Use mole ratios to calculate reacting quantities, percent yield, limiting reagents and solution concentrations

Learning goal: Interpret the coefficients of a balanced equation as ratios of particles and of moles; convert a given quantity of one substance in a reaction (moles, mass, particles, gas volume at STP or solution volume) to the quantity of another; compare theoretical and actual yields as a percent yield; identify the limiting reagent conceptually and with a before-change-after table; and calculate molarity, prepare dilutions and use titration data.

What the coefficients mean. A balanced equation is a recipe in particles. In 2H₂ + O₂ → 2H₂O the coefficients say two molecules of hydrogen react with one molecule of oxygen to make two molecules of water, and, because a mole is just a fixed count, two moles of hydrogen react with one mole of oxygen to make two moles of water. The coefficients do not give mass ratios: 2 g of hydrogen does not react with 1 g of oxygen. Every stoichiometric calculation therefore has the same shape: convert whatever you are given into moles, use the coefficient ratio to move to moles of the substance you want, and convert those moles into whatever unit is asked for. The mole ratio is the only step that is specific to the reaction; everything else is the conversion map from the quantifying-matter lesson, plus the molar volume of 22.4 L/mol for a gas at STP and molarity for a solution.

A worked model: mass to mass. Zinc reacts with hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. What mass of hydrogen forms from 3.27 g of zinc with excess acid, and what volume would it occupy at STP? Step 1, grams to moles of the given: 3.27 g Zn × (1 mol / 65.38 g) = 0.0500 mol Zn. Step 2, the ratio: the equation shows 1 mol H₂ per 1 mol Zn, so 0.0500 mol H₂. Step 3, moles to the unit asked: 0.0500 mol × 2.02 g/mol = 0.101 g of hydrogen; and 0.0500 mol × 22.4 L/mol = 1.12 L at STP. Writing the three steps as one chain with units canceling, g Zn → mol Zn → mol H₂ → g H₂, is dimensional analysis, and the units catch a ratio written upside down.

A second model: how much reactant is needed. Copper displaces silver from silver nitrate: Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag. To produce silver from 1.00 g of copper, how much silver nitrate is needed and how much silver should form? Moles of copper: 1.00 / 63.55 = 0.01574 mol. Ratio: 2 mol AgNO₃ per 1 mol Cu, so 0.03147 mol AgNO₃, which is 0.03147 × 169.88 = 5.35 g. Silver: 2 mol Ag per 1 mol Cu, so 0.03147 mol × 107.87 g/mol = 3.39 g of silver expected. Silver nitrate stains skin and is a teacher-handled reagent; the calculation is safe for anyone.

Theoretical yield, actual yield, percent yield. The mass calculated from the balanced equation is the theoretical yield, the most that could form if the limiting reactant were completely converted and every particle of product collected. The actual yield is what you weigh. Percent yield = actual / theoretical × 100. If the silver reaction above gave 3.10 g of silver, the percent yield is 3.10 / 3.39 × 100 = 91.3 percent (carrying the unrounded 3.395 g through the division). A yield below 100 percent has causes to name: product lost in transfer or filtering, an incomplete reaction, competing side reactions, or product left dissolved in the solution. A yield above 100 percent is a signal of error, usually a wet or impure product, never a bonus. In a multi-step preparation the overall percent yield is found by multiplying the fractions from each step, so three steps at 80 percent each give only 51 percent overall. Whenever the theoretical and actual amounts differ, provide possible explanations for the discrepancy rather than adjusting the numbers. Yield is also not the only measure of a good process: when several chemical routes to a product exist, chemists compare their efficiency, cost and environmental impacts, and sustainability and green chemistry should be considered, favoring routes that use less energy, produce less hazardous waste and rely on renewable starting materials.

