Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

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Q. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

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Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.

Atoms and molecules are unimaginably tiny. A single water molecule ( $H_2O$ ) weighs about $3 \times 10^{-23}$ grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, $6.022 \times 10^{23}$ molecules, which is called one mole.

Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have $6.022 \times 10^{23}$ molecules of this compound, how much would they weigh on a lab balance?"

Note

The number $6.022 \times 10^{23}$ is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.

The Precise Definition

Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.

For example:

Water ( $H_2O$ ): 2 hydrogen atoms + 1 oxygen atom
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molecular mass of $H_2O$ = $2(1.008) + 16.00 = 18.016$ g/mol

\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass of that element})

How to Calculate It — Step by Step

Let's take glucose, $C_6H_{12}O_6$ , as a worked example.

Step 1: Identify each element and its count

Carbon (C): 6 atoms
Hydrogen (H): 12 atoms
Oxygen (O): 6 atoms

Step 2: Look up atomic masses (from the periodic table)

C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol

Step 3: Multiply and add

\text{Molecular mass} = 6(12.01) + 12(1.008) + 6(16.00)

= 72.06 + 12.096 + 96.00

= 180.156 \text{ g/mol}

Tip

Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.

Why This Matters

Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:

Convert grams to moles: $\text{moles} = \frac{\text{mass in grams}}{\text{molecular mass}}$
Convert moles to grams: $\text{mass} = \text{moles} \times \text{molecular mass}$
Determine the number of molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

Watch out

Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like $CO_2$ = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).

Common Exam Pitfalls

Forgetting to multiply by the subscript. In $H_2SO_4$ , there are 2 hydrogens, not 1.
Using atomic number instead of atomic mass. Atomic number (protons) is not mass.
Rounding too early. Keep 2-3 decimal places until the final answer.
Confusing molecular mass with molecular weight. They mean the same thing — both are in g/mol.

Quick Reference Table

Substance	Formula	Calculation	Molecular Mass (g/mol)
Oxygen gas	$O_2$	$2(16.00)$	32.00
Carbon dioxide	$CO_2$	$12.01 + 2(16.00)$	44.01
Methane	$CH_4$	$12.01 + 4(1.008)$	16.042
Sodium chloride	$NaCl$	$22.99 + 35.45$	58.44

The last one is a formula mass (ionic compound), but the calculation is identical.

The Big Picture

Molecular mass is not a property you measure directly — it's a calculated value from the periodic table. Every molecule of a given compound has the same molecular mass. When you weigh out that many grams, you know exactly how many moles (and therefore how many molecules) you have. That's the foundation of all stoichiometry.

Searches like "molecular mass calculation formula chemistry" and "mole concept class 11 chemistry" are extremely common, since this is one of the very first skills taught in the Some Basic Concepts of Chemistry chapter of the NCERT/CBSE Class 11 curriculum. Molecular mass calculations underpin virtually every stoichiometry question in board exams, JEE Main, and NEET.

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.

1. The Core Question: What is a Mole?

A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains $6.022 \times 10^{23}$ particles (Avogadro's number, $N_A$ ).

Why this number?

It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.

Example: 1 atom of carbon-12 has mass 12 amu.
1 mole of carbon-12 has mass 12 grams.

This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).

2. The Fundamental Relationship

The key formula is:

n = \frac{m}{M}

Where:

$n$ = number of moles
$m$ = mass of substance (in grams)
$M$ = molar mass (in g/mol)

Why does this work?

Think of it as a conversion factor:

If 1 mole of a substance weighs $M$ grams, then $m$ grams contains $\frac{m}{M}$ moles.

Derivation logic:

Molar mass $M$ tells you: "1 mol = $M$ g"
So the conversion factor is $\frac{1 \text{ mol}}{M \text{ g}}$
Multiply mass $m$ by this factor: $m \times \frac{1}{M} = \frac{m}{M}$ moles

3. Connecting to Number of Particles

n = \frac{N}{N_A}

Where:

$N$ = number of particles (atoms, molecules, ions)
$N_A = 6.022 \times 10^{23}$ particles/mol

Why?

1 mole = $N_A$ particles
So $N$ particles = $\frac{N}{N_A}$ moles

Combine both formulas:

\frac{m}{M} = \frac{N}{N_A}

This single equation ties mass, molar mass, number of particles, and Avogadro's number together.

