“Everything through moles”
How to start any numerical
Grams, litres and particle counts cannot be compared directly. Convert each to moles first, use the balanced equation, then convert back to whatever the question wants.
Chemistry, Class 11 and 12 · Gujarat Board, CBSE, GUJCET, JEE and NEET
Mole Concept and Stoichiometry formulas
The mole is the chapter everything else in chemistry leans on. Physical chemistry numericals almost always begin by converting what you were given into moles, so the few conversions here are worth more marks than their size suggests.
16 formulas, 3 memory hooks, and the 5 mistakes that cost the most marks.
Counting particles
| What | Formula | Symbols and notes |
|---|---|---|
| Avogadro's number | NA = 6.022 × 10²³ mol−1 | particles in one mole |
| Moles from mass | n = given mass / molar mass | the first line of most numericals |
| Moles from particles | n = N / NA | N is the number of atoms or molecules |
| Moles of a gas at STP | n = volume / 22.4 L | at 273 K and 1 bar, for an ideal gas |
| Number of particles | N = n × NA | atoms, molecules or ions, so say which |
Concentration
| What | Formula | Symbols and notes |
|---|---|---|
| Molarity | M = moles of solute / volume of solution in litres | changes with temperature, because volume does |
| Molality | m = moles of solute / mass of solvent in kg | independent of temperature |
| Mole fraction | x = ncomponent / ntotal | the fractions of all components add to 1 |
| Normality | N = gram equivalents / volume in litres | N = M × n-factor |
| Dilution | M₁V₁ = M₂V₂ | moles of solute do not change on adding water |
| Parts per million | ppm = (mass of solute / mass of solution) × 10⁶ | used for very dilute solutions |
Formulas and yield
| What | Formula | Symbols and notes |
|---|---|---|
| Empirical formula | the simplest whole number ratio of atoms | found from percentage composition |
| Molecular formula | molecular formula = n × empirical formula | n = molar mass / empirical formula mass |
| Percentage composition | % of element = (mass of element / molar mass) × 100 | per mole of compound |
| Percentage yield | (actual yield / theoretical yield) × 100 | theoretical yield comes from the limiting reagent |
| Limiting reagent | the reactant giving the fewest moles of product | divide moles by the coefficient and take the smallest |
Taught by Darshit Bhatt, founder and lead faculty at Darshit Sir Institute. Instrumentation and Control engineer, with a master's degree and a chartered degree, teaching in Paldi, Ahmedabad since 2006.
Know the formula but not when to use it?
Darshit Sir has taught this in Paldi since 2006, in the classroom and live online.
Memory hooks
How to remember this chapter.
“Molarity has volume, molality has mass”
Telling the two apart
Molarity is per litre of solution, molality per kilogram of solvent. Only molality is safe when the temperature changes.
“Smallest share wins”
Finding the limiting reagent
Divide each reactant's moles by its coefficient in the balanced equation. The smallest answer is the limiting reagent, and it sets the yield.
Where marks go
The mistakes that cost the most.
- Dividing moles by the wrong volume. Molarity uses the volume of the whole solution, not of the solvent.
- Using 22.4 litres per mole for a gas that is not at STP, or for a liquid or solid.
- Forgetting to balance the equation before comparing mole ratios.
- Assuming the reactant present in the smallest mass is the limiting reagent. It is the smallest mole-to-coefficient ratio that decides.
- Reporting the empirical formula when the molecular formula was asked for.
Test yourself
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