Moles, formulas and reaction calculations

Moles, formulas and reaction calculations

Chemists count enormous numbers of small particles using the mole. One mole contains Avogadro’s number of specified entities, such as atoms, molecules, ions or formula units. Always identify the entities being counted. A mole is an amount of substance, not a fixed mass: a mole of carbon atoms and a mole of water molecules contain the same number of entities but have different masses.

Molar mass is the mass per mole and is usually expressed in grams per mole. To find a compound’s molar mass, add the contributions of all atoms in its formula. Using H = 1 and O = 16, water has molar mass 2(1) + 16 = 18 g/mol. The relationship connects amount n, mass m and molar mass M.

A balanced chemical equation states the relative numbers of reacting and produced particles. Its coefficients therefore also give mole ratios. In moles of hydrogen react with one mole of oxygen to form two moles of water. Balancing changes coefficients in front of formulas; changing subscripts would change the identities of the substances themselves.

For a mass-to-mass calculation, convert the known mass to moles, apply the balanced-equation ratio and convert the required moles back to mass. In a limiting-reactant problem, compare the available mole amounts with the coefficients. The reactant used up first limits the amount of product. The larger starting mass is not automatically the excess reactant because molar masses and reaction ratios differ.

Solution concentration in mol/dm³ equals moles of solute divided by volume of solution in dm³. Convert cm³ to dm³ by dividing by 1,000. A solution containing 0.20 mol in 0.50 dm³ has concentration 0.40 mol/dm³. Mass concentration, often stated in g/dm³, is a different quantity; conversion between them uses the solute’s molar mass.

At a stated temperature and pressure, a supplied molar gas volume links gas amount and volume. Use the value given in the question instead of assuming all conditions have the same molar volume. Check significant figures, units and the balanced equation at the end. Many errors arise from using grams as though they were moles or from applying an equation that has not been balanced first.

Avogadro counting and formula interpretation

Use the specified particle type when converting amount to number, with NA approximately 6.02 × 10²³ mol⁻¹ in school calculations. One mole of O₂ molecules contains two moles of oxygen atoms; one mole of NaCl formula units contains one mole each of sodium and chloride ions in the ionic model. The number in front of a formula multiplies the whole formula. Thus 2Ca(OH)₂ represents two calcium atoms, four oxygen atoms and four hydrogen atoms in symbolic particle counting. A subscript within parentheses multiplies every atom inside them. Careful formula reading comes before any arithmetic with molar masses.

Empirical and molecular formulas

An empirical formula gives the simplest whole-number atom ratio, while a molecular formula gives the actual atom count in a molecule. To find an empirical formula from percentage composition, imagine a 100 g sample, divide each element's mass by its relative atomic mass and divide all resulting amounts by the smallest. Multiply ratios if necessary to remove simple fractional values such as 1.5. Do not round 1.5 directly to 2. A molecular formula is an integer multiple of the empirical formula, found by dividing molecular molar mass by empirical-formula mass. Ionic compounds normally use formula units rather than molecular formulas in this sense.

Balancing and the laws of combination

The law of conservation of mass is reflected in equal atom counts for each element on both sides of a balanced chemical equation. The law of definite proportions concerns a fixed mass ratio in a particular pure compound. The law of multiple proportions compares simple whole-number ratios when the same elements form different compounds. For gaseous reactants and products at the same temperature and pressure, balanced coefficients also give volume ratios under the ideal-gas approximation. Water must be in the gaseous state if its gas-volume coefficient is used this way. State symbols help prevent a volume comparison that silently treats a liquid as an ideal gas.

Limiting reactants and excess

Calculate available moles of every reactant when the quantities of more than one are supplied. Divide each mole amount by its equation coefficient; the smallest result identifies the limiting extent of reaction. In and 2 mol O₂ give ratios 3/2 = 1.5 and 2/1 = 2. Hydrogen limits, producing 3 mol water and leaving 0.5 mol oxygen. Comparing only gram amounts is unreliable. After finding product, calculate the amount of excess reagent consumed using the same balanced ratio, then subtract from its initial amount if asked for what remains.

Percentage yield, purity and atom economy

Percentage yield is actual isolated product divided by theoretical product under the stated assumptions, multiplied by 100. A low yield can arise from incomplete reaction, side reactions or product losses during separation. Percentage purity is mass of the desired substance divided by sample mass, multiplied by 100. Use pure-reactant mass in a stoichiometric calculation when an impure sample is supplied. Atom economy compares the formula mass of desired product with total formula mass of reactants according to the balanced equation; it is different from experimental yield. An apparent yield above 100% often signals wet or impure product, measurement error or an incorrect theoretical calculation.

Solutions, dilution and titration

For a solute amount n in volume V, concentration. Dilution adds solvent without changing solute moles, so c₁V₁ = c₂V₂ when no reaction or loss occurs. Use the same volume unit on both sides of that ratio equation, but convert to dm³ when multiplying mol/dm³ by a volume to obtain moles. In titration, write the balanced equation before using a ratio. Sulfuric acid and sodium hydroxide react 1:2, so equal volumes do not necessarily mean equal concentrations. The convenient expression follows from the stoichiometric coefficients, but explicitly calculating moles is often easier to explain and check.

Using units and checking experimental data

Molar mass in g/mol matches mass in grams; molar mass in kg/mol matches kilograms. A unit cancellation can reveal an error before the final number is obtained. Record titre as final burette reading minus initial reading, not automatically the final reading itself. Use concordant suitable titres according to the laboratory method rather than averaging a clearly rough or inconsistent result with precise trials. Carry extra digits during intermediate mole ratios. A theoretical product mass should respect the limiting reagent and balanced equation. If a reaction appears to create much more total mass than all supplied reactants, check whether another reactant such as oxygen from air was omitted from the account.

Self-check with reasoned solutions

A compound containing 40.0% carbon, 6.7% hydrogen and 53.3% oxygen gives approximate mole ratios 40/12: 6.7/1: 53.3/16 = 3.33: 6.7: 3.33, or 1:2:1, so its empirical formula is CH₂O. If its molar mass is 180 g/mol, the multiplier is 180/30 = 6 and molecular formula C₆H₁₂O₆. In another problem, 25.0 cm³ of 0.100 mol/dm³ HCl contains 0.00250 mol. It needs the same amount of NaOH under a 1:1 equation, so 20.0 cm³ of that alkali has concentration. Every number comes from a stated ratio and unit conversion.