The mole is the chemist's counting unit; every stoichiometry question is a conversion to moles, a ratio from the balanced equation, and a conversion back.
How it's asked: Limiting reagent with mass data, number of atoms or molecules in a given mass, molarity/molality/mole fraction conversions, and empirical-formula problems from percentage composition.
Why this is true
A balanced equation is a statement about numbers of particles, not masses, because atoms are conserved one by one during a reaction. Since 1 mol of anything contains the same number of particles (6.022 × 10²³), the coefficients can be read directly as mole ratios. Masses differ only because particles have different molar masses, so you must convert grams to moles before using the coefficients. The reactant that runs out first — the one giving the smaller amount of product — fixes the yield, and is called the limiting reagent. Concentration terms are just different ways of reporting moles of solute against volume of solution (molarity), mass of solvent (molality) or total moles (mole fraction); molality and mole fraction do not change with temperature because masses do not.
Key points
Key formulae
Moles
n = m / M = N / NA
m in g, M in g mol⁻¹
Molarity
M = n(solute) / V(solution in L)
Molality
m = n(solute) / W(solvent in kg)
Dilution
M₁V₁ = M₂V₂
moles of solute unchanged
Mole fraction
xA = nA / (nA + nB)
xA + xB = 1
Concentration terms at a glance
| Term | Definition | Unit | Depends on temperature? |
|---|---|---|---|
| Molarity (M) | mol solute per L solution | mol L⁻¹ | Yes (volume changes) |
| Molality (m) | mol solute per kg solvent | mol kg⁻¹ | No |
| Mole fraction (x) | mol of component / total mol | none | No |
| Mass percent | g solute per 100 g solution | % | No |
| ppm | parts of solute per 10⁶ parts solution | none | No (by mass) |
Common traps
- Using the given masses directly in the coefficient ratio instead of converting to moles first.
- Counting molecules when the question asks for atoms (0.5 mol NH₃ has 0.5 NA molecules but 2 NA atoms).
Worked example
4 g of H₂ is mixed with 16 g of O₂ and the mixture is exploded to form water. What mass of water forms and which gas is left over?
High-yield
Always go grams → moles → coefficient ratio → moles → grams; the reactant giving less product is limiting.
Quick check
Q1.4 g of NaOH (M = 40 g mol⁻¹) is dissolved to make 250 mL of solution. The molarity is:
Q2.Total number of atoms in 0.5 mol of NH₃ is: