ChemistryCBSE Class 11 · Some Basic Concepts of Chemistry

Mole Concept, Molar Mass and Formula Calculations Class 11

Updated 17 October 2026 · Fact-checked

One mole contains 6.022 × 10²³ entities (Avogadro constant) and has a mass equal to the molar mass in grams. Moles = mass/molar mass = number/N_A. The empirical formula is found from percentage (or mass) composition by dividing by atomic masses and then by the smallest ratio; the molecular formula is n times the empirical formula, with n = molar mass/empirical formula mass.

Moles are the middle step: divide mass by M to get moles, multiply moles by N_A to count particles; 18 g water is 1 mole.

Illustration drawn for QuestionBuddy.

Understand Mole Concept, Molar Mass and Formula Calculations

The mole. A mole is the amount of substance containing as many elementary entities (atoms, molecules, ions) as there are atoms in 12 g of carbon-12; this number, the Avogadro constant, is $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ . The molar mass (g mol⁻¹) is numerically equal to the atomic, molecular or formula mass in u.
$ $n = \frac{m}{M} = \frac{N}{N _{A}} .$ $
Number of atoms of an element = $n \times$ (atoms per formula unit) $\times N_{A}$ .

Average atomic mass. For isotopes of masses $m_{i}$ and fractional abundances $f_{i}$ , $\overset{m}{ˉ} = \sum f_{i} m_{i}$ .

Percentage composition. Mass % of an element $= \frac{mass of the element in one mole}{molar mass} \times 100$ .

Empirical and molecular formula.
1. Write the percentage (or mass) of each element.
2. Divide by the atomic mass to get moles.
3. Divide by the smallest number to get the simplest ratio; multiply by 2, 3, … if the ratios are like 1.5 or 2.33.
4. The molecular formula is $(empirical formula)_{n}$ with $n = M / M_{e m p i r i c a l}$ .
For a combustion analysis of a compound of C, H and O, mass of C $= \frac{12}{44} \times m_{C O_{2}}$ , mass of H $= \frac{2}{18} \times m_{H_{2} O}$ , and the oxygen is found by difference.

Hydrates. A formula such as Na₂CO₃·xH₂O: $x = \frac{moles of water lost}{moles of anhydrous salt}$ .

Significant figures. Round the final answer to the least number of significant figures in the data.

Mole Concept, Molar Mass and Formula Calculations: formulas and key facts

Moles: $n = \frac{m}{M} = \frac{N}{N _{A}}$; $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ .
Average atomic mass: $\overset{m}{ˉ} = \sum f_{i} m_{i}$; Fractional abundances.
Mass percentage: $% element = \frac{mass in one mole}{M} \times 100$; Percentage composition.
Molecular formula: $n = \frac{M}{M _{e m p i r i c a l}}$; Multiplier.
C and H from combustion: $m_{C} = \frac{12}{44} m_{C O_{2}}, m_{H} = \frac{2}{18} m_{H_{2} O}$; Oxygen by difference.
Hydrate water of crystallisation: $x = \frac{n ( H _{2} O )}{n ( anhydrous )}$; Heating to constant mass.

How to solve Mole Concept, Molar Mass and Formula Calculations questions

Convert every mass to moles first; ratios of moles give formulas, and moles times N_A gives numbers of particles.

1Find the molar mass from the formula using atomic masses.
2Convert the given mass (or number) to moles.
3For formulas, divide each element's mass (or %) by its atomic mass and divide by the smallest result.
4Scale the ratio to whole numbers and write the empirical formula.
5Find n = M/M_emp and write the molecular formula.
6Report the result to the correct significant figures with units.

Mole Concept, Molar Mass and Formula Calculations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Salt in a pinch

Suppose a pinch of table salt weighs 0.585 g (NaCl, M = 58.5 g/mol).

Find the number of formula units and the number of ions in the pinch.

Try it first, then show the solutionSolution

n = 0.585/58.5 = 0.0100 mol.
Formula units = 0.0100 × 6.022 × 10²³ = 6.02 × 10²¹; ions = 2 × 6.02 × 10²¹ = 1.20 × 10²².

Answer: 6.02 × 10²¹ formula units; 1.20 × 10²² ions.

Memory hook: Each formula unit gives two ions.

A sugar cube

Suppose a sugar cube is 3.42 g of sucrose, C₁₂H₂₂O₁₁ (M = 342 g/mol).

Find the moles of sucrose and the number of carbon atoms.

Try it first, then show the solutionSolution

n = 3.42/342 = 0.0100 mol.
Carbon atoms = 0.0100 × 12 × 6.022 × 10²³ = 7.23 × 10²².

