Stoichiometry and solution chemistry

Stoichiometry (from the Greek for “element” and “measure”) is the calculation of quantities in chemical reactions. How much ore yields a tonne of iron? How much oxygen does a rocket need to burn its fuel? Which reactant runs out first, and how much product can we expect? A balanced equation answers all these questions, because its coefficients give the ratios in which substances react. This lesson builds the standard stoichiometry calculation, adds limiting reactants and yields, and extends the method to solutions, ending with titrations.

Mole ratios from a balanced equation

The coefficients in a balanced equation give mole ratios. For the synthesis of ammonia,

1 mol of N₂ reacts with 3 mol of H₂ to give 2 mol of NH₃. Each pair of coefficients is a conversion factor: (2 mol NH₃ / 1 mol N₂), (3 mol H₂ / 2 mol NH₃), and so on. This reaction, carried out industrially by the Haber–Bosch process at high pressure over an iron catalyst, makes the nitrogen fertilizers that feed roughly half the world's people. Alberta and the US Gulf Coast both host large ammonia plants, built near cheap natural gas, which supplies the hydrogen.

The standard stoichiometry calculation

Almost every stoichiometry problem follows one path: convert the given quantity to moles, use the mole ratio from the balanced equation, and convert moles of the target substance to the units wanted. Writing the conversion factors in a single chain makes the logic visible and lets the units check the work.

The same chain works for any pair of substances in the equation, reactants or products. The code below wraps it in a small function and applies it to the combustion of propane, the fuel in barbecue cylinders.

def convert(mass_given, M_given, coef_given, coef_target, M_target):
    """Mass of target from mass of given substance, via moles and the mole ratio."""
    mol_given = mass_given / M_given
    mol_target = mol_given * coef_target / coef_given
    return mol_target * M_target

# C3H8 + 5 O2 -> 3 CO2 + 4 H2O
M = {"C3H8": 44.10, "O2": 32.00, "CO2": 44.01, "H2O": 18.02}
coef = {"C3H8": 1, "O2": 5, "CO2": 3, "H2O": 4}

propane_g = 9000   # a full 20 lb (9 kg) cylinder
for target in ["O2", "CO2", "H2O"]:
    m = convert(propane_g, M["C3H8"], coef["C3H8"], coef[target], M[target])
    print(f"burning 9.0 kg of propane involves {m / 1000:5.1f} kg of {target}")

burning 9.0 kg of propane involves  32.7 kg of O2
burning 9.0 kg of propane involves  26.9 kg of CO2
burning 9.0 kg of propane involves  14.7 kg of H2O

The result surprises many people: burning 9 kg of propane produces about 27 kg of carbon dioxide. The extra mass comes from atmospheric oxygen, which combines with the carbon. Conservation of mass holds: 9.0 kg of propane plus 32.7 kg of oxygen (41.7 kg) equals 26.9 kg of carbon dioxide plus 14.7 kg of water (41.6 kg, with the difference due only to rounding).

Limiting reactants

Reactants are rarely mixed in exact stoichiometric proportions. The limiting reactant is the one that runs out first; it determines how much product can form. The other reactants are in excess, and some of each remains when the reaction stops. An everyday analogy: with 10 slices of bread and 3 slices of cheese, you can make only 3 cheese sandwiches (2 bread + 1 cheese each), no matter how much bread remains. Cheese limits.

To identify the limiting reactant, calculate how much product each reactant could make on its own; the reactant that gives the least product is limiting. Equivalently, divide the moles of each reactant by its coefficient: the smallest result identifies the limiting reactant.

def limiting(moles, coefs):
    """Return the limiting reactant and the 'reaction extent' (mol of reaction)."""
    extents = {s: moles[s] / coefs[s] for s in moles}
    lim = min(extents, key=extents.get)
    return lim, extents[lim]

# 2 Al + 3 Cl2 -> 2 AlCl3 ; start with 10.0 g Al and 35.0 g Cl2
M = {"Al": 26.98, "Cl2": 70.90, "AlCl3": 133.33}
moles = {"Al": 10.0 / M["Al"], "Cl2": 35.0 / M["Cl2"]}
coefs = {"Al": 2, "Cl2": 3}
lim, xi = limiting(moles, coefs)
product_g = xi * 2 * M["AlCl3"]
print(f"limiting reactant: {lim}")
print(f"AlCl3 formed: {product_g:.1f} g")
for s in moles:
    left = moles[s] - xi * coefs[s]
    print(f"{s} remaining: {max(left, 0) * M[s]:.2f} g")

limiting reactant: Cl2
AlCl3 formed: 43.9 g
Al remaining: 1.12 g
Cl2 remaining: 0.00 g

Theoretical, actual and percent yield

The theoretical yield is the maximum amount of product calculated from the limiting reactant. The actual yield, what is really obtained, is almost always less: some reactant forms side products, the reaction may not go to completion, and product is lost during separation and purification. The percent yield compares the two.

