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Mole Concept and Stoichiometry formulas

7 formulas, 1 reference table and 10 common traps for NDA Chemistry Mole Concept and Stoichiometry, grouped by subtopic.

Full notes with worked examples

The Mole, Avogadro's Law and Molar Calculations

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The mole and Avogadro's number

Avogadro's number

N_A = 6.022 \times 10^{23}\ \text{particles per mole}

Molar mass and moles from mass

Moles from mass

n = \dfrac{m}{M}

$n$ number of moles
$m$ given mass (g)
$M$ molar mass (g/mol)

Avogadro's law and molar volume at STP

Moles from gas volume at STP

n = \dfrac{V}{22.4}

$n$ number of moles
$V$ volume of gas at STP (litres)
$22.4$ molar volume at STP (L/mol)

Counting particles from moles

Particles from moles

N = n \, N_A = \dfrac{m}{M}\, N_A

$N$ number of particles (molecules/atoms)
$n$ number of moles
$N_A$ Avogadro's number, $6.022 \times 10^{23}$

Mass-percent composition

Mass percent of an element

\%\,\text{element} = \dfrac{a \times A}{M} \times 100

$a$ number of atoms of the element in the formula
$A$ atomic mass of the element
$M$ molar mass of the whole compound

Common traps

Molecules and atoms differ for diatomic gases

One mole of

\text{H}_2

\text{O}_2

\text{N}_2

holds

6.022 \times 10^{23}

molecules but twice that many atoms. Read whether the question asks for molecules or atoms.

Use the molar mass of the WHOLE molecule

For 0.5 mol of

\text{N}_2

the molar mass is 28 g/mol (a nitrogen molecule), not 14. So mass

= 0.5 \times 28 = 14

g — the 14 comes from the calculation, not from using the atomic mass of one N.

22.4 L only at STP, and only for gases

The molar volume of 22.4 L per mole applies to gases at STP only. It does not apply to liquids or solids, and it does not apply to a gas at room temperature or other pressures.

Half a mole of a gas = 11.2 L (this is correct)

In a NOT-correct statement question, 'half mole of nitrogen measures 11.2 L at STP' is a true statement (

0.5 \times 22.4 = 11.2

) — so it is not the wrong one. Check each option's arithmetic separately.

4 g of H2 is TWO Avogadro numbers, not one

\text{H}_2

has molar mass 2, so 4 g

= 2

mol

= 2 \times 6.022 \times 10^{23}

molecules. The classic NOT-correct option claims '4 g of hydrogen contains

6.022 \times 10^{23}

molecules' — that is the false statement.

Mass-percent ratio ignores the oxygen count

For

\text{C}_6\text{H}_{12}\text{O}_4

versus

\text{C}_6\text{H}_{12}\text{O}_6

, the %C : %H ratio is the same (= 6) because the oxygen mass cancels from the ratio. Changing n only changes the individual percents, not their ratio.

Stoichiometry and the Laws of Chemical Combination

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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}}

$E$ equivalent weight
$M$ molar mass
$n\text{-factor}$ replaceable H+ (acid) or OH- (base) per molecule

Laws of chemical combination

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.

Common traps

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.

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).

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).

Mole Concept and Stoichiometry formulas

The Mole, Avogadro's Law and Molar Calculations

The mole and Avogadro's number

Molar mass and moles from mass

Avogadro's law and molar volume at STP

Counting particles from moles

Mass-percent composition

Common traps

Stoichiometry and the Laws of Chemical Combination

Mole ratios from a balanced equation

Equivalent weight and n-factor

Laws of chemical combination

Common traps

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