Question

Question #c88fc

Answer 1

There are four monocarboxylic acids that contain 37.2 % oxygen.

The general formula for a monocarboxylic acid is RCOOH or `"C_nH_(2n+1)COOH`.

The two O atoms have a total mass of 32.00 u.

So the molecular mass of the acid A is

32.00 u O × `(100"u A")/(37.2"u O")` = 86.0 u A

By trial and error, we find that if n = 3, the molecular mass of `C_nH_(2n+1)COOH` or C₃H₇COOH ≈ (36 + 7 + 12 + 32 + 1) u = 88 u

This is 2 u too high. The molecule must contain a double bond in the R group. Each double bond removes two H atoms and reduces the molecular mass by 2 u.

The formula of the acid must be `C_3H_5COOH`. This has a molecular mass of 86.09 u.

Various possibilities are

1. Z-but-2-enoic acid, CH₃CH=CHCOOH (isocrotonic acid)

2. E-but-2-enoic acid, CH₃CH=CHCOOH (crotonic acid)

3. but-3-enoic acid, CH₂=CHCH₂COOH

4. 2-methylpropenoic acid, CH₂=C(CH₃)COOH

Hope this helps.