A compound has a molecular mass of 90. If it contains 40% Carb... | General Science MCQ | Examoogle

General ScienceChemistry: Atoms and Molecules

A compound has a molecular mass of 90. If it contains 40% Carbon, 6.7% Hydrogen, and 53.3% Oxygen, what is its molecular formula?

$CH_{2}O$

$C_{2}H_{4}O_{2}$

$C_{3}H_{6}O_{3}$

$C_{4}H_{8}O_{4}$

Correct Answer

🔬 SC • Science Concept & PrincipleGeneral ScienceChemistry: Atoms and Molecules

Option C

$C_{3}H_{6}O_{3}$

Quick Summary:

The empirical formula represents the simplest whole-number ratio of atoms of elements in a compound. By calculating the mole ratios of C, H, and O from their given percentages and multiplying the resulting empirical formula by the ratio of molecular mass to empirical formula mass, we determine the molecular formula.

🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of the empirical formula as the 'recipe' ratio (e.g., 1 cup flour to 2 cups water), and the molecular formula as the actual amount used to make the total batch based on the total weight of the cake.

🧠 Memory Mnemonic / Shortcut Aid

MOLE: Mass over atomic mass (M / Atomic mass) gives moles, then Divide by smallest for ratio.

🔢 Key Formulas

$\text{Moles} = \frac{\text{Mass in grams}}{\text{Atomic mass}}$

$\text{Molecular Formula} = n \times \text{Empirical Formula}$ , where $n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}$

⚙️ Working Principle

First, assume $100\text{ g}$ of the compound to find the moles of each element: $\text{Moles of C} = \frac{40}{12} \approx 3.33$ , $\text{Moles of H} = \frac{6.7}{1} \approx 6.7$ , and $\text{Moles of O} = \frac{53.3}{16} \approx 3.33$ . Dividing by the smallest value ( $3.33$ ) gives the ratio $1:2:1$ , resulting in an empirical formula of $CH_{2}O$ . Since the empirical mass is $30$ and the given molecular mass is $90$ , the ratio $n = \frac{90}{30} = 3$ indicates the molecular formula is $(CH_{2}O) \times 3 = C_{3}H_{6}O_{3}$ .

📌 Key Points

▸
Empirical formula represents the simplest ratio, while molecular formula represents the actual number of atoms.
▸
The factor $n$ must always be a positive integer.
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Atomic masses used: $C=12, H=1, O=16$ .

✅ Advantages

▸
Provides structural insight for further chemical analysis.

❌ Disadvantages / Limitations

▸
Does not reveal the spatial arrangement of atoms (isomers can have the same molecular formula).

🛠️ Applications / Uses

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Identifying unknown organic compounds in analytical chemistry.
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Determining purity in pharmaceutical formulations.

📄 Additional Information

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Molecular mass is $90\text{ u}$ (e.g., Lactic acid).
▸
Option A ( $CH_{2}O$ ) has a molar mass of $30\text{ u}$ . Option B ( $C_{2}H_{4}O_{2}$ ) has $60\text{ u}$ . Option D ( $C_{4}H_{8}O_{4}$ ) has $120\text{ u}$ .

📊 Diagram / Illustration

✅

C is correct — The molecular formula $C_{3}H_{6}O_{3}$ corresponds to a molecular mass of $90\text{ g/mol}$ and matches the percentage composition provided.

Core Concepts Used

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Stoichiometry Empirical formula Molar mass calculation

💡 EXAM TIP

If the empirical formula mass equals the molecular mass, $n=1$ . Always verify that your final formula adds up to the provided molecular mass to avoid calculation errors.

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