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Moles, balancing and stoichiometry

Stoichiometry is chemistry's bookkeeping: with a balanced equation and the molar masses you know how much of each substance reacts and how much forms. This topic explains the mole, how to balance an equation (including redox and ionic ones) and how to find the limiting reactant and the yield.

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The mole and molar mass

A mole is a fixed number of particles: exactly $N_{A} = 6.02214076 \times 1 0^{23}$ (Avogadro's constant, exact by definition of the SI). The molar mass $M$ of a substance is the mass of one mole, in g/mol, and is obtained by adding the atomic masses of the atoms in its formula:

M = i \sum n_{i} A_{i} n = \frac{m}{M} N = n N_{A} % X = \frac{n _{X} A _{X}}{M} \cdot 100

When counting atoms, a subscript multiplies the symbol or parenthesis right before it, and in a hydrate the number before the water multiplies the whole molecule: CuSO₄·5H₂O has 10 H and 9 O. The atomic masses are IUPAC's standard ones.

Balancing an equation

Atoms are neither created nor destroyed in a reaction (law of conservation of mass): each element must appear the same number of times on each side. Only the coefficients change; the subscripts do not, because they would change the substance. There are three ways to find them:

Trial and error: set the most complex species to 1 and adjust the elements in order (metals, nonmetals, H and finally O). If fractions appear, multiply everything by the least common multiple. It is quick for simple equations, but gets stuck when an element is in several species without a coefficient.
Algebraic: a letter is put on each coefficient and each element gives a linear equation. There is one more unknown than independent equations, so one is fixed (for example at 1) and the rest are solved.
Matrix: the same equations in a matrix (rows = elements, columns = species, with a minus sign on the products); the coefficients are the null-space vector, scaled to integers.

In an ionic equation charge is also conserved: the sum of the charges must be equal on both sides. In redox equations the physical idea is that the electrons lost by the species that is oxidized (its oxidation number goes up) are the ones gained by the one that is reduced (its oxidation number goes down):

Oxidation-number rule	Value
Uncombined element (Cu, O₂, N₂)	0
Single-atom ion	its charge (Fe³⁺: +3)
Oxygen in compounds	−2 (except peroxides)
Hydrogen with nonmetals	+1
Sum in a molecule or ion	0, or the ion's charge

From one substance to another

The coefficients of the balanced equation are ratios between moles, not grams. Every stoichiometric calculation goes through them: from grams (or liters of gas, or particles) to moles of the given, from those moles to moles of the unknown with the coefficient ratio, and back to grams.

Map of conversions: the only bridge between two substances is the coefficient ratio.

For a gas the ideal-gas equation is used, $P V = n RT$ with $R = 0.082057$ L·atm/(mol·K). At 0 °C and 1 atm one mole takes up $RT / P = 22.41$ L.

Limiting reactant, excess and yield

If amounts of two or more reactants are given, they are almost never in the exact proportion: one runs out first (the limiting one) and sets how much product forms; some of the others is left over (they are in excess). Two equivalent ways to find it:

Divide the moles of each reactant by its coefficient: the smallest quotient is the limiting one.
Compute how much product each reactant would give if it were used up: the one that gives less is the limiting one.

With 2 N₂ and 3 H₂ only one N₂ reacts: H₂ is the limiting reactant and one N₂ is left over.

rendimiento porcentual = \frac{masa real obtenida}{rendimiento te o ˊ rico} \cdot 100

The theoretical yield is the mass of product the limiting reactant gives if it reacts completely. In practice less is obtained (incomplete reactions, side reactions, losses when separating), and the ratio between what is obtained and the theoretical amount is the percent yield.

Common mistakes

Using the unbalanced equation: every ratio comes out wrong.
Applying the coefficient ratio to grams instead of moles.
Taking as limiting the reactant with fewer grams or fewer moles, without dividing by its coefficient.
Changing subscripts to balance (H₂O₂ is not H₂O).
Confusing CO (carbon monoxide) with Co (cobalt): capitals matter.
Forgetting the charge when balancing an ionic equation.

Worked examples

Example 1Molar mass of a hydrate

Find the molar mass of copper sulfate pentahydrate, CuSO₄·5H₂O, its percent composition and how many moles and formula units there are in 10 g.

Solution

Atoms: Cu 1, S 1, O 4 + 5 = 9, H 10.
$M = 63.546 + 32.06 + 9 \cdot 15.999 + 10 \cdot 1.008 = 249.677$ g/mol.
Composition: Cu 63.546/249.677 = 25.45%, S 12.84%, O 143.991/249.677 = 57.67%, H 4.037%. The water of hydration ( $5 \cdot 18.015 = 90.08$ g/mol) is 36% of the mass.
In 10 g: $n = 10/249.677 = 0.040052$ mol and $N = 0.040052 \cdot 6.02214 \times 1 0^{23} = 2.412 \times 1 0^{22}$ formula units.

M = 249.677

g/mol; 10 g are 0.040052 mol (

2.412 \times 1 0^{22}

formula units).

Solve it in Simuveliq

In Chemistry → Molar mass, type CuSO4·5H2O (or CuSO4.5H2O) in Chemical formula, with 10 g in Sample mass. The table gives atoms, atomic mass, contribution and %, and the four steps count the atoms, add, get the percentages and convert to moles.

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Example 2A redox equation by the algebraic method

Balance Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O.

