Q.A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Concept: Molecular Mass Calculation from percentage composition

We find the empirical formula from mass ratios, then scale it to match the given molar mass.

Step 1: Convert percentages to moles by dividing by atomic masses.

• Hydrogen: $\frac{4.07}{1} = 4.07$ mol
• Carbon: $\frac{24.27}{12} = 2.02$ mol
• Chlorine: $\frac{71.65}{35.5} = 2.02$ mol

Step 2: Divide by the smallest (2.02) to get the simplest ratio.

• H : C : Cl = $\frac{4.07}{2.02} : \frac{2.02}{2.02} : \frac{2.02}{2.02}$ = 2 : 1 : 1

The empirical formula is $\ce{CH2Cl}$ with mass $= 12 + 2(1) + 35.5 = 49.5$ g/mol. …

The percentage composition tells us how much of each element is present by mass, but chemistry happens in terms of atoms and molecules. The empirical formula captures the simplest whole-number ratio of atoms, while the molecular formula shows the actual number of each atom in one molecule. The bridge between mass and atoms is the mole concept: dividing mass by atomic mass gives moles, which directly count particles.

Finding the Empirical Formula

Assume we have exactly 100 g of the compound. Then the percentages become masses directly:

• Hydrogen: 4.07 g
• Carbon: 24.27 g
• Chlorine: 71.65 g

1. Convert each mass to moles using atomic masses

Using $\mathrm{H} = 1.008$ g/mol, $\mathrm{C} = 12.01$ g/mol, $\mathrm{Cl} = 35.45$ g/mol:

2. Find the simplest whole-number ratio

Divide each mole value by the smallest (2.02):

The ratio is $\mathrm{C : H : Cl} = 1 : 2 : 1$.

Empirical formula: $\mathrm{CH_2Cl}$

Finding the Molecular Formula

The molecular formula is a whole-number multiple of the empirical formula: $(\mathrm{CH_2Cl})_n$. …

Method: Percent Composition → Empirical Formula → Molecular Formula

This is a two-step stoichiometry method used when you have percentage composition and molar mass.

Step 1: Assume 100 g of compound

Percentages become grams directly:

• H: 4.07 g
• C: 24.27 g
• Cl: 71.65 g

Step 2: Convert grams to moles

Use atomic masses (g/mol):

H = 1.008, C = 12.01, Cl = 35.45

Step 3: Find the simplest whole-number ratio

Divide each by the smallest number of moles (2.02):

• H: $\frac{4.04}{2.02} = 2$
• C: $\frac{2.02}{2.02} = 1$
• Cl: $\frac{2.02}{2.02} = 1$

Empirical formula: $\mathbf{CH_2Cl}$

Step 4: Calculate empirical formula mass

Step 5: Find the molecular formula …

Here are the most common mistakes students make on this empirical and molecular formula problem, along with how to avoid each.

1. Forgetting to convert percentages to grams (or assuming 100 g is optional)

The Mistake:

Students try to work directly with percentages (e.g., 4.07% H, 24.27% C, 71.65% Cl) without first assuming a 100 g sample. This leads to confusion in the mole calculation.

How to Avoid:

Always assume 100 g of the compound. Then the percentages become masses directly:

• $4.07 \, \text{g H}$
• $24.27 \, \text{g C}$
• $71.65 \, \text{g Cl}$

This is the standard first step — never skip it.

2. Using wrong atomic masses

The Mistake:

Using rounded atomic masses (e.g., H = 1, C = 12, Cl = 35.5) when the problem expects more precise values, or vice versa.

How to Avoid:

Use the atomic masses given in the problem or the standard values from your textbook/board:

• H = 1.008 g/mol
• C = 12.01 g/mol
• Cl = 35.45 g/mol

For this problem, the molar mass is given as 98.96 g/mol, so use precise masses to match.

3. Dividing by the wrong number in the mole ratio step

The Mistake:

After calculating moles, students divide each mole value by the largest number instead of the smallest number to find the simplest ratio.

Example of error:

If moles are: H = 4.04, C = 2.02, Cl = 2.02, dividing by 4.04 gives 1, 0.5, 0.5 — which is correct only if you then multiply by 2. But dividing by the largest is not the standard method.

How to Avoid:

Always divide each mole value by the smallest mole value among the elements. This directly gives the simplest whole-number ratio.

4. Rounding ratios incorrectly

The Mistake:

Rounding a ratio like 1.98 to 2 is fine, but rounding 1.33 to 1 is wrong — it should be multiplied by 3 to get 4.

How to Avoid:

• If the ratio is within 0.1 of a whole number (e.g., 1.98, 2.02), round it.
• If it’s a fraction like 0.33, 0.5, 0.67, 1.33, 1.5, etc., multiply all ratios by the smallest integer that clears the fraction (e.g., multiply by 3 for 0.33, by 2 for 0.5).

5. Confusing empirical formula mass with molar mass

The Mistake:

Students think the empirical formula mass is the same as the given molar mass, or they forget to compare them.

How to Avoid:

• Calculate the empirical formula mass (sum of atomic masses in the empirical formula).
• Then find $n$ using:

• If $n = 1$, the molecular formula = empirical formula.
• If $n = 2, 3, \dots$, multiply all subscripts in the empirical formula by $n$.

6. Forgetting to check that percentages add to 100% …