9th Class Chemistry Chapter 4 Notes PDF 2026 – Stoichiometry

Definition

A mole is a counting unit for atoms/molecules, equal to Avogadro's number: $6.022 \times 1 0^{23}$ particles.

Molar mass (g/mol) numerically equals an element's atomic mass, or a compound's formula mass — this is what makes mole calculations possible from a simple weighing.

Number of moles $n = \frac{given mass}{molar mass}$

Number of particles $= n \times 6.022 \times 1 0^{23}$

Example 1 — moles from mass

Find the number of moles in $25$ g of $H_{2} S O_{4}$ (molar mass $= 98$ g/mol).

$n = \frac{25}{98}$ .

Answer: $\approx 0.255$ mol

Example 2 — molecules from mass

How many molecules are present in $15$ g of $H_{2} O$ (molar mass $= 18$ g/mol)?

$n = \frac{15}{18} \approx 0.833$ mol.
Molecules $= 0.833 \times 6.022 \times 1 0^{23}$ .

Answer: $\approx 5.02 \times 1 0^{23}$ molecules

Definition

Empirical formula: the simplest whole-number ratio of atoms in a compound. Molecular formula: the actual number of each atom in one molecule — always a whole-number multiple of the empirical formula.

Example 3 — molecular formula from empirical formula

A compound's empirical formula is $C H_{2} O$ (empirical mass $= 30$ ) and its molar mass is $180$ g/mol. Find its molecular formula.

Multiplier $= \frac{180}{30} = 6$ .
Multiply every subscript in $C H_{2} O$ by $6$ .

Answer: $C_{6} H_{12} O_{6}$

Rule

A balanced equation gives the mole ratio between reactants and products — use it to convert moles of one substance to moles of another, then convert to mass using molar mass.

Example 4 — mass from a balanced equation

How many grams of $C O_{2}$ form when $10$ g of $C H_{4}$ reacts with excess $O_{2}$ ? ( $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ ; molar masses: $C H_{4} = 16$ , $C O_{2} = 44$ )

Moles of $C H_{4} = \frac{10}{16} = 0.625$ mol.
Mole ratio $C H_{4} : C O_{2} = 1 : 1$ , so moles of $C O_{2} = 0.625$ mol.
Mass of $C O_{2} = 0.625 \times 44$ .

Answer: $27.5$ g

Quick Recap

● $1$ mole $= 6.022 \times 1 0^{23}$ particles (Avogadro's number)
● $n = \frac{mass}{molar mass}$ ; particles $= n \times 6.022 \times 1 0^{23}$
●Empirical formula: simplest ratio. Molecular formula $=$ empirical formula $\times$ (molar mass $\div$ empirical mass)
●Balanced equation $\to$ mole ratio $\to$ convert to mass with molar mass

9th Class Chemistry Chapter 4 Notes

The Mole and Avogadro's Number

Empirical and Molecular Formulas

Chemical Equations and Mass Calculations

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