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Loading home…One mole of a substance contains Avogadro’s number of particles, NA = 6.022 × 10²³ mol⁻¹. Molar mass is the mass of one mole of the substance, numerically equal to the relative formula mass in grams. Number of moles n = mass / molar mass. Number of particles = n × NA. The mole links countable particles to weighable mass in the laboratory. Stoichiometric calculations almost always begin by converting given data into moles.
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. Atoms are only rearranged; their total number of each type remains constant. Therefore a chemical equation must be balanced with equal numbers of each atom on both sides. Balanced equations give the mole ratios needed for stoichiometric calculations. Without balancing, predicted product masses and volumes would be incorrect. The law underpins all quantitative chemistry at this level.
The empirical formula shows the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of each atom in a molecule. Molecular formula = (empirical formula) × n, where n = molecular mass / empirical formula mass. For example, glucose has empirical formula CH₂O and molecular formula C₆H₁₂O₆. Ionic compounds are usually represented by empirical formulae. Combustion or percentage composition data first yield the empirical formula.
The limiting reactant is the reactant that is completely consumed first in a reaction. It determines the maximum amount of product that can form. Other reactants present in excess remain partly unused. To identify it, convert all given amounts to moles and compare with the stoichiometric ratio. Theoretical yield is calculated from the moles of limiting reactant. Percentage yield = (actual yield / theoretical yield) × 100.
Molarity (M) is moles of solute per litre of solution (mol dm⁻³). Molality (m) is moles of solute per kilogram of solvent. Molarity depends on volume, which changes slightly with temperature, so molarity is temperature-dependent. Molality uses mass of solvent, which does not change with temperature, so molality is temperature-independent. Both are used to express concentration in quantitative work. For dilute aqueous solutions the numerical values of M and m are often similar but not identical.
Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Therefore gas volumes in a balanced equation are in the same ratio as moles, under the same conditions. For example, N₂ + 3H₂ → 2NH₃ means 1 volume N₂ reacts with 3 volumes H₂ to give 2 volumes NH₃. This simplifies many industrial and laboratory gas calculations. At STP, one mole of an ideal gas occupies about 22.4 dm³. Combined with mole ratios, volume data give amounts of reactants and products.
The mole is the SI unit for amount of substance. One mole contains exactly Avogadro’s number of elementary entities (atoms, molecules, ions, or formula units). Molar mass M converts between mass m and amount n by n = m/M. For particles, N = n × NA. For an ideal gas at STP, V ≈ n × 22.4 dm³ mol⁻¹ (using the conventional school value).
These three relationships allow any measurable quantity—mass, particle count, or gas volume—to be expressed in moles. Once all data are in moles, a balanced equation supplies mole ratios for stoichiometric conversion from reactants to products. Finally, moles are converted back to the desired unit.
Example pathway: given mass of reactant → moles of reactant → moles of product (by equation coefficients) → mass or volume of product. The same logic applies to titration, where molarity × volume gives moles of acid or base.
Students must always check units: grams with g mol⁻¹, volumes in dm³ with mol dm⁻³, and consistent T and P for gas laws. Significant figures should reflect the precision of the data. The mole concept is therefore the central bridge of quantitative chemistry.
Percentage composition gives the mass of each element in 100 g of compound. Divide each mass by the atomic mass of that element to obtain moles of atoms. Divide all mole values by the smallest number to get a simple ratio. If necessary, multiply by a small integer to clear fractions (e.g. 1.5 × 2). The resulting whole-number ratio is the empirical formula.
To obtain the molecular formula, the molar mass of the compound must be known from experiment (e.g. vapour density or mass spectrometry). Compute the empirical formula mass, then n = molar mass / empirical formula mass. Multiply all subscripts in the empirical formula by n.
Combustion analysis of organic compounds yields CO₂ and H₂O masses, from which C and H percentages are calculated; oxygen is often found by difference. Hydrated salts are analysed by finding water of crystallisation from mass loss on heating.
Common errors include incorrect rounding of ratios and forgetting to convert percentages to moles. Careful arithmetic and chemical judgement about plausible formulae are essential. Empirical and molecular formulae are foundational for writing correct chemical equations.
In many reactions reactants are not mixed in exact stoichiometric amounts. The limiting reactant is completely used up and stops the reaction. Excess reactant remains after the limiting reactant is consumed.
Procedure: write a balanced equation; convert all given masses or volumes to moles; for each reactant compute how many moles of product it could produce; the reactant that produces the least product is limiting. Theoretical yield is the mass of product calculated from the limiting reactant assuming 100% conversion. Actual yield is the mass obtained experimentally. Percentage yield = (actual / theoretical) × 100%.
Yields below 100% arise from incomplete reaction, side reactions, equilibrium, losses during transfer, and impure products. Sometimes yields above 100% indicate wet or impure product and need correction.
Industrial chemistry maximises percentage yield and atom economy while controlling cost and waste. School calculations usually assume ideal stoichiometry unless told otherwise. Always identify the limiting reactant before quoting a theoretical yield.
Molarity (c) = moles of solute / volume of solution in dm³. It is convenient for titrations and reactions in solution because volumes are easy to measure. Because volume changes with temperature, molarity is slightly temperature-dependent.
Molality (b) = moles of solute / mass of solvent in kg. It is preferred in colligative-property work because it does not depend on temperature. Mole fraction (x) = moles of component / total moles of all components; the sum of mole fractions is 1. Mole fraction is useful in vapour-pressure and ideal-solution relations (Raoult’s law).
Parts per million (ppm) expresses very dilute concentrations, often as mg of solute per kg of solution (approximately mg dm⁻³ in dilute aqueous solutions). Environmental and trace analysis commonly use ppm or ppb.
Converting between units requires density when relating volume-based and mass-based concentrations. Students should choose the unit that matches the experimental method and the formula being used. Clear labelling of M versus m avoids confusion in calculations.
A balanced chemical equation is a mole map of the reaction. Coefficients give the ratio in which reactants combine and products form. Mass–mass problems convert given mass → moles → use ratio → moles of unknown → mass of unknown.
For gases at the same T and P, volume ratios equal mole ratios (Avogadro’s law). At STP, moles can also be found from V/22.4 dm³ mol⁻¹. For reactions with solutions, moles = molarity × volume (in dm³).
Mixed problems may combine a solid reactant with a gas or solution product. Keep a clear table of given, moles, and required quantities. Check that the equation is balanced before any arithmetic.
Significant figures and unit cancellation reduce mistakes. If two reactants are given, determine the limiting reactant first. These methods apply to synthesis, combustion, and industrial process calculations alike.