Solution

Coefficients $a$ Cu + $b$ HNO₃ → $c$ Cu(NO₃)₂ + $d$ NO + $e$ H₂O. One balance per element: Cu: $a = c$ ; H: $b = 2 e$ ; N: $b = 2 c + d$ ; O: $3 b = 6 c + d + e$ .
With $e = 1$ : $b = 2$ , and from N and O, $d = 2 - 2 c$ and $6 = 6 c + 2 - 2 c + 1$ → $c = 3/4$ , $d = 1/2$ , $a = 3/4$ . Multiplying by 4: $a = 3$ , $b = 8$ , $c = 3$ , $d = 2$ , $e = 4$ .
Redox reading: Cu goes from 0 to +2 (loses 2 e⁻) and N goes from +5 in HNO₃ to +2 in NO (gains 3 e⁻). Electrons lost = gained: $3 \cdot 2 = 2 \cdot 3$ , so 3 Cu and 2 NO; the other 6 nitrates do not change oxidation number.

3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O.

Solve it in Simuveliq

In Chemistry → Equation balancing, in Examples, choose «Cu + HNO3 -> Cu(NO3)2 + NO + H2O». The balanced equation comes out with its check table (Cu 3, H 8, N 8, O 24 on each side).
In Method, «Trial and error» shows how it gets stuck (N is in three species) and tries values up to 8/3; «Algebraic» writes the four equations and fixes e = 1, as above.

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Example 3Ionic equation: charge and electrons

Balance in acid medium MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O.

Solution

Oxidation numbers: Mn goes from +7 to +2 (gains 5 e⁻) and each Fe from +2 to +3 (loses 1 e⁻). To match electrons: 1 MnO₄⁻ with 5 Fe²⁺.
Oxygen: the 4 O of permanganate leave as 4 H₂O, which need 8 H⁺.
Charge check: left $- 1 + 5 (+ 2) + 8 (+ 1) = + 17$ ; right $+ 2 + 5 (+ 3) = + 17$ .

MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O (charge +17 on each side).

Solve it in Simuveliq

In Examples, «MnO4^- + Fe^2+ + H^+ -> Mn^2+ + Fe^3+ + H2O» (ions are written with ^). The check adds a «charge» row with 17 and 17, and trial and error tries Fe²⁺ = 1, 2, 3, 4 until the charge balances with 5.

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Example 4Limiting reactant in the synthesis of ammonia

28 g of N₂ and 10 g of H₂ are mixed: N₂ + 3 H₂ → 2 NH₃. Which is limiting, how much NH₃ forms and how much of the other is left over?

Solution

Moles: N₂ $28/28.014 = 0.9995$ mol; H₂ $10/2.016 = 4.960$ mol.
Divided by the coefficients: N₂ $0.9995/1 = 0.9995$ ; H₂ $4.960/3 = 1.653$ . The smaller is N₂'s: it is limiting, even though there are more grams of N₂ than of H₂.
NH₃: $0.9995 \cdot 2 = 1.999$ mol $\times 17.031$ g/mol $= 34.04$ g.
H₂ that reacts: $3 \cdot 0.9995 = 2.999$ mol; $4.960 - 2.999 = 1.962$ mol $= 3.955$ g are left. Mass check: $28 + 10 = 34.04 + 3.955$ (38 g on each side).

N₂ is limiting; 34.04 g of NH₃ form and 3.955 g of H₂ are left over.

Solve it in Simuveliq

In Chemistry → Stoichiometry, equation «N2 + H2 -> NH3» (it balances itself) with 28 g of N2 and 10 g of H2 (the button sets them).
The cards give Limiting reactant N2 and Theoretical yield 34.04 g of NH3; the species table shows initial, reacting and final amounts, and the steps find the limit by both methods and the excess.

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Example 5Percent yield in a blast furnace

500 g of Fe₂O₃ react with excess CO: Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂. 300 g of iron are obtained. What is the percent yield?

Solution

$M (Fe_{2} O_{3}) = 2 \cdot 55.845 + 3 \cdot 15.999 = 159.687$ g/mol, so $n = 500/159.687 = 3.131$ mol.
Theoretical yield: $3.131 \cdot 2 = 6.262$ mol of Fe $\times 55.845 = 349.7$ g.
Percent yield: $300/349.7 \cdot 100 = 85.78%$ .

Theoretical 349.7 g of Fe; yield 85.78%.

Solve it in Simuveliq

The button loads «Fe2O3 + CO -> Fe + CO2» with 500 g of Fe2O3 and, in Actual yield of, Fe with Mass obtained = 300 g. Step 4 computes the percent yield, 85.78%.

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Example 6A reactant given as a gas volume

5 L of methane measured at 0 °C and 1 atm are burned: CH₄ + 2 O₂ → CO₂ + 2 H₂O. How many grams of CO₂ and water form?

Solution

$n = P V / RT = (1) (5) / (0.082057 \cdot 273.15) = 0.2231$ mol of CH₄ (5/22.41).
Only the methane is given: oxygen is assumed in excess.
CO₂: $0.2231 \cdot 1 \cdot 44.009 = 9.817$ g. H₂O: $0.2231 \cdot 2 \cdot 18.015 = 8.037$ g.

0.2231 mol of CH₄ → 9.817 g of CO₂ and 8.037 g of H₂O.

Solve it in Simuveliq

In Stoichiometry, «CH4 + O2 -> CO2 + H2O · 5 L CH4 (gas)»: with the «L of gas» unit, Gas temperature (0 °C) and Gas pressure (1 atm) appear. The Starting reactant card gives 0.2231 mol of CH4 and the theoretical yield, 9.817 g of CO2.

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Moles, balancing and stoichiometry

The mole and molar mass

Balancing an equation

From one substance to another

Limiting reactant, excess and yield

Common mistakes

Worked examples

Example 1Molar mass of a hydrate

Solution

Solve it in Simuveliq

Example 2A redox equation by the algebraic method

Solution

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Example 3Ionic equation: charge and electrons

Solution

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Example 4Limiting reactant in the synthesis of ammonia

Solution

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Example 5Percent yield in a blast furnace

Solution

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Example 6A reactant given as a gas volume

Solution

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