Answer: 0.0100 mol; 7.23 × 10²² C atoms.

Memory hook: Multiply by the atoms per molecule.

Tricks to remember Mole Concept, Molar Mass and Formula Calculations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Mass ÷ M, then × N_A
Moles first, then particles.
Use it when: Counting particles.
Percent to moles
Treat the percentage as grams in 100 g.
Use it when: Empirical formula.
C = 12/44, H = 2/18
Fractions of CO₂ and H₂O that are C and H.
Use it when: Combustion analysis.
n = M/M_emp
Whole-number multiplier for molecular formula.
Use it when: Molecular formula.

Mole Concept, Molar Mass and Formula Calculations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Moles, molecules and atoms · 2 marks · Very short answerBoard level

Find the number of moles and molecules in 18 g of glucose, C₆H₁₂O₆ (H = 1, C = 12, O = 16), and the number of hydrogen atoms in it.

Try it first, then show the solutionSolution

M = 180 g/mol; n = 18/180 = 0.10 mol; molecules = 0.10 × 6.022 × 10²³ = 6.02 × 10²². (1)
H atoms = 12 × 6.02 × 10²² = 7.23 × 10²³. (1)

Answer: 0.10 mol; 6.02 × 10²² molecules; 7.23 × 10²³ H atoms.

Example 2: Percentage composition · 3 marks · Short answerBoard level

Find the percentage composition of calcium carbonate (Ca = 40, C = 12, O = 16) and the mass of calcium in 1.00 kg of pure CaCO₃.

Try it first, then show the solutionSolution

M = 40 + 12 + 48 = 100 g/mol. (1)
Ca = 40%, C = 12%, O = 48%. (1)
Mass of Ca = 0.40 × 1000 g = 400 g. (1)

Answer: Ca 40%, C 12%, O 48%; 400 g of Ca.

Example 3: Combustion analysis · 3 marks · Short answerHarder board level

0.46 g of an organic compound containing C, H and O gives 0.88 g of CO₂ and 0.54 g of H₂O on complete combustion. Find its empirical formula; if its molar mass is 46 g/mol, find the molecular formula.

Try it first, then show the solutionSolution

C = (12/44) × 0.88 = 0.24 g; H = (2/18) × 0.54 = 0.06 g; O = 0.46 − 0.30 = 0.16 g. (1)
Moles: C 0.020, H 0.060, O 0.010; ratio 2 : 6 : 1, so empirical formula C₂H₆O. (1)
Empirical mass = 46, so n = 1 and the molecular formula is C₂H₆O. (1)

Answer: C₂H₆O.

Example 4: How big is an atom of gold? · 4 marks · Long answerHarder board level

A gold ring has a mass of 3.94 g (Au = 197 g/mol) and gold has density 19.3 g/cm³. Find (a) the number of moles and atoms in the ring, (b) the volume occupied per atom, and (c) the edge of a cube of this volume.

Try it first, then show the solutionSolution

(a) n = 3.94/197 = 0.0200 mol; atoms = 0.0200 × 6.022 × 10²³ = 1.20 × 10²². (1)
(b) Volume of ring = 3.94/19.3 = 0.204 cm³. (1)
Volume per atom = 0.204/1.20 × 10²² = 1.70 × 10⁻²³ cm³. (1)
(c) Edge = (1.70 × 10⁻²³)^(1/3) cm = 2.57 × 10⁻⁸ cm = 0.257 nm. (1)

Answer: (a) 0.0200 mol, 1.20 × 10²² atoms; (b) 1.70 × 10⁻²³ cm³; (c) 0.257 nm.

Example 5: Isotopes and average mass · 4 marks · Long answerHOTS / competency

An element X has three isotopes: 78.99% of mass 23.985 u, 10.00% of mass 24.986 u and 11.01% of mass 25.983 u. (a) Find its average atomic mass. (b) Find the percentage of X in its chloride XCl₂ (Cl = 35.5). (c) How many atoms of the heaviest isotope are present in 1.00 g of X?

Try it first, then show the solutionSolution

(a) 0.7899 × 23.985 + 0.1000 × 24.986 + 0.1101 × 25.983 = 18.945 + 2.499 + 2.861 = 24.30 u. (1)
(b) M(XCl₂) = 24.30 + 71.0 = 95.30; % X = 24.30/95.30 = 25.5%. (1)
(c) n = 1.00/24.30 = 0.04115 mol; atoms = 0.04115 × 6.022 × 10²³ = 2.478 × 10²². (1)
Heaviest isotope: 0.1101 × 2.478 × 10²² = 2.73 × 10²¹ atoms. (1)

Answer: (a) 24.30 u; (b) 25.5%; (c) 2.73 × 10²¹ atoms.