A percent yield above 100 % does not mean extra product was created. It signals an error, most often a product that is still wet with solvent or contaminated with another substance.

In multistep syntheses, yields multiply. A pharmaceutical made in ten steps, each with a 90 % yield, has an overall yield of only 0.90¹⁰ = 35 %. This is one reason chemists work hard to shorten synthetic routes.

Atom economy

Percent yield measures how efficiently a reaction was carried out, but not how wasteful the reaction is by design. Atom economy, a central idea of green chemistry, is the fraction of the mass of all reactants that ends up in the desired product, assuming 100 % yield.

An addition reaction such as ethene + water → ethanol has 100 % atom economy, since every atom ends up in the product. A substitution that produces a by-product, such as the classic aspirin synthesis that also produces acetic acid, has lower atom economy (about 75 %). Regulators and industry increasingly track metrics like these; the US Environmental Protection Agency's Green Chemistry Challenge Awards and Canadian green chemistry networks reward processes that design waste out.

Solution concentration: molarity

Many reactions are carried out in solution, where it is easier to measure volumes than masses. The most common concentration unit is molarity, M, the moles of solute per litre of solution (not per litre of solvent). BIOL-107 introduced molarity and dilution for laboratory solutions; here we use them for stoichiometry.

Solution stoichiometry

In a solution reaction, volume times molarity gives moles, and from there the standard stoichiometry chain applies unchanged.

Titration

A titration measures the concentration of a solution by reacting it with a solution of known concentration, the titrant, delivered from a burette. The equivalence point is reached when the titrant has supplied exactly the stoichiometric amount. An indicator, a substance that changes colour near the equivalence point, signals the end point; phenolphthalein, colourless in acid and pink in base, is the classic choice for strong acid–strong base titrations. Modern laboratories often use a pH meter or an automatic titrator instead.

Because sodium hydroxide absorbs water and carbon dioxide from the air, its solutions are standardized against a primary standard, a pure, stable solid that can be weighed accurately. Potassium hydrogen phthalate (KHP, KHC₈H₄O₄, 204.22 g/mol) is the usual choice; it reacts with NaOH in a 1 : 1 ratio.

# Standardize NaOH with KHP, then use it to find the concentration of an unknown HCl solution.
M_KHP = 204.22

trials = [   # (mass KHP in g, initial burette mL, final burette mL)
    (0.5103, 0.45, 25.52),
    (0.5048, 1.10, 25.88),
    (0.5121, 0.20, 25.37),
]
concs = []
for m, v0, v1 in trials:
    mol = m / M_KHP
    c = mol / ((v1 - v0) / 1000)
    concs.append(c)
    print(f"KHP {m:.4f} g, titre {v1 - v0:5.2f} mL -> c(NaOH) = {c:.4f} M")
c_naoh = sum(concs) / len(concs)
spread = max(concs) - min(concs)
print(f"mean c(NaOH) = {c_naoh:.4f} M (range {spread:.4f} M)")

# 25.00 mL of HCl needs 21.36 mL of this NaOH (1 : 1 ratio)
c_hcl = c_naoh * 21.36 / 25.00
print(f"c(HCl) = {c_hcl:.4f} M")

KHP 0.5103 g, titre 25.07 mL -> c(NaOH) = 0.0997 M
KHP 0.5048 g, titre 24.78 mL -> c(NaOH) = 0.0998 M
KHP 0.5121 g, titre 25.17 mL -> c(NaOH) = 0.0996 M
mean c(NaOH) = 0.0997 M (range 0.0001 M)
c(HCl) = 0.0852 M

Titrations appear well beyond the teaching laboratory: in quality control of vinegar and wine, in measuring water hardness, in clinical chemistry and in environmental monitoring. CHEM-102 develops titration curves and indicators in detail, and CHEM-211 treats titrations as precise analytical methods.