Stoichiometry and Quantative Chemistry

Mole concept = → counting system used in chemistry

1 mole contains → 6.02 × 10²³ particles

6.02 × 10²³ = → Avogadro’s constant

Avogadro’s constant symbol → Nₐ

One mole can represent → atoms, molecules, ions, formula units

Molar mass = → mass of one mole of substance

Molar mass units → g mol⁻¹

Molar mass calculated using → atomic masses from periodic table

Equation linking moles, mass, molar mass → n = m / M

In n = m/M → n means → number of moles

In n = m/M → m means → mass

In n = m/M → M means → molar mass

To find moles from mass → divide by molar mass

To find mass from moles → multiply by molar mass

Molar mass of H₂O → 18 g mol⁻¹

Molar mass of CO₂ → 44 g mol⁻¹

Molar mass of NaCl → 58.5 g mol⁻¹

Gas molar volume at STP → 22.4 dm³ mol⁻¹

Gas molar volume at RTP → 24 dm³ mol⁻¹

STP means → 0°C and 1 atm pressure

RTP means → room temperature and pressure

Equation for gas moles at RTP → n = V / 24

Equal volumes of gases at same conditions contain → equal number of particles

Law of conservation of mass → mass cannot be created or destroyed

Total mass of reactants = → total mass of products

Chemical equations must be balanced because → atoms must be conserved

Balancing equations changes → coefficients only

Balancing equations never changes → subscripts

Coefficient = → number placed before formula in equation

Subscript = → small number within chemical formula

Stoichiometry = → calculation of quantities in reactions

Stoichiometric calculations use → balanced equations

Balanced equations compare → mole ratios

Mole ratio comes from → coefficients in balanced equation

Why must mass be converted to moles in stoichiometry → equations compare particles/moles not grams

General stoichiometry workflow → grams → moles → mole ratio → moles → grams

Example equation → 2H₂ + O₂ → 2H₂O

In 2H₂ + O₂ → 2H₂O → H₂ : O₂ ratio = → 2 : 1

In 2H₂ + O₂ → 2H₂O → H₂O : O₂ ratio = → 2 : 1

Limiting reactant = → reactant used up first

Limiting reactant determines → maximum product formed

Excess reactant = → reactant left over after reaction

To identify limiting reactant → convert all reactants to moles and compare mole ratios

Theoretical yield = → maximum possible product predicted by stoichiometry

Actual yield = → amount of product actually obtained

Percentage yield equation → % yield = (actual ÷ theoretical) × 100

Low percentage yield caused by → side reactions, product loss, incomplete reaction

100% yield rarely achieved because → reactions are rarely perfectly efficient

Solution = → homogeneous mixture of solute and solvent

Solute = → substance dissolved