Difficulty: Introductory to Intermediate | Prerequisites: Molar mass calculations, balancing chemical equations
Stoichiometry is the maths of chemical reactions: how much of each substance you need, and how much product you get. When two reagents react, one runs out first (the limiting reagent) and caps the amount of product. Percent yield compares what you collected in the lab to what the stoichiometry predicted, and it is almost always below 100%.
Stoichiometry
The quantitative relationship between reactants and products in a balanced chemical equation. In simple terms, it tells you how much of each substance is involved in a reaction.
Limiting reagent (limiting reactant)
The reactant that is completely consumed first in a reaction, determining the maximum amount of product that can form. Think of it as the ingredient you run out of first when cooking.
Excess reagent (excess reactant)
The reactant that remains after the limiting reagent has been fully consumed. Some of it is left over in the beaker when the reaction finishes.
Theoretical yield
The maximum mass of product predicted by stoichiometry, assuming the reaction goes perfectly to completion with no losses. This is the number you calculate on paper.
Actual yield
The mass of product you collect in the lab. It is nearly always less than the theoretical yield because of side reactions, incomplete reactions, and transfer losses.
Percent yield
The ratio of actual yield to theoretical yield, expressed as a percentage. It measures how efficient the reaction was in practice.
Law of conservation of mass
Mass is neither created nor destroyed in a chemical reaction; the total mass of reactants equals the total mass of products. In simple terms, everything that goes in must come out somewhere.
Law of definite proportions (law of constant composition)
A given chemical compound always contains the same elements in the same mass ratio, regardless of how much of it you have or where it came from. Think of it as a fixed recipe that never changes.
Mole ratio
The ratio of moles of one substance to moles of another in a balanced equation. This is the bridge you use to convert between different substances in a reaction.
The total mass before a reaction equals the total mass after. If you weigh Fe + S going in, the mass of FeS coming out (plus any unreacted excess) should match.
In Lab 8 ("Foiled Again" Part B), the iron(II) sulfide reaction demonstrated this: as the combined mass m(Fe + S) increased, more FeS formed, and the total mass was conserved.
A percent yield above 100% (the lab reported 103.48%) signals experimental error, not a violation of conservation. Possible causes include incomplete drying, absorbed moisture, or weighing errors.
A compound always contains its elements in the same mass ratio. For FeS, the mass ratio Fe:S is always approximately 1.74:1 (55.85 g/mol divided by 32.06 g/mol).
Graphical evidence: a plot of mass of iron reacted vs. mass of sulfur reacted should yield a straight line through the origin. The slope reflects the fixed mass ratio.
The mole ratio for Fe:S in FeS is 1:1 (confirmed experimentally at roughly 0.999:1 in the lab data).
Convert every reactant mass to moles using its molar mass.
Divide each mole value by its coefficient in the balanced equation.
The reactant with the smallest result is the limiting reagent; the others are in excess.
In the aluminium + copper(II) sulfate reaction, aluminium was limiting because it was fully consumed. Copper remained in the beaker after the solution evaporated.
Start from the moles of the limiting reagent.
Use the mole ratio from the balanced equation to find moles of product.
Multiply by the product's molar mass to get grams.
Example: for 2 Fe + 3 Cl₂ → 2 FeCl₃, if Cl₂ is limiting at 0.0071 mol, the theoretical yield of FeCl₃ is 0.0071 mol × (2/3) × 162.2 g/mol = 0.77 g.
Percent yield = (actual yield / theoretical yield) × 100.
The lab obtained 0.65 g of FeCl₃ against a theoretical 0.77 g, giving 84.4%.
Values below 100% are normal. Values above 100% indicate contamination or measurement error, not extra product from nowhere.
Once you know the limiting reagent, calculate how many moles of the excess reagent are consumed (via the mole ratio).
Subtract consumed moles from starting moles, then convert back to grams.
Example: 5 g Fe and 2 g S reacting to form FeS. S is limiting (0.0624 mol). Fe consumed = 0.0624 mol. Excess Fe = (0.0895 - 0.0624) mol × 55.85 g/mol = 1.52 g.
