Question

What is the molecular formula of a compound, whose empirical formula is \(\mathrm{CH}_{2} \mathrm{O}\) and molecular weight is 90 ? (a) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{3}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}\)

Short answer

The molecular formula of the compound is \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \), which corresponds to option (c).

Step-by-step solution

Determine the Empirical Formula Mass

Calculate the molar mass of the empirical formula \( \mathrm{CH}_{2} \mathrm{O} \). \( \mathrm{C} \) has an atomic mass of 12, \( \mathrm{H} \) is 1, and \( \mathrm{O} \) is 16. Thus, the empirical formula mass is \( 12 + (2 \times 1) + 16 = 30 \).

Determine the Multiplication Factor

Divide the molar mass of the compound (90) by the empirical formula mass (30): \( \frac{90}{30} = 3 \). Thus, the multiplication factor is 3.

Calculate the Molecular Formula

Multiply each subscript in the empirical formula \( \mathrm{CH}_{2} \mathrm{O} \) by 3 to get the molecular formula: \( \left(\mathrm{C}_{1}\right)_{3} \left(\mathrm{H}_{2}\right)_{3} \left(\mathrm{O}_{1}\right)_{3} = \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \).

Identify the Correct Molecular Formula

Among the options provided, (c) \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \) matches the calculated molecular formula.

Key concepts

Empirical Formula

The empirical formula is a good starting point for understanding the composition of a compound. It represents the simplest whole-number ratio of the elements present in a compound. For a compound with the empirical formula \( \text{CH}_2\text{O} \), we can see that for every carbon atom, there are two hydrogen atoms and one oxygen atom. This formula does not give us the actual number of atoms within a molecule, but it does provide a simplified version of its composition. Knowing how to interpret empirical formulas is essential when progressing to deduce the molecular formula of a compound.

Molar Mass Calculation

To move from an empirical formula to a molecular formula, we first calculate the empirical formula mass. This process involves adding up the atomic masses of all the atoms in the empirical formula. Let's break it down for \( \text{CH}_2\text{O} \):

• The atomic mass of Carbon (\( \mathrm{C} \)) is 12.
• Hydrogen (\( \mathrm{H} \)), which is found as \( \mathrm{H}_2 \), contributes \( 2 \times 1 = 2 \).
• Oxygen (\( \mathrm{O} \)) has a mass of 16.

Adding these together, we find that the empirical formula mass of \( \text{CH}_2\text{O} \) is 30. This is a critical step, as it sets the stage for identifying how many times this empirical unit is repeated in the molecular structure of the compound.

Multiplication Factor

The multiplication factor helps us understand how many times the empirical formula fits into the molecular formula. To find this factor, we divide the given molecular weight of the compound by the empirical formula mass. In this problem, dividing 90 (the molecular weight) by 30 (the empirical mass) provides us with a multiplication factor of 3. This multiplication factor indicates that the entire empirical formula is repeated three times in the actual molecular structure. By understanding this recurring unit, you have the key to discovering the molecular formula.

Step-by-Step Solution

Applying the multiplication factor, we can determine the molecular formula. With our example of \( \text{CH}_2\text{O} \) and a multiplication factor of 3, we multiply all subscripts in the empirical formula by this factor:

• Carbon goes from \( \mathrm{C}_1 \) to \( \mathrm{C}_3 \).
• Hydrogen goes from \( \mathrm{H}_2 \) to \( \mathrm{H}_6 \).
• Oxygen goes from \( \mathrm{O}_1 \) to \( \mathrm{O}_3 \).

Assembling these results, we derive \( \text{C}_3\text{H}_6\text{O}_3 \) as the molecular formula. By linking each step methodically, you clarify how empirical data transforms into a complete picture of molecular composition, verifying that option (c) is indeed the correct answer.