10th Class Chemistry Chapter 2 Notes PDF 2026 – Stoichiometry

Definition

Room Temperature and Pressure (RTP) is defined as $2 5^{\circ}$ C and $1$ atmosphere pressure — a fixed reference condition used so gas volumes can be compared meaningfully.

Molar volume is the volume occupied by one mole of any gas at RTP — it's the same for every gas, regardless of its identity, which follows directly from Avogadro's Law: equal volumes of different gases at the same temperature and pressure contain equal numbers of molecules (and so equal moles), meaning one mole of any gas must occupy the same fixed volume under fixed conditions.

At RTP, molar volume $= 24 dm^{3} / mol$ — knowing this lets you convert directly between a gas's volume and the number of moles it contains, without needing its mass or identity at all.

$n = \frac{V}{24}$ at RTP ( $V$ in dm $^{3}$ )

$n = \frac{m}{M}$ ( $m$ = mass, $M$ = molar mass)

Example 1 — volume from moles

Find the volume occupied by $15$ mol of ammonia gas at RTP.

$V = n \times 24 = 15 \times 24$ .

Answer: $V = 360 dm^{3}$

Example 2 — moles from volume

How many moles of methane ( $C H_{4}$ ) are in a cylinder of volume $18 dm^{3}$ at RTP?

$n = \frac{V}{24} = \frac{18}{24}$ .

Answer: $n = 0.75$ mol

Example 3 — moles, volume, and number of molecules

Calculate the number of moles, volume, and number of molecules in $4.0$ g of $C H_{4}$ at RTP ( $M_{C H_{4}} = 16$ g/mol, Avogadro's number $= 6.02 \times 1 0^{23}$ /mol).

$n = \frac{m}{M} = \frac{4.0}{16} = 0.25$ mol.
$V = n \times 24 = 0.25 \times 24 = 6$ dm $^{3}$ .
Molecules $= n \times N_{A} = 0.25 \times 6.02 \times 1 0^{23}$ .

Answer: $n = 0.25$ mol, $V = 6$ dm $^{3}$ , molecules $\approx 1.505 \times 1 0^{23}$

Definition

Concentration describes how much solute is dissolved per unit volume of solution — the chapter uses two units: $mol / dm^{3}$ (molar concentration, moles of solute per dm $^{3}$ of solution) and $g / dm^{3}$ (mass concentration, grams of solute per dm $^{3}$ of solution).

Concentration ( $mol / dm^{3}$ ) $= \frac{n}{V}$ ( $V$ in dm $^{3}$ )

Concentration ( $g / dm^{3}$ ) $= \frac{m}{V}$

Example 4 — molar concentration from mass

Find the molar concentration of a potassium hydroxide (KOH) solution if $2.0$ g is dissolved in $100 cm^{3}$ of solution ( $M_{K O H} = 56$ g/mol).

$n = \frac{m}{M} = \frac{2.0}{56} \approx 0.0357$ mol.
$V = 100 cm^{3} = 0.1 dm^{3}$ .
Concentration $= \frac{n}{V} = \frac{0.0357}{0.1}$ .

Answer: $\approx 0.357 mol / dm^{3}$

Example 5 — mass needed for a given concentration

Find the mass of NaCl needed to make $15 dm^{3}$ of solution at a concentration of $0.7 g / dm^{3}$ .

$m = concentration \times V = 0.7 \times 15$ .

Answer: $m = 10.5$ g

Example 6 — concentration from a titration

$10 cm^{3}$ of $H_{2} S O_{4}$ reacts exactly with $20 cm^{3}$ of NaOH solution of concentration $0.1 mol / dm^{3}$ . Find the concentration of the $H_{2} S O_{4}$ . ( $H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O$ )

Moles NaOH $= 0.1 \times 0.020 = 0.002$ mol.
Mole ratio $H_{2} S O_{4} : N a O H = 1 : 2$ , so moles $H_{2} S O_{4} = \frac{0.002}{2} = 0.001$ mol.
Concentration $= \frac{0.001}{0.010}$ (volume $10 cm^{3} = 0.010 dm^{3}$ ).

Answer: $0.1 mol / dm^{3}$

Definition

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

Two different compounds can share the same empirical formula because empirical formula only captures the simplest ratio, not the actual molecule size — e.g. both glucose ( $C_{6} H_{12} O_{6}$ ) and formaldehyde ( $C H_{2} O$ ) share the empirical formula $C H_{2} O$ , since $6 : 12 : 6$ simplifies to $1 : 2 : 1$ .

Multiplier $n = \frac{molecular mass}{empirical formula mass}$ ; molecular formula $=$ (empirical formula) $\times n$

Example 7 — molecular formula from empirical formula

A compound has molar mass $178$ g/mol and empirical formula $C_{3} H_{5} O_{3}$ (empirical formula mass $= 3 (12) + 5 (1) + 3 (16) = 89$ ). Find its molecular formula.

$n = \frac{178}{89} = 2$ .
Molecular formula $= (C_{3} H_{5} O_{3}) \times 2$ .

Answer: $C_{6} H_{10} O_{6}$

Example 8 — empirical formula from masses

A sulphide of iron contains $1.926$ g sulphur and $2.333$ g iron. Find its empirical formula ( $M_{S} = 32$ , $M_{F e} = 56$ ).

Moles $S = \frac{1.926}{32} \approx 0.0602$ ; moles $F e = \frac{2.333}{56} \approx 0.0417$ .
Divide by the smaller value: $F e : S = \frac{0.0417}{0.0417} : \frac{0.0602}{0.0417} = 1 : 1.44$ .
Multiply both by $2$ to reach whole numbers: $2 : 2.89 \approx 2 : 3$ .

