Is it correct to say: Confusing molecular mass (u) with molar mass (g/mol)?

Numerically same, units different. In mole calculations always use g/mol.

Is it correct to say: Using 6.02 × 10^22 for Avogadro's number?

Correct value: 6.022 × 10^23. Wrong exponent = wrong answer by factor 10.

Is it correct to say: Forgetting to multiply atomic mass by number of atoms per molecule in % composition?

For H_2O, mass of H = 2 × 1 = 2, not 1. Always use the subscript.

Is it correct to say: Rounding empirical ratios too early (e.g. 1.33 → 1)?

1.33 ≈ 4/3 — multiply all ratios by 3 to get whole numbers instead.

Is it correct to say: Applying 22.4 L/mol to non-STP conditions?

22.4 L is valid only at STP (0 °C, 1 atm). At room T use PV = nRT.

Is it correct to say: Mixing up n (moles) and N (particles)?

n is in mol, N is a pure count. N = n × N_A.

Introduction to Matter — Mole Concept and Chemical Calculations

Mole concept, molar mass, Avogadro's number, empirical/molecular formulae, mass % — SSC Science Part 1 foundational calculations

📐 9 formulas✏️ 3 examples🎯 5 practice⚖️ 3-4 marks🏫 Maharashtra State Board📚 Class 10✓ 2025–26 syllabus

💡

Board Exam Tips

→SSC almost always asks one numerical: convert mass ↔ moles ↔ particles. Practise unit tracking.
→State Avogadro's number with correct exponent: 6.022 × 10^23. Marks cut for wrong power.
→Marathi-medium term: mol = 'रेणुसंख्या एकक'. Molar mass = 'मोलर वस्तुमान' (g/mol).
→Always write the unit — mol, g/mol, atoms — even in one-step answers.
→For empirical formula: divide by smallest mole value first, then round to whole numbers.

📐 Formulas(9)

Mole from Mass★ Board fav

n = \frac{m}{M}

Symbol	Meaning
$n$	Number of moles (mol)
$m$	Mass of substance (g)
$M$	Molar mass (g/mol)

Number of Particles★ Board fav

N = n \times N_{A}

Symbol	Meaning
$N_{A}$	Avogadro's number = 6.022 × 10^23 per mol

Avogadro's Number

N_{A} = 6.022 \times 1 0^{23} mol^{- 1}

Molar Volume at STP

V_{m} = 22.4 L/mol at STP

Mass Percentage★ Board fav

% w / w = \frac{m _{so l u t e}}{m _{so l u t i o n}} \times 100

Volume Percentage

% v / v = \frac{V _{so l u t e}}{V _{so l u t i o n}} \times 100

Percentage Composition

% element = \frac{n \times atomic mass}{M _{co m p o u n d}} \times 100

Empirical Formula Ratio

ratio = \frac{% mass}{atomic mass}

Molecular Formula from Empirical

Mol. = n \times Emp., n = \frac{M _{m o l}}{M _{e m p}}

✏️ Solved Examples

1Solved Exampleeasy2 steps

Calculate the number of moles in 18 g of water.

Molar mass of water

M (H_{2} O) = 2 (1) + 16 = 18 g/mol

2Solved Exampleboard4 steps

Find the percentage composition of carbon and hydrogen in methane (CH_4).

Molar mass of CH_4

M = 12 + 4 (1) = 16 g/mol

3Solved ExampleHOTS5 steps

A compound has 40% C, 6.7% H and 53.3% O by mass. Its molar mass is 180 g/mol. Find the molecular formula.

Mole ratio: divide % by atomic mass

C : H : O = \frac{40}{12} : \frac{6.7}{1} : \frac{53.3}{16} = 3.33 : 6.7 : 3.33

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Confusing molecular mass (u) with molar mass (g/mol)
✓Numerically same, units different. In mole calculations always use g/mol.
2
Using 6.02 × 10^22 for Avogadro's number
✓Correct value: 6.022 × 10^23. Wrong exponent = wrong answer by factor 10.
3
Forgetting to multiply atomic mass by number of atoms per molecule in % composition
✓For H_2O, mass of H = 2 × 1 = 2, not 1. Always use the subscript.
4
Rounding empirical ratios too early (e.g. 1.33 → 1)
✓1.33 ≈ 4/3 — multiply all ratios by 3 to get whole numbers instead.
5
Applying 22.4 L/mol to non-STP conditions
✓22.4 L is valid only at STP (0 °C, 1 atm). At room T use PV = nRT.
6
Mixing up n (moles) and N (particles)
✓n is in mol, N is a pure count. N = n × N_A.

🎯 Practice Yourself

🎯Practice Yourself5 questions

Q1
How many moles are there in 44 g of CO_2?
Q2
Number of oxygen atoms in 0.5 mol of O_2?
Q3
Find the % of nitrogen in NH_3.
Q4
Mass of 2 moles of NaOH (M = 40 g/mol)?
Q5
Volume occupied by 0.5 mol of H_2 gas at STP?

📝 Notes

Matter, Moles and Chemical Calculations

The mole is chemistry's counting unit — it lets us jump between the macroscopic world (grams we can weigh) and the microscopic world (atoms and molecules we cannot see). Maharashtra SSC builds on this idea throughout Metallurgy (Ch 8) and Carbon Compounds (Ch 9).

SSC angle

Maharashtra Board Science Part 1 introduces the mole concept as a foundation for stoichiometry:

One 2-mark definition (mole, molar mass, Avogadro's number).
One 3-mark numerical involving mass ↔ mole ↔ particle conversion.
Occasional HOTS on empirical/molecular formula.

The three golden equations

Mass → Moles: n = m/M
Moles → Particles: N = n × N_A
Moles → Volume (gas, STP): V = n × 22.4 L

Percentage composition strategy

Write formula, compute molar mass M.
For each element: %E = (atoms × atomic mass) / M × 100.
Check sum ≈ 100% (tiny rounding OK).

Empirical vs molecular formula

Empirical = simplest whole-number ratio (e.g. CH_2O).
Molecular = actual formula in the molecule (e.g. C_6H_12O_6).
Molecular = n × Empirical, where n = M_molecular / M_empirical.

Board vs CBSE

Both boards teach identical mole-concept content. Maharashtra SSC leans on percentage-composition numericals — practise unit tracking (g, mol, mol/L, L).