The relative molecular mass of an alkanoic acid with the empirical formula of CH$_2$O is [ H = 1, C = 12, O = 16 ]

Analysis

To find the relative molecular mass of an alkanoic acid given its empirical formula, we need to understand the general formula for alkanoic acids (carboxylic acids) and how empirical formulas relate to molecular formulas.

Concept

1. Empirical Formula: This is the simplest whole-number ratio of atoms in a compound. Here, it is $CH_2O$.
2. General Formula of Alkanoic Acids: Alkanoic acids have the general formula $C_nH_{2n}O_2$.
3. Relationship: The molecular formula is always a multiple of the empirical formula ($n \times \text{Empirical Formula}$). We need to find the value of $n$ that fits the general formula of an alkanoic acid.

Step-by-Step Solution

Step 1: Determine the empirical mass. Using the atomic masses provided ($H=1, C=12, O=16$): Mass of $CH_2O = 12 + (2 \times 1) + 16$ Mass of $CH_2O = 12 + 2 + 16 = 30\text{ g/mol}$

Step 2: Relate to the general formula. The empirical formula $CH_2O$ has a ratio of $1:2:1$. The general formula for alkanoic acids is $C_nH_{2n}O_2$. If we multiply the empirical formula $CH_2O$ by $2$, we get: $2 \times (CH_2O) = C_2H_4O_2$

Step 3: Check if $C_2H_4O_2$ is an alkanoic acid. $C_2H_4O_2$ is Ethanoic acid ($CH_3COOH$). This fits the definition of an alkanoic acid.

Step 4: Calculate the molecular mass. Molecular Mass $= 2 \times (\text{Empirical Mass})$ Molecular Mass $= 2 \times 30 = 60\text{ g/mol}$

(Note: If we multiplied by 1, we would get $CH_2O$ (Methanal, an aldehyde, not an acid). If we multiplied by 3, we would get $C_3H_6O_3$, which is not the standard form for a simple alkanoic acid.)

Conclusion

The relative molecular mass of the alkanoic acid (Ethanoic acid) is 60.

Exam Tip: Always remember that for alkanoic acids, the number of Oxygen atoms is always 2. If your empirical formula has only one oxygen, you must multiply the entire formula by 2 to satisfy the requirement for an alkanoic acid ($C_nH_{2n}O_2$).