NDA Chemistry · Mole Concept and Stoichiometry

Stoichiometry and the Laws of Chemical Combination

A balanced equation is a recipe in moles: its coefficients give the ratio in which substances react and form, while the named laws (conservation of mass, definite and multiple proportions) state why those ratios are fixed.

All Mole Concept and Stoichiometry notes NDA Chemistry strategy

Why this matters

Four PYQs here, split two ways: a couple of calculation questions (mass of CO2 from 1 kg of carbon, equivalent weight of oxalic acid) and a couple of name-the-law questions (which law a given reaction or statement illustrates). The calculation half rests on the mole bridges from the previous subtopic; the name-the-law half is pure recall of one short table — so this subtopic mixes one formula technique with one reference table.

Concept 1 of 3

Mole ratios from a balanced equation

Intuition

The big coefficients in a balanced equation are a ratio of moles, not of grams. So the recipe is always the same: turn the known mass into moles, scale by the coefficient ratio to get moles of the product, then turn those moles back into grams.

Definition

Stoichiometry in three steps:

Balance the equation; the coefficients are the mole ratio of reactants to products.
For $\text{C} + \text{O}_2 \to \text{CO}_2$ the ratio C : $\text{CO}_2$ is 1 : 1, so 1 mole of carbon makes 1 mole of $\text{CO}_2$ .
Mass of product $= (\text{moles of known}) \times (\text{ratio}) \times (\text{molar mass of product})$ .
Always go through moles — a gram-to-gram shortcut only works by accident.

Mass of product from mass of reactant

m_{\text{product}} = \dfrac{m_{\text{reactant}}}{M_{\text{reactant}}}\times \text{(mole ratio)} \times M_{\text{product}}

Worked example

How much CO2 is produced on burning 1 kg of carbon? C + O2 -> CO2 (C = 12, CO2 = 44).

Moles of carbon $= 1000/12$ mol (1 kg = 1000 g).
Ratio C : $\text{CO}_2$ is 1 : 1, so moles of $\text{CO}_2 = 1000/12$ mol.
Mass of $\text{CO}_2 = (1000/12) \times 44 = 44000/12 = 11000/3$ g $= 11/3$ kg.

Answer:

\dfrac{11}{3}

kg of

\text{CO}_2

(about 3.67 kg).

Practice this conceptself-check · 3 quick reps

Try it yourself

How many grams of water form when 4 g of hydrogen burns completely? 2H2 + O2 -> 2H2O (H2 = 2, H2O = 18).

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
In C + O2 -> CO2, how many moles of CO2 come from 1 mole of carbon?
2.
Mass of CO2 from 12 g (1 mol) of carbon?
3.
In 2H2 + O2 -> 2H2O, moles of water from 1 mole of O2?

From the bank · past-year question

Example 1Mole Concept and StoichiometryMODERATE

How much

\text{CO}_2

is produced on heating of 1 kg of carbon?

[Q115 · Sep · 2017]

Coefficients are moles, not grams

The 1 : 1 in

\text{C} + \text{O}_2 \to \text{CO}_2

means 1 mole carbon gives 1 mole

\text{CO}_2

— but 12 g of carbon gives 44 g of

\text{CO}_2

, because their molar masses differ. Never assume equal masses.

Concept 2 of 3

Equivalent weight and n-factor

Intuition

Equivalent weight is the molar mass scaled down by how many reactive units the substance contributes per molecule — for an acid, how many replaceable hydrogen ions it has. Find the molar mass, find the n-factor, divide.

Definition

The equivalent-weight rule:

Equivalent weight $= \dfrac{\text{molar mass}}{n\text{-factor}}$ .
For an acid, the n-factor is its basicity — the number of replaceable $\text{H}^+$ ions (1 for HCl, 2 for $\text{H}_2\text{SO}_4$ , 2 for oxalic acid).
For a base, the n-factor is its acidity — the number of $\text{OH}^-$ ions.
Oxalic acid dihydrate $\text{C}_2\text{H}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$ has molar mass 126 and is dibasic, so equivalent weight $= 126/2 = 63$ .

Equivalent weight

E = \dfrac{M}{n\text{-factor}}

Eequivalent weight
Mmolar mass
n\text{-factor}replaceable H+ (acid) or OH- (base) per molecule

Worked example

Find the equivalent weight of sulphuric acid, H2SO4. (M = 98)

$\text{H}_2\text{SO}_4$ has two replaceable $\text{H}^+$ ions, so n-factor = 2 (dibasic).
Equivalent weight $= M/n = 98/2$ .

Answer:49.

Practice this conceptself-check · 3 quick reps

Try it yourself

Find the equivalent weight of oxalic acid dihydrate, C2H2O4.2H2O.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Equivalent weight of HCl (M = 36.5, monobasic)?
2.
n-factor of oxalic acid?
3.
Equivalent weight of NaOH (M = 40, monoacidic base)?

From the bank · past-year question

Example 2Mole Concept and StoichiometryMODERATE

The equivalent weight of oxalic acid in

\text{C}_2\text{H}_2\text{O}_4\cdot 2\text{H}_2\text{O}

[Q86 · Apr · 2019]

Include the water of crystallisation in the molar mass

Oxalic acid dihydrate is

\text{C}_2\text{H}_2\text{O}_4 \cdot 2\text{H}_2\text{O}

with molar mass 126, not 90 (the anhydrous value). Forgetting the

2\text{H}_2\text{O}

gives the wrong equivalent weight (45 instead of 63).