Limiting reagents. Reactants are seldom mixed in exactly the ratio the equation demands. The reactant that runs out first is the limiting reagent; it decides how much product can form, and the other reactant is in excess. A particle diagram makes the idea concrete: draw four H₂ molecules and three O₂ molecules; pairing them by the recipe 2H₂ : 1O₂ uses all four H₂ and only two O₂, leaving one O₂ unreacted, so hydrogen limits and four H₂O form. A before-change-after (BCA) table does the same in moles. Write the equation across the top; in the Before row put the starting moles; in the Change row subtract for reactants and add for products using the coefficient ratio, scaled so that the limiting reagent reaches zero; in the After row add the rows. For 4.0 mol H₂ and 3.0 mol O₂: Before 4.0, 3.0, 0; Change −4.0, −2.0, +4.0; After 0, 1.0, 4.0. Hydrogen is limiting, 1.0 mol of oxygen is left over and 4.0 mol of water forms. To decide which reactant limits, ask how much of the second reactant the first would need: 4.0 mol H₂ needs 2.0 mol O₂ and 3.0 is available, so oxygen is in excess. The concept is what matters in this course; the calculation is a tool to make it visible. The same idea applies outside the laboratory: a pond's algae growth is limited by whichever nutrient, nitrate or phosphate, runs out first, and an industrial process is designed so that the expensive reactant limits and the cheap one is in excess.

Solutions and molarity. A solution is a homogeneous mixture: the solute is the substance dissolved, usually the smaller amount, and the solvent is what it dissolves in. Molarity, M, is the concentration in moles of solute per liter of solution, mol/L, and it is the unit that lets a volume of solution stand in for an amount of substance. To make 250.0 mL of a sodium chloride solution from 5.85 g of salt: moles = 5.85 / 58.44 = 0.100 mol; molarity = 0.100 mol / 0.2500 L = 0.400 mol/L. Preparation matters: the salt is dissolved in less than the final volume and water is then added to reach exactly 250.0 mL in a volumetric flask, because the final volume is of the solution, not of the water added. A dilution keeps the moles of solute fixed while adding solvent, so M₁V₁ = M₂V₂: to make 100.0 mL of 0.100 mol/L solution from the 0.400 mol/L stock, V₁ = 0.100 × 100.0 / 0.400 = 25.0 mL of stock, diluted to 100.0 mL. Molality and normality are reserved for later study.

Molarity in stoichiometry and titration. Because moles = molarity × liters, a solution volume converts to moles as easily as a mass does. How many grams of silver chloride form when 50.0 mL of 0.200 mol/L silver nitrate reacts with excess sodium chloride? Moles AgNO₃ = 0.200 × 0.0500 = 0.0100 mol; the ratio AgNO₃ : AgCl is 1 : 1; mass = 0.0100 × 143.32 = 1.43 g. In a titration the same logic runs backwards: a measured volume of a standard solution of known molarity reacts completely with an unknown, the mole ratio links them, and the unknown's molarity is moles divided by its volume. Testing water for a pollutant often works this way, by reacting a sample with a standard reagent or by comparing it with a series of standard solutions of known concentration, a calibration.

Where the rest of the standards is taught. Measurement, significant figures, average atomic mass and empirical formulas from C.PM.5 are taught in the quantifying matter lesson; this lesson uses its mole conversions.

Investigation: predict and measure a gas yield. Materials: baking soda, white vinegar, a cup, a balance reading to 0.01 g, a spoon, goggles. The reaction is NaHCO₃ + CH₃COOH → NaCH₃COO + H₂O + CO₂, and most of the carbon dioxide escapes, so the mass lost measures the gas that left the cup; some carbon dioxide stays dissolved in the liquid, so the loss is a little less than the gas made. Steps: weigh 2.00 g of baking soda; predict the carbon dioxide: 2.00 / 84.01 = 0.0238 mol NaHCO₃, ratio 1 : 1, so 0.0238 × 44.01 = 1.05 g CO₂; place 60 mL of 5 percent vinegar in the cup (about 0.05 mol of acetic acid, roughly twice the 0.0238 mol needed, so it is in excess) on the balance and record the total mass with the baking soda beside it; add the baking soda slowly, wait until fizzing stops and the surface is still, and record the final mass; repeat three times. A result that supports the claim: the mass loss is close to, and usually somewhat below, 1.05 g; dividing it by 1.05 g gives an apparent percent yield based on the carbon dioxide that escaped (no particular percentage is guaranteed, because the amount left dissolved depends on temperature, stirring and time); a loss well over 1.05 g suggests spray left the cup. What the model cannot show: the balance cannot tell carbon dioxide from splashed liquid or evaporated water, so the method assumes all mass lost is gas, and it misses the carbon dioxide still dissolved in the liquid; and it does not show the limiting reagent directly, which is why the vinegar is deliberately in excess. Wear goggles; the reaction can spit.