Where students lose marks in Mole Concept, Molar Mass and Formula Calculations

The slips examiners see most often, and what to write instead.

Mistake: Using mass instead of moles for the formula ratio.
Skipping the division by atomic mass.
Do this: Divide each mass by its atomic mass first.
Usually costs: The formula marks.
Mistake: Forgetting atoms per molecule when counting atoms.
Counting molecules only.
Do this: Multiply by the number of atoms of that element in the formula.
Usually costs: 1 mark.
Mistake: Rounding the ratio 1.5 to 2.
Not scaling.
Do this: Multiply all ratios by 2 to get whole numbers.
Usually costs: The formula mark.
Mistake: Quoting too many significant figures.
Calculator digits.
Do this: Round to the least number of significant figures in the data.
Usually costs: The final mark.

Exam tips

Write the molar mass computation on its own line.
Show the table: element, mass or %, moles, ratio, whole-number ratio.
State N_A and the atomic masses you use.
Give the formula in the correct form (subscripts) and units on all quantities.
Check that percentages add to 100.

Practice questions on Mole Concept, Molar Mass and Formula Calculations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The mass of 0.25 mol of sodium hydroxide (Na = 23, O = 16, H = 1) is
(a) 5 g (b) 10 g (c) 20 g (d) 40 g
Hint
M(NaOH) = 40 g/mol.
Answer
(b) 10 g.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): One mole of any substance contains the same number of entities.
Reason (R): A mole is defined as the amount containing as many entities as there are atoms in 12 g of carbon-12.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Definition of the mole.
Answer
(a).
Q3 · 2 marks · Very short answerHarder board level
Find the number of ions in 0.50 mol of aluminium sulphate, Al₂(SO₄)₃, and its molar mass (Al = 27, S = 32, O = 16).
Hint
Each formula unit gives 5 ions.
Answer
M = 54 + 96 + 192 = 342 g/mol; ions = 0.5 × 5 × 6.022 × 10²³ = 1.51 × 10²⁴.
Q4 · 3 marks · Short answerHarder board level
Urea, CO(NH₂)₂ (M = 60 g/mol), and ammonium nitrate, NH₄NO₃ (M = 80 g/mol), are used as nitrogen fertilisers. Find the percentage of nitrogen in each and the mass of each needed to supply 28 kg of nitrogen.
Hint
N per mole: 28 g in each.
Answer
Urea: 28/60 = 46.7% N; mass = 28/0.467 = 60 kg. NH₄NO₃: 28/80 = 35% N; mass = 28/0.35 = 80 kg.
Q5 · 4 marks · Long answerHOTS / competency
2.86 g of a crystalline sodium carbonate, Na₂CO₃·xH₂O (Na = 23, C = 12, O = 16), loses water on heating and leaves 1.06 g of the anhydrous salt. Find x, the percentage of water in the crystals, and the number of oxygen atoms in the 2.86 g sample.
Hint
Moles of water ÷ moles of salt.
Answer
Water lost = 1.80 g = 0.100 mol; Na₂CO₃ = 1.06/106 = 0.0100 mol; x = 10. Water = 1.80/2.86 = 62.9%. Oxygen atoms per formula unit = 3 + 10 = 13, so atoms = 0.0100 × 13 × 6.022 × 10²³ = 7.83 × 10²².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Avogadro constant?Show answer
6.022 × 10²³ mol⁻¹.
2.Moles formula?Show answer
Mass/molar mass.
3.Empirical formula?Show answer
Simplest whole-number ratio of atoms.
4.n for molecular formula?Show answer
M / M(empirical).
5.C in CO₂ fraction?Show answer
12/44.
6.H in H₂O fraction?Show answer
2/18.

Mole Concept, Molar Mass and Formula Calculations: frequently asked questions

Is a mole a mass or a number?

It is an amount of substance equal to a number of entities (6.022 × 10²³), with its own SI base unit.

Is the molar mass the same as the molecular mass?

Numerically yes; the molar mass is in g/mol and the molecular mass in u.

Why divide by the smallest number?

To get the simplest whole-number ratio of moles of atoms.

What if the molecular formula equals the empirical formula?

Then n = 1, as for water or ethanol.

How is a hydrate different from an anhydrous salt?

A hydrate contains water of crystallisation in its structure.