Concept 3 of 3

Laws of chemical combination

Intuition

These are the named rules the bank asks you to recognise from a one-line description or a worked example. Learn the name, its one-line statement, and a stock example of each.

Definition

Five named laws govern how elements combine. Each is tested by either a definition or a 'which law is shown' example:

Conservation of mass — the most-asked: total mass of reactants equals total mass of products.
Definite (constant) proportions — a pure compound always has the same elements in the same fixed mass ratio.
Multiple proportions — when two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in small whole-number ratios.

Law	Statement	Stock example
Law of conservation of mass	Matter can neither be created nor destroyed in a chemical reaction; total mass of reactants = total mass of products.	1.7 g AgNO3 + 0.585 g NaCl produce 1.435 g AgCl + 0.85 g NaNO3 (masses balance both sides).Q By far the most-asked law in this chapter; any reaction where the two sides' masses add up to the same total is illustrating this law.
Law of definite (constant) proportions	A given pure compound always contains the same elements in the same fixed proportion by mass.	Water is always 1 : 8 hydrogen to oxygen by mass, whatever its source.
Law of multiple proportions	If two elements form more than one compound, the masses of one combining with a fixed mass of the other are in a ratio of small whole numbers.	Carbon + oxygen: CO and CO2 — the oxygen masses per fixed carbon are in a 1 : 2 ratio.
Avogadro's law	Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.	22.4 L of any gas at STP contains one mole (6.022 x 10^23 molecules). Also the basis for the 22.4 L molar volume used in the previous subtopic.

Recognise the law from either its definition or a worked mass-balance example.

Practice this conceptself-check · 4 quick reps

Try it yourself

When 1.7 g of silver nitrate reacts with 0.585 g of sodium chloride to give 1.435 g of silver chloride and 0.85 g of sodium nitrate, which law of chemical combination is illustrated?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Which law states matter can neither be created nor destroyed?
2.
Which law says a pure compound always has the same fixed mass ratio of elements?
3.
CO and CO2 (oxygen masses in a 1 : 2 ratio per fixed carbon) illustrate which law?
4.
Which law underlies the 22.4 L molar volume of a gas at STP?

From the bank · past-year question

Example 3Mole Concept and StoichiometryEASY

1.7 g of

\text{AgNO}_3

dissolved in 100 g water + 0.585 g

\text{NaCl}

in 100 g water gives 1.435 g

\text{AgCl}

+ 0.85 g

\text{NaNO}_3

. Which law of chemical combination is shown?

[Q83 · Apr · 2026]

Mass balancing means conservation of mass, not definite proportions

When a question gives reactant and product masses that add to the same total, the law shown is conservation of mass. Definite proportions is about one compound's fixed internal ratio, not about both sides of a reaction balancing.

Definite vs multiple proportions

Definite proportions = one compound, one fixed ratio. Multiple proportions = two different compounds of the same two elements, ratios in small whole numbers. The give-away for multiple proportions is two compounds being compared (CO vs CO2).

Drill 1 more on laws of chemical combination

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Mole ratios from a balanced equation
Mass of product from mass of reactant
$m_{\text{product}} = \dfrac{m_{\text{reactant}}}{M_{\text{reactant}}}\times \text{(mole ratio)} \times M_{\text{product}}$
Equivalent weight and n-factor
Equivalent weight
$E = \dfrac{M}{n\text{-factor}}$

Reference tables (1)

Laws of chemical combination4 rows

Law	Statement	Stock example
Law of conservation of mass	Matter can neither be created nor destroyed in a chemical reaction; total mass of reactants = total mass of products.	1.7 g AgNO3 + 0.585 g NaCl produce 1.435 g AgCl + 0.85 g NaNO3 (masses balance both sides).Q By far the most-asked law in this chapter; any reaction where the two sides' masses add up to the same total is illustrating this law.
Law of definite (constant) proportions	A given pure compound always contains the same elements in the same fixed proportion by mass.	Water is always 1 : 8 hydrogen to oxygen by mass, whatever its source.
Law of multiple proportions	If two elements form more than one compound, the masses of one combining with a fixed mass of the other are in a ratio of small whole numbers.	Carbon + oxygen: CO and CO2 — the oxygen masses per fixed carbon are in a 1 : 2 ratio.
Avogadro's law	Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.	22.4 L of any gas at STP contains one mole (6.022 x 10^23 molecules). Also the basis for the 22.4 L molar volume used in the previous subtopic.

Recognise the law from either its definition or a worked mass-balance example.

Watch out for (4)

Coefficients are moles, not grams
→ Mole ratios from a balanced equation
Include the water of crystallisation in the molar mass
→ Equivalent weight and n-factor
Mass balancing means conservation of mass, not definite proportions
→ Laws of chemical combination
Definite vs multiple proportions
→ Laws of chemical combination

Mastery check — 1 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Mole Concept and StoichiometryEASY

Which one of the following statements about the law of conservation of mass is correct?

[Q107 · Sep · 2018]

Drill every past-year question on this subtopic

4 questions from the bank — paginated, with cart and Word-export support.

Drill the 4 questions