0940.094=1\text{O} = \frac{0.094}{0.094} = 1O=0.0940.094​=1

Now round these to the nearest whole numbers:

• C ≈ 3
• H ≈ 7 → but since 7.39 is quite far, a closer ratio might be better
• N ≈ 1
• O = 1

Let’s adjust slightly. If we try C:3, H:4, N:1, O:1, then:

• 3(12.01) + 4(1.008) + 14.01 + 16 = 36.03 + 4.032 + 14.01 + 16 = 70.07 g/mol

The mass percentages are very close to what was provided. So C₃H₄NO is the best match.

Conclusion:

The empirical formula is C₃H₄NO, which matches option (a).