Chemistry · General chemistry I · Concept

Stoichiometry and theoretical yield

Stoichiometry predicts how much product a reaction can form, or how much of a reactant it needs. Convert the given mass to moles, use the mole ratio from the balanced equation, then convert back to grams. When two reactants are given, the one that runs out first sets the theoretical yield.

A concrete case

Making ammonia, counted two ways

Ammonia, ${NH}_{3}$ , is made from nitrogen and hydrogen, $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ . The equation counts molecules. One $N_{2}$ molecule and three $H_{2}$ molecules make two ${NH}_{3}$ molecules.

Scale every count up by Avogadro’s number and the same equation reads in moles. $1 mol$ of $N_{2}$ reacts with $3 mol$ of $H_{2}$ to make $2 mol$ of ${NH}_{3}$ . In grams, $28.02 g$ of nitrogen and $6.048 g$ of hydrogen make $34.07 g$ of ammonia.

Those masses are nowhere near the ratio $1$ : $3$ : $2$ . A nitrogen molecule is about $14$ times heavier than a hydrogen molecule, so equal numbers of them do not weigh the same. The coefficients count particles, and only moles count particles.

N_{2} + 3 H_{2} \to 2 {NH}_{3}

The rule

The mole ratio

The coefficients of a balanced equation give the mole ratio between any two substances in it. For $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ , the ratio of hydrogen to ammonia is $3 mol H_{2}$ to $2 mol {NH}_{3}$ . Written as a factor, a mole ratio turns moles of one substance into moles of another.

A mass-to-mass problem is a chain of three factors. The molar mass of the given substance turns its grams into moles. The mole ratio, wanted over given, turns those into moles of the wanted substance. Its molar mass turns them into grams.

n_{wanted} = n_{given} \times \frac{coefficient of wanted}{coefficient of given}

\begin{array}{c} n_{wanted} = n_{given} \\ \times \frac{coeff. of wanted}{coeff. of given} \end{array}

Why it works

Why the ratio works in moles and not in grams

A mole is a fixed number of particles, $6.022 \times 10^{23}$ . Three molecules for every two is therefore three moles for every two, since both counts are multiplied by the same number.

A gram holds no fixed number of particles. One gram of $H_{2}$ holds about $14$ times as many molecules as one gram of $N_{2}$ . A ratio of grams would change with every pair of substances, so it cannot come from the coefficients.

Mass still balances, because atoms are conserved. In the ammonia case, $28.02 g + 6.048 g$ $= 34.07 g$ on each side of the arrow.

$N_{2}$ $+ 3 H_{2}$ $\to 2 {NH}_{3}$ read three ways
	$N_{2}$	$H_{2}$	${NH}_{3}$
Molecules	$1$	$3$	$2$
Moles	$1$	$3$	$2$
Grams	$28.02$	$6.048$	$34.07$

The method

From grams of one substance to grams of another

The chain runs grams, moles, moles, grams, and each step below adds one link.

1. Write the balanced equation, with states.

2. Find the molar masses of the given substance and the wanted one.

3. Convert the given mass to moles.

4. Multiply by the mole ratio, with the wanted substance on top.

5. Convert to the unit the question asks for. Use the molar mass for grams, or $V = n / M$ for the volume of a solution.

6. Write the whole thing as one chain, and round once, at the end.

The factors in a stoichiometry chain
From	To	Factor
grams of $A$	moles of $A$	$1 mol$ $A / M (A) g$ $A$
moles of $A$	moles of $B$	coefficient of B / coefficient of A
moles of $B$	grams of $B$	$M (B) g B / 1 mol B$
volume of a solution of $A$	moles of $A$	$n = M V$ , with $V$ in liters

When two amounts are given

If the amounts of two reactants are given, only one of them can run out first. Convert both to moles and divide each by its coefficient. The smaller result belongs to the limiting reactant, and the theoretical yield is calculated from it alone.

For $5.00 g$ of ${CaCO}_{3}$ and $0.200 mol$ of $HCl$ , the carbonate is $5.00 \div 100.09$ $= 0.04996 mol$ , so ${CaCO}_{3}$ gives $0.04996 \div 1$ and $HCl$ gives $0.200 \div 2 = 0.100$ . The carbonate runs out first.

reaction extent = min (\frac{n ({CaCO}_{3})}{1}, \frac{n (HCl)}{2})

Where it breaks

A theoretical yield assumes four things. The written reaction is the only one, it goes to completion, the reactants are pure, and nothing is lost. A real yield is lower. The ammonia reaction itself stops at an equilibrium and never converts all of the gas.

