Core Mole Equations

$n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

$N_{\text{particles}} = n \times N_A = n \times 6.022 \times 10^{23}$

$V_{\text{STP}} = n \times 22.4 \ \text{L/mol} \quad \text{(0 °C, 1 atm)}$

Percentage Composition

$\% \text{ element} = \frac{n \times \text{atomic mass}}{\text{molar mass}} \times 100$

Molecular Formula from Empirical Formula

$n = \frac{M_r}{\text{Empirical Formula Mass}}$

$\text{Molecular Formula} = n \times \text{Empirical Formula}$

Concentration Interconversions

$M = \frac{1000 \times d \times w\%}{M_r \times 100}$

$m = \frac{1000 \times w\%}{M_r \times (100 - w\%)}$

$N = M \times n\text{-factor}$

$\text{Equivalent weight} = \frac{M_r}{n\text{-factor}}$

Key Reactions with Conditions

$Na_{2}CO_{3}$ + 2 HCl → 2 NaCl + $H_{2}O$ + $CO_{2}$ (aq, room temp)

$CaCO_{3}$ --[heat, >840 °C]--> CaO + $CO_{2}$

2 $H_{2}$ + $O_{2}$ --[spark / Pt catalyst]--> 2 $H_{2}O$

H_{2}$$SO_{4} + 2 NaOH → Na_{2}$$SO_{4} + 2 $H_{2}O$ (neutralisation, aq)

Limiting Reagent Method

1. Convert all given masses to moles using molar masses.
2. Divide each by its stoichiometric coefficient.
3. The reactant with the smallest such value is the limiting reagent.
4. Theoretical yield = moles of limiting reagent × stoichiometric ratio to product × molar mass of product.