4. The Gas Volume Connection (for gases at STP)

For gases only:

n = \frac{V}{22.4 \text{ L/mol}}

Why 22.4 L?

From the ideal gas law: $PV = nRT$

At STP (Standard Temperature and Pressure: 0°C, 1 atm):

$P = 1$ atm
$T = 273.15$ K
$R = 0.0821$ L·atm/(mol·K)

For $n = 1$ mole:

V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273.15}{1} \approx 22.4 \text{ L}

So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.

5. The Stoichiometry Chain: From One Substance to Another

In a balanced chemical equation like:

aA + bB \rightarrow cC + dD

The coefficients tell you the mole ratio:

\frac{\text{moles of A}}{\text{moles of B}} = \frac{a}{b}

Why this works:

Coefficients represent relative numbers of molecules (or moles of molecules)
If $a$ molecules of A react with $b$ molecules of B, then $a$ moles of A react with $b$ moles of B
The ratio is fixed by the balanced equation

The complete problem-solving chain:

\text{Mass of A} \xrightarrow{\div M_A} \text{Moles of A} \xrightarrow{\times \frac{c}{a}} \text{Moles of C} \xrightarrow{\times M_C} \text{Mass of C}

Each step uses one of the relationships above.

Summary: The Logical Flow

What you know	Formula	Why it works
Mass of substance	$n = m/M$	Molar mass is the conversion factor between grams and moles
Number of particles	$n = N/N_A$	Avogadro's number is the conversion factor between particles and moles
Volume of gas (STP)	$n = V/22.4$	Derived from ideal gas law at standard conditions
Moles of one reactant	$n_C = n_A \times (c/a)$	Balanced equation gives fixed mole ratios

The mole is the universal translator — it converts between mass, particle count, and gas volume, allowing you to move seamlessly through a chemical reaction.

Convert mass percentages to moles, find the simplest whole-number ratio of Fe to O atoms, and reduce it. The empirical formula is $\mathrm{Fe_2O_3}$ .

Why this approach works

An empirical formula tells us the simplest whole-number ratio of atoms in a compound. Mass percentages alone don't reveal this ratio because different elements have different atomic masses—69.9 g of iron contains far fewer atoms than 69.9 g of oxygen. We convert mass to moles (which count particles) using molar masses, then scale to the smallest integers.

Step-by-step solution

1. Assume a 100 g sample

This makes the arithmetic transparent: 69.9% iron means 69.9 g Fe, and 30.1% oxygen means 30.1 g O in our sample.

2. Convert each mass to moles

The molar mass of iron is $M_{\mathrm{Fe}} = 56 \, \mathrm{g/mol}$ , and for oxygen $M_{\mathrm{O}} = 16 \, \mathrm{g/mol}$ .

n_{\mathrm{Fe}} = \frac{69.9}{56} = 1.248 \, \mathrm{mol}

n_{\mathrm{O}} = \frac{30.1}{16} = 1.881 \, \mathrm{mol}

3. Find the mole ratio

Divide both by the smaller number of moles to get the ratio:

\frac{n_{\mathrm{Fe}}}{1.248} : \frac{n_{\mathrm{O}}}{1.248} = 1 : 1.507

4. Convert to whole numbers

The ratio $1 : 1.507$ is close to $1 : 1.5 = 1 : \frac{3}{2}$ . Multiply both sides by 2 to clear the fraction:

2 \times 1 : 2 \times 1.5 = 2 : 3

This tells us there are 2 iron atoms for every 3 oxygen atoms.

Watch out

A common mistake is to round 1.507 to 2 immediately. Always check if the decimal is close to a simple fraction ( $\frac{1}{2}, \frac{1}{3}, \frac{2}{3}$ , etc.) before rounding, or you'll miss formulas like $\mathrm{Fe_2O_3}$ .

5. Write the empirical formula

The simplest whole-number ratio Fe : O = 2 : 3 gives us $\mathrm{Fe_2O_3}$ .

Element	Mass (g)	Molar mass (g/mol)	Moles	Ratio	×2
Fe	69.9	56	1.248	1	2
O	30.1	16	1.881	1.507	3

✓Final answer

\boxed{\mathrm{Fe_2O_3}}

Here are the common mistakes students make when solving this classic empirical formula problem, along with how to avoid each.