The coefficients are not a ratio of grams. $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ does not mean $3 g$ of hydrogen for every gram of nitrogen. $28.02 g$ of $N_{2}$ takes $6.048 g$ of $H_{2}$ , about $0.216 g$ per gram.

The ratio comes only from a balanced equation. With an unbalanced one, every number downstream is wrong, however careful the arithmetic.

Watch the particle. Hydrogen gas is $H_{2}$ at $2.016 g / mol$ , not H at $1.008 g / mol$ , and a mole ratio refers to the formulas as written in the equation.

Check your answer

Check the direction of the ratio. $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ needs more moles of $H_{2}$ than it makes of ${NH}_{3}$ , so a chain from ammonia to hydrogen should increase the moles by half.

Check mass. For a complete reaction, the masses of the reactants used add up to the masses of the products. In the ammonia example, $8.88 g$ of $H_{2}$ and $41.1 g$ of $N_{2}$ make $50.0 g$ of ${NH}_{3}$ .

Run your answer through the chain in reverse, and you should return to the given amount.

In a Chemistry box, write the balanced equation on the first row and each chain on a row below it. The Checker names every factor. For the ammonia ratio it says “3 mol H₂ / 2 mol NH₃: matches the coefficients in row 1 (3 H₂ : 2 NH₃).”

A ratio turned upside down reads ✗ with “row 1 has 3 H₂ for every 2 NH₃.”

Without the equation above it, a mole ratio reads ? with “no reaction is written above this row, so this mole ratio can’t be checked.”

The extent of reaction

IUPAC describes a reaction’s progress with one number, the extent of reaction, $ξ$ , in moles. Each substance changes by its stoichiometric number times $ξ$ . Stoichiometric numbers are negative for reactants and positive for products, so $n (B) = n_{0} (B) + ν (B) ξ$ .

A reactant runs out when its amount reaches zero, at $ξ = n_{0}$ divided by the size of its stoichiometric number. The reaction stops at the smallest such $ξ$ . That is why dividing each amount by its coefficient finds the limiting reactant.

For the chalk example, $ξ$ stops at $0.04996 mol$ , set by the carbonate. ${CO}_{2}$ has $ν = 1$ , so $0.04996 mol$ of it forms.

n_{B} = n_{B, 0} + ν_{B} ξ

Common mistakes

Reading the coefficients as grams. $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ does not mean $3 g$ of hydrogen per gram of nitrogen. The coefficients count moles, so $28.02 g$ of $N_{2}$ takes $6.048 g$ of $H_{2}$ .
Skipping the mole ratio. Converting $50.0 g$ of ${NH}_{3}$ to $2.94 mol$ and calling that $2.94 mol$ of $H_{2}$ assumes a $1$ : $1$ ratio. The equation gives $3 mol H_{2}$ per $2 mol {NH}_{3}$ , so the answer is $4.40 mol$ , or $8.88 g$ .
Turning the mole ratio over. $2 mol H_{2}$ over $3 mol {NH}_{3}$ gives $1.96 mol$ of hydrogen instead of $4.40 mol$ . The wanted substance goes on top, so the given substance cancels.
Working from an unbalanced equation. Reading ${CaCO}_{3} + HCl$ as $1$ : $1$ says $5.00 g$ of carbonate needs $0.0500 mol$ of acid. The balanced equation has 2HCl, so it needs $0.0999 mol$ .
Using the atomic mass for a diatomic gas. Hydrogen gas is $H_{2}$ at $2.016 g / mol$ . Using $1.008 g / mol$ makes the ammonia example’s answer $4.44 g$ , half the true $8.88 g$ .
Reporting a theoretical yield as a measurement. $2.20 g$ of ${CO}_{2}$ is the most the reaction can make under its assumptions. The mass actually collected is the actual yield, and actual over theoretical is the percent yield.

Key terms

Limiting reactant: The reactant that runs out first. It sets the most product the reaction can make. The other reactants are left over, in excess.
Theoretical yield: The most product a reaction can make, calculated from the limiting reactant, assuming the reaction goes to completion with no losses or side reactions.
Molar mass: The mass of one mole of a substance, in $g / mol$ . It converts between grams and moles: $moles$ $= mass \div molar mass$ .
Mole ratio: A ratio of moles of two substances taken from the coefficients of a balanced equation, such as $2 mol H_{2}$ : $1 mol O_{2}$ . It converts moles of one substance into moles of another.