Worked example 1 — moles to moles to mass. How many grams of oxygen are needed to burn 0.750 mol of propane, C₃H₈ + 5O₂ → 3CO₂ + 4H₂O? Ratio: 5 mol O₂ per 1 mol C₃H₈, so 0.750 × 5 = 3.75 mol O₂. Mass: 3.75 × 32.00 = 120 g of oxygen. The ratio step is where a student who skips to mass goes wrong; 0.750 mol of propane is 33.1 g, and nothing about 33.1 leads to 120 without the moles in between.

Worked example 2 — gas volume at STP. What volume of carbon dioxide at STP forms when 10.0 g of calcium carbonate decomposes, CaCO₃ → CaO + CO₂? Moles CaCO₃ = 10.0 / 100.09 = 0.0999 mol; ratio 1 : 1, so 0.0999 mol CO₂; volume = 0.0999 × 22.4 = 2.24 L at STP. At room conditions the volume would be larger, and the ideal gas law would be used instead of 22.4 L/mol.

Worked example 3 — limiting reagent with a BCA table. 0.300 mol of aluminum is placed in a solution containing 0.600 mol of copper(II) chloride: 2Al + 3CuCl₂ → 2AlCl₃ + 3Cu. Does aluminum limit? 0.300 mol Al needs 0.300 × 3/2 = 0.450 mol CuCl₂, and 0.600 is available, so aluminum is limiting and copper(II) chloride is in excess. BCA table: Before 0.300, 0.600, 0, 0; Change −0.300, −0.450, +0.300, +0.450; After 0, 0.150, 0.300, 0.450. Copper formed: 0.450 mol × 63.55 = 28.6 g. Copper(II) chloride left: 0.150 mol.

Worked example 4 — molarity, dilution and a solution reaction. A student dissolves 4.00 g of sodium hydroxide (40.00 g/mol) to make 500.0 mL of solution: 0.100 mol / 0.5000 L = 0.200 mol/L. To make 50.0 mL of 0.0500 mol/L from it: V₁ = 0.0500 × 50.0 / 0.200 = 12.5 mL of stock diluted to 50.0 mL. Then 20.0 mL of the 0.200 mol/L base neutralizes how many moles of sulfuric acid? Moles NaOH = 0.200 × 0.0200 = 0.00400 mol; H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so moles H₂SO₄ = 0.00400 / 2 = 0.00200 mol.

Representation task: a particle diagram and a BCA table. For 2H₂ + O₂ → 2H₂O starting with six H₂ and two O₂ molecules, draw the before and after particle pictures, circle the leftover molecules, and name the limiting reagent. Build the matching BCA table in moles (6.0, 2.0, 0). Then draw the stoichiometry road map: grams, particles, liters of gas at STP and liters of solution all converting to moles of the given substance, the mole ratio bridge, and moles of the wanted substance converting back to any of the four units, with the conversion factor on every arrow.

Independent performance: yield investigation and problem set. Carry out the baking soda and vinegar mass-loss investigation in triplicate and write a report. Success criteria: the balanced equation; the theoretical yield of carbon dioxide from your measured mass of baking soda with the mole ratio shown; the measured mass loss for each trial and the average; the apparent percent yield with at least two explanations for the shortfall, including carbon dioxide left dissolved, distinguishing losses of product from errors of measurement; an explanation of why vinegar was used in excess and which reactant was limiting; and a problem set from your teacher containing one mass-to-mass, one mass-to-gas-volume, one limiting reagent with a BCA table, one molarity and one dilution calculation, each with units shown at every step and the answer reported to the correct significant figures. Finally, create a solution and a dilution: dissolve 5.85 g of table salt in water and make the volume up to 250 mL in the most precise measuring vessel you have (0.400 mol/L), then dilute 25.0 mL of it to 100 mL (0.100 mol/L); record every mass and volume with the precision of the equipment and say which measurement limits the accuracy of each concentration.