1. Using the Wrong Atomic Masses

The Mistake:

Using atomic masses like $Fe = 55$ or $O = 16.0$ when the problem expects precise values (e.g., $Fe = 55.85$ , $O = 16.00$ ). This shifts the mole ratio and can lead to a wrong formula.

How to Avoid:

Always use the standard atomic masses given in your textbook or exam data booklet. For this problem:

Iron: $55.85 \, \text{g/mol}$
Oxygen: $16.00 \, \text{g/mol}$

2. Confusing "Dioxygen" with "Oxygen Atom"

The Mistake:

Treating "30.1% dioxygen" as 30.1% oxygen atoms. Dioxygen ( $O_2$ ) means the mass given is already for $O_2$ molecules, but in the oxide, oxygen is present as atoms.

How to Avoid:

Remember: dioxygen = $O_2$ . The percentage is the mass of oxygen atoms (since the oxide contains O atoms, not $O_2$ molecules). So you directly use 30.1 g of oxygen atoms per 100 g of compound.

3. Dividing by Atomic Mass Instead of Molar Mass

The Mistake:

Dividing the mass of oxygen by $32.00 \, \text{g/mol}$ (molar mass of $O_2$ ) instead of $16.00 \, \text{g/mol}$ (atomic mass of O).

How to Avoid:

Always convert mass of element to moles of atoms, not molecules.

Correct step:

Moles of Fe = $\frac{69.9}{55.85}$
Moles of O = $\frac{30.1}{16.00}$

4. Rounding Mole Ratios Too Early

The Mistake:

Rounding 1.25 to 1 or 1.3 before dividing by the smallest number, leading to a wrong ratio like $FeO$ instead of $Fe_2O_3$ .

How to Avoid:

Keep at least 3 decimal places until the final step. Only round after dividing by the smallest mole value and checking if the ratio is close to a whole number.

Example:

Moles of Fe = $69.9 / 55.85 = 1.251$
Moles of O = $30.1 / 16.00 = 1.881$
Divide by smallest (1.251):
- Fe: $1.251 / 1.251 = 1.00$
- O: $1.881 / 1.251 = 1.504$
Multiply by 2 to get whole numbers: $Fe_2O_3$

5. Forgetting to Multiply to Get Whole Numbers

The Mistake:

Stopping at $Fe_{1}O_{1.5}$ and writing the formula as $FeO_{1.5}$ .

How to Avoid:

Empirical formulas must have whole-number subscripts. If a ratio ends in .5, .33, or .25, multiply all subscripts by the smallest integer that clears the decimal:

$1.5 \rightarrow$ multiply by 2
$1.33 \rightarrow$ multiply by 3
$1.25 \rightarrow$ multiply by 4

6. Writing the Formula Backwards

The Mistake:

Writing $O_3Fe_2$ instead of $Fe_2O_3$ .

How to Avoid:

In empirical formulas, write the metal first, then the non-metal. Standard convention: $Fe_2O_3$ , not $O_3Fe_2$ .

Quick Checklist to Avoid All Mistakes

Step	Do This	Don't Do This
Atomic masses	Use $Fe = 55.85$ , $O = 16.00$	Use $Fe = 56$ , $O = 32$
Mass to moles	Divide by atomic mass	Divide by molecular mass of $O_2$
Ratio	Keep 3+ decimals	Round early
Whole numbers	Multiply if needed	Leave as decimals
Final formula	Metal first	Non-metal first

Final correct answer: $\boxed{Fe_2O_3}$

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Q.Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

The Precise Definition

How to Calculate It — Step by Step

Why This Matters

Common Exam Pitfalls

Quick Reference Table

The Big Picture

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

1. The Core Question: What is a Mole?

2. The Fundamental Relationship

Why does this work?

3. Connecting to Number of Particles

Why?

4. The Gas Volume Connection (for gases at STP)

Why 22.4 L?

5. The Stoichiometry Chain: From One Substance to Another

Why this works:

Summary: The Logical Flow

Why this approach works

Step-by-step solution

Method: Percentage Composition to Empirical Formula

Steps

1. Using the Wrong Atomic Masses

2. Confusing "Dioxygen" with "Oxygen Atom"

3. Dividing by Atomic Mass Instead of Molar Mass

4. Rounding Mole Ratios Too Early

5. Forgetting to Multiply to Get Whole Numbers

6. Writing the Formula Backwards

Quick Checklist to Avoid All Mistakes

Q.Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Why this approach works

Step-by-step solution

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