Moles from mass
moles = mass (g) / molar mass (g/mol)
Mass from moles
mass (g) = moles × molar mass (g/mol)
Stoichiometric conversion (mole ratio bridge)
moles of product = moles of limiting reagent × (coefficient of product / coefficient of limiting reagent)
Percent yield
percent yield = (actual yield / theoretical yield) × 100
Excess reagent remaining
mass of excess remaining = (moles of excess initially - moles of excess consumed) × molar mass of excess reagent
Key molar masses from this lab
Fe: 55.85 g/mol
S: 32.06 g/mol
FeS: 87.91 g/mol
Cl₂: 70.90 g/mol
FeCl₃: 162.2 g/mol
Stoichiometry is how pharmaceutical companies calculate the exact quantities of raw materials needed to produce a batch of medication, and how chemical engineers size reactors and feedstocks for industrial processes. Percent yield matters in any manufacturing setting: a lower-than-expected yield means wasted reagent and higher cost per unit of product.
Students often assume the reagent present in the largest mass is the excess reagent. Mass alone does not determine which reagent is limiting; you must convert to moles and compare using the balanced equation's coefficients.
A percent yield above 100% does not mean extra product was created. It means something went wrong with the measurement: the product was wet, impure, or the balance was off.
Students sometimes confuse mole ratio with mass ratio. The mole ratio comes from the balanced equation coefficients (unitless). The mass ratio involves molar masses and has units of g/g.
The law of conservation of mass applies to the entire system. If you measure only the solid product and ignore gases that escaped, the mass will appear not to be conserved, but it is.
Limiting reagent identification is tested frequently. Expect a problem giving masses of two reactants and asking which runs out first.
Percent yield calculations appear on nearly every general chemistry exam. Know the formula cold.
Be prepared to explain why yield can be below 100% (side reactions, incomplete reactions, transfer losses, impurities) and why a value above 100% signals error rather than extra product.
The law of definite proportions may be tested graphically: you should be able to explain why a plot of mass of one element reacted vs. mass of product is linear and what the slope represents.
Excess reagent calculations (how much is left over) are a common follow-up to limiting reagent problems.
True or false: The limiting reagent is always the one with the smallest mass.
False. You must compare moles divided by stoichiometric coefficients, not raw mass.
Fill in the blank: Percent yield = (_____ / theoretical yield) × 100.
Actual yield.
True or false: If a reaction has a percent yield of 105%, the reaction produced more product than theoretically possible.
False. A yield above 100% indicates measurement error or impurities, not extra product.
Fill in the blank: The law of _____ states that a compound always contains the same elements in the same mass ratio.
Definite proportions (also called constant composition).
True or false: After a reaction, the mass of the excess reagent left over equals the starting mass of the excess reagent.
False. Some of the excess reagent is consumed; only the portion beyond what the limiting reagent required remains.
Q: 5.00 g of aluminium reacts with excess copper(II) sulfate. The balanced equation is 2 Al + 3 CuSO₄ → Al₂(SO₄)₃ + 3 Cu. How many grams of copper are produced?
A: Moles of Al = 5.00 / 26.98 = 0.1853 mol. From the equation, 2 mol Al produces 3 mol Cu. Moles of Cu = 0.1853 × (3/2) = 0.2780 mol. Mass of Cu = 0.2780 × 63.55 = 17.67 g.
Q: In a reaction between Fe and S to form FeS, you start with 5 g of Fe and 2 g of S. Which is the limiting reagent?
A: Moles of Fe = 5 / 55.85 = 0.0895 mol. Moles of S = 2 / 32.06 = 0.0624 mol. The balanced equation is Fe + S → FeS (1:1 ratio). S has fewer moles, so S is the limiting reagent.
Q: Using the data above, calculate the theoretical yield of FeS.
A: Moles of FeS = moles of limiting reagent (S) = 0.0624 mol. Mass of FeS = 0.0624 × 87.91 = 5.49 g.
Q: You collect 5.10 g of FeS from the reaction above. What is the percent yield?
A: Percent yield = (5.10 / 5.49) × 100 = 92.9%.
Q: A student obtains a percent yield of 108%. Give two possible explanations.
A: The product was not fully dried (retained moisture added mass), or the product was contaminated with unreacted reagent or another substance that increased the measured mass.
Limiting reagent analysis builds directly on mole-to-mass conversions and balanced equations from earlier in the course. The percent yield concept reappears in organic chemistry synthesis, where multi-step reactions multiply each step's yield and overall efficiency drops quickly. The law of definite proportions connects to empirical and molecular formula determination: if you know a compound's mass ratio, you can work backwards to its formula.
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