Chemistry (university) · Part 04
Stoichiometry and reactions in solution
Moles, formulas, limiting reagents, concentration, precipitation and redox reactions.
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- Mole and molar massA mole is 6.022×10²³ particles (Avogadro's number). Molar mass is the mass of one mole in grams per mole. It equals the atomic or formula mass in atomic mass units. Moles = mass ÷ molar mass.
- Empirical and molecular formulaThe empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula is the actual number of atoms. Molecular formula = empirical formula × n, where n = molar mass ÷ empirical mass.
- Mole ratio from a balanced equationThe coefficients in a balanced equation give the ratio of moles reacting. Use that ratio as a conversion factor to go from the moles of one substance to the moles of another. Atoms are conserved, so both sides must balance.
- Limiting reagentThe limiting reagent is the reactant that runs out first and so decides how much product forms. The other reactants are in excess. To find it, compare the moles of each reactant with the balanced ratio.
- Percent yieldTheoretical yield is the maximum product the equation predicts. Actual yield is what you really collect. Percent yield = (actual ÷ theoretical) × 100. It is usually less than 100 percent.
- Molarity and dilutionMolarity (M) is moles of solute per litre of solution. For dilution, the moles of solute stay the same, so M₁V₁ = M₂V₂. Add solvent and the concentration falls.
- Precipitation reaction and net ionic equationIn a precipitation reaction, two soluble salts react and an insoluble solid, the precipitate, forms. A net ionic equation shows only the ions that change, leaving out the spectator ions that stay dissolved.
- Oxidation state and redoxRedox reactions move electrons. Oxidation is loss of electrons and reduction is gain. Oxidation state is the charge an atom would have if all bonds were ionic. Oxidation raises it, reduction lowers it.
- Balancing redox equationsThe half-reaction method splits a redox reaction into an oxidation and a reduction half. Balance atoms other than H and O, then O using H₂O, then H using H⁺, then charge using electrons. Match the electrons and add.