Here's how to do it:
1. Determine the empirical formula:
* Calculate the moles of each element:
- Molar mass of N = 14.01 g/mol
- Molar mass of H = 1.01 g/mol
- Molar mass of Cl = 35.45 g/mol
- Moles of N = (51.5 g NH2Cl) * (14.01 g N/mol NH2Cl) = 3.61 mol N
- Moles of H = (51.5 g NH2Cl) * (2 * 1.01 g H/mol NH2Cl) = 103.1 g H
- Moles of Cl = (51.5 g NH2Cl) * (35.45 g Cl/mol NH2Cl) = 18.25 mol Cl
* Find the simplest whole-number ratio of moles:
- Divide each number of moles by the smallest number of moles (3.61 mol N):
- N: 3.61 mol / 3.61 mol = 1
- H: 103.1 mol / 3.61 mol ≈ 28.5 ≈ 29
- Cl: 18.25 mol / 3.61 mol ≈ 5.05 ≈ 5
* The empirical formula is NH29Cl5
2. Determine the molecular formula:
* Calculate the molar mass of the empirical formula:
- Molar mass of NH29Cl5 ≈ (14.01 + 29 * 1.01 + 5 * 35.45) g/mol ≈ 231.5 g/mol
* Divide the given molar mass by the empirical formula molar mass:
- 51.5 g/mol / 231.5 g/mol ≈ 0.22
* Since this value is close to 1/5, multiply the subscripts in the empirical formula by 5 to get the molecular formula:
The molecular formula of NH2Cl is NH145Cl25.
Important Note: It is highly unusual for a compound to have such a large number of hydrogen and chlorine atoms. The provided molar mass (51.5 g) may be incorrect. Please double-check the information provided.
