Q13 V20.0438 mol of an organic compound has a mass of 1.84 g. The compound is likely to be an:

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The question provides: 0.0438 mol of an organic compound has a mass of 1.84 g, and asks what class the compound is likely to belong to.

First, compute the molar mass: mass / moles = 1.84 g / 0.0438 mol ≈ 41.96 g/mol, which is about 42 g/mol.

Option a: alkene. General formula for many alkenes is CnH2n. For n = 3, you get C3H6 with a molar mass of 3(12.01) + 6(1.008) ≈ 36.03 + 6.05 ≈ 42.08 g/mol. This matches the calculated ~42 g/mol, making an alkene a plausible identification.

Option b: amine. Amines have nitrogen in addition to carbon and hydrogen, with typical molar masses depending on substituents. A simple small amine like propylamine (C3H9N) has a molar mass around 57 g/mol, which is notably higher than 42 g/mol, making this a less likely match.

Option c: aromatic. Aromatic hydrocarbons like benzene (C6H6) have a molar mass of about 78 g/mol. Other substituents could vary, but achieving a molar mass near 42 g/mol with an aromatic framework is unlikely, since the smallest common aromatic unit is already around 78 g/mol.

Option d: alkane. Alkanes follow CnH2n+2. The smallest straightforward match near 42 g/mol would be C3H8 (propane) with a molar mass of about 44 g/mol. While 44 g/mol is close to 42 g/mol, it is still a bit higher than our calculated value, and the exact match to 42 g/mol aligns better with C3H6 (an alkene) than with an alkane.

In summary, the calculated molar mass points to a formula like C3H6, which is characteristic of an alkene, whereas the other classes would generally have notably different molar masses or structures that don’t fit the data as closely.

MUF0042 Chemistry Unit 2 - Semester 2, 2025

Q13 V20.0438 mol of an organic compound has a mass of 1.84 g. The compound is likely to be an:

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