Answer: $F e_{2} S_{3}$

Example 9 — molecular formula from percentage composition

A compound is $40.0%$ C, $6.71%$ H, $53.3%$ O by mass; $0.320$ mol of it weighs $28.8$ g. Find its molecular formula.

In $100$ g: moles $C = \frac{40.0}{12} \approx 3.33$ ; moles $H = \frac{6.71}{1} \approx 6.71$ ; moles $O = \frac{53.3}{16} \approx 3.33$ .
Ratio $C : H : O \approx 1 : 2 : 1 \Rightarrow$ empirical formula $C H_{2} O$ (mass $= 30$ ).
Molar mass $= \frac{28.8}{0.320} = 90$ g/mol.
$n = \frac{90}{30} = 3$ .

Answer: $C_{3} H_{6} O_{3}$

Example 10 — percentage composition

Find the percentage of sodium in $N a H C O_{3}$ ( $M_{N a} = 23$ , $M_{N a H C O_{3}} = 84$ ).

$% Na = \frac{23}{84} \times 100$ .

Answer: $\approx 27.4%$

Definition

The limiting reactant is the reactant that runs out first in a reaction, stopping the reaction and determining the maximum amount of product that can form — every other reactant is left over (in excess) once it's used up.

To identify the limiting reactant: convert each reactant's given amount to moles, then compare the mole ratio actually available to the mole ratio required by the balanced equation — whichever reactant falls short of the required ratio is limiting.

In a reversible reaction, there's no true limiting reactant in the same sense, since the reaction doesn't go to completion in one direction — instead it reaches equilibrium, with all reactants and products still present.

Example 11 — finding the limiting reactant

For $3 H_{2} (g) + N_{2} (g) \to 2 N H_{3} (g)$ , if $12$ g of $H_{2}$ reacts with $64$ g of $N_{2}$ , find the limiting reactant ( $M_{H_{2}} = 2$ , $M_{N_{2}} = 28$ ).

Moles $H_{2} = \frac{12}{2} = 6$ mol; moles $N_{2} = \frac{64}{28} \approx 2.29$ mol.
Required ratio is $3 H_{2} : 1 N_{2}$ . For $2.29$ mol $N_{2}$ , need $3 \times 2.29 \approx 6.86$ mol $H_{2}$ — but only $6$ mol $H_{2}$ is available.
Since the available $H_{2}$ falls short of what's needed, it runs out first.

Answer: $H_{2}$ is the limiting reactant

Definition

Theoretical yield is the maximum possible mass of product predicted by the balanced equation, assuming the limiting reactant fully converts to product with no losses. Actual yield is what's really obtained in practice — always less than or equal to the theoretical yield, due to incomplete reactions, side reactions, or losses during handling.

Percentage yield matters because it tells a chemist or manufacturer how efficient a reaction actually is in practice, which is essential for judging cost and feasibility on an industrial scale.

Percentage purity tells us what fraction of a sample is actually the substance of interest, rather than other (impurity) material mixed in — this matters enormously for a compound used as a medicine, since impurities in a drug could be ineffective at best or actively harmful at worst, so purity must be verified before use.

Percentage yield $= \frac{actual yield}{theoretical yield} \times 100$

Percentage purity $= \frac{mass of pure substance}{total mass of sample} \times 100$

Example 12 — percentage yield

The actual yield of a compound is $0.198$ g while the theoretical yield is $0.217$ g. Find the percentage yield.

$% yield = \frac{0.198}{0.217} \times 100$ .

Answer: $\approx 91.2%$

Example 13 — percentage yield from a reaction

$305$ g of $A g N O_{3}$ ( $M = 170$ ) reacts with excess $M g C l_{2}$ , producing $23.7$ g of $M g (N O_{3})_{2}$ ( $M = 148$ ). Find the percentage yield. ( $2 A g N O_{3} + M g C l_{2} \to M g (N O_{3})_{2} + 2 A g C l$ )

Moles $A g N O_{3} = \frac{305}{170} \approx 1.794$ mol.
Mole ratio $A g N O_{3} : M g (N O_{3})_{2} = 2 : 1$ , so theoretical moles $M g (N O_{3})_{2} = \frac{1.794}{2} \approx 0.897$ mol.
Theoretical mass $= 0.897 \times 148 \approx 132.8$ g.
$% yield = \frac{23.7}{132.8} \times 100$ .

Answer: $\approx 17.8%$

Example 14 — percentage purity

A metal sample has total mass $3.66$ kg and contains $2.45$ kg of gold. Find the percentage purity of gold.

$% purity = \frac{2.45}{3.66} \times 100$ .

Answer: $\approx 66.9%$

Quick Recap

●RTP: $2 5^{\circ}$ C, $1$ atm; molar volume $= 24 dm^{3} / mol$ for any gas (follows from Avogadro's Law)
● $n = V /24$ (RTP), $n = m / M$ ; concentration $= n / V$ (mol/dm $^{3}$ ) or $m / V$ (g/dm $^{3}$ )
●Empirical formula: simplest ratio. Molecular formula $=$ empirical $\times n$ , $n = \frac{molar mass}{empirical mass}$
●Limiting reactant: compare available mole ratio to the equation's required ratio — the one that falls short runs out first
● $% yield = \frac{actual}{theoretical} \times 100$ ; $% purity = \frac{pure mass}{total mass} \times 100$

10th Class Chemistry Chapter 2 Notes

The Mole and Molar Volume

Concentration of Solutions

Empirical and Molecular Formula

Limiting Reactant

Percentage Yield and Percentage Purity

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