Stoichiometry, Limiting Reactants, and Percent Yield in General Chemistry

Stoichiometry and Limiting Reactants

Introduction to Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction based on balanced chemical equations.

• Stoichiometric coefficients in a balanced equation indicate the relative number of moles of each substance involved.

• Calculations often involve converting between grams, moles, and molecules using molar mass and Avogadro's number.

Limiting Reactant Concept

In many reactions, one reactant is used up before the others, limiting the amount of product that can be formed. This reactant is called the limiting reactant (or limiting reagent).

• Limiting Reactant: The reactant that produces the fewest moles of product; determines the maximum amount of product possible.

• Excess Reactant: Any reactant that remains after the limiting reactant is consumed.

• To identify the limiting reactant, calculate the amount of product formed from each reactant; the smallest value indicates the limiting reactant.

Example: Ammonia Synthesis

Given the reaction:

$\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$

• Suppose you have 3.0 mol $\mathrm{N_2}$ and 6.0 mol $\mathrm{H_2}$.

• Calculate moles of $\mathrm{NH_3}$ from each reactant:

  • From $\mathrm{N_2}$: $3.0\ \mathrm{mol\ N_2} \times \frac{2\ \mathrm{mol\ NH_3}}{1\ \mathrm{mol\ N_2}} = 6.0\ \mathrm{mol\ NH_3}$

  • From $\mathrm{H_2}$: $6.0\ \mathrm{mol\ H_2} \times \frac{2\ \mathrm{mol\ NH_3}}{3\ \mathrm{mol\ H_2}} = 4.0\ \mathrm{mol\ NH_3}$

• Limiting reactant: $\mathrm{H_2}$ (produces less $\mathrm{NH_3}$)

Theoretical Yield and Percent Yield

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from the limiting reactant, calculated using stoichiometry.

• Calculated by converting moles of product (from limiting reactant) to grams using molar mass.

Example:

• $4.0\ \mathrm{mol\ NH_3} \times 17.0\ \mathrm{g/mol} = 68\ \mathrm{g\ NH_3}$ (theoretical yield)

Percent Yield

The percent yield measures the efficiency of a reaction, comparing the actual amount of product obtained to the theoretical yield.

• Formula:

$\mathrm{Percent\ Yield} = \frac{\mathrm{Actual\ Yield}}{\mathrm{Theoretical\ Yield}} \times 100$

• Example: If 52 g $\mathrm{NH_3}$ is collected, percent yield is $\frac{52}{68} \times 100 = 76\%$

Stepwise Stoichiometry Calculations

General Steps

1. Balance the chemical equation.

2. Convert grams to moles for all reactants.

3. Calculate moles of product from each reactant using stoichiometry.

4. The lowest amount of product determines the limiting reactant and theoretical yield.

Example: Water Formation

Reaction: $2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O(g)}$

• Given: 150 g $\mathrm{H_2}$ (molar mass = 2.0 g/mol), 1500 g $\mathrm{O_2}$ (molar mass = 32.0 g/mol)

• Moles: $\mathrm{H_2}: 75\ \mathrm{mol}$, $\mathrm{O_2}: 47\ \mathrm{mol}$

• Product from $\mathrm{H_2}$: $75\ \mathrm{mol\ H_2} \times \frac{2\ \mathrm{mol\ H_2O}}{2\ \mathrm{mol\ H_2}} = 75\ \mathrm{mol\ H_2O}$

• Product from $\mathrm{O_2}$: $47\ \mathrm{mol\ O_2} \times \frac{2\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ O_2}} = 94\ \mathrm{mol\ H_2O}$

• Limiting reactant: $\mathrm{H_2}$

• Mass of water: $75\ \mathrm{mol\ H_2O} \times 18\ \mathrm{g/mol} = 1350\ \mathrm{g}$

• If 800 g $\mathrm{H_2O}$ is collected: $\frac{800}{1350} \times 100 = 59.3\%$ yield

Empirical and Percent Composition Calculations

Percent Composition by Mass

Percent composition expresses the mass percentage of each element in a compound.

• Formula: $\mathrm{Percent\ by\ mass} = \frac{\mathrm{mass\ of\ element\ in\ 1\ mol\ compound}}{\mathrm{molar\ mass\ of\ compound}} \times 100$

Example: Calcium nitrate, $\mathrm{Ca(NO_3)_2}$

• Molar mass: $40.1 + 2 \times 14.0 + 6 \times 16.0 = 164.1\ \mathrm{g/mol}$

• Mass of N: $2 \times 14.0 = 28.0\ \mathrm{g}$

• Percent N: $\frac{28.0}{164.1} \times 100 = 17.1\%$

Empirical Formula from Percent Composition

• Convert percent composition to grams (assume 100 g sample).

• Convert grams to moles for each element.

• Divide by the smallest number of moles to get the simplest ratio.

Example: Compound with 84.2 g, 85.6% C, 14.4% H

• Mass C: $0.856 \times 84.2 = 72.1\ \mathrm{g}$, moles C: $\frac{72.1}{12.0} = 6.01$

• Mass H: $0.144 \times 84.2 = 12.1\ \mathrm{g}$, moles H: $\frac{12.1}{1.0} = 12.1$

• Empirical formula: $\mathrm{CH_2}$

Additional Stoichiometry Examples

Example: Titanium Tetrachloride Synthesis

Reaction: $\mathrm{Ti(s) + 2Cl_2(g) \rightarrow TiCl_4(s)}$

• Given: 3.00 g Ti ($47.9\ \mathrm{g/mol}$), 6.00 g $\mathrm{Cl_2}$ ($70.9\ \mathrm{g/mol}$)

• Moles: Ti: $0.0626$, $\mathrm{Cl_2}$: $0.0846$

• Stoichiometry: $0.0626\ \mathrm{mol\ Ti} \rightarrow 0.0626\ \mathrm{mol\ TiCl_4}$

• $0.0846\ \mathrm{mol\ Cl_2} \times \frac{1\ \mathrm{mol\ TiCl_4}}{2\ \mathrm{mol\ Cl_2}} = 0.0423\ \mathrm{mol\ TiCl_4}$

• No limiting reagent if both produce the same amount.

• Mass: $0.0423\ \mathrm{mol\ TiCl_4} \times 189.7\ \mathrm{g/mol} = 8.02\ \mathrm{g}$

• If 7.7 g is collected: $\frac{7.7}{8.02} \times 100 = 96\%$ yield

Example: Aluminum Chloride Synthesis

Given: 1.50 kg $\mathrm{Al_2O_3}$ ($102\ \mathrm{g/mol}$), $133.5\ \mathrm{g/mol}$ $\mathrm{AlCl_3}$

• Moles $\mathrm{Al_2O_3}$: $\frac{1500}{102} = 14.7$

• Stoichiometry: $14.7\ \mathrm{mol\ Al_2O_3} \times \frac{2\ \mathrm{mol\ AlCl_3}}{1\ \mathrm{mol\ Al_2O_3}} = 29.4\ \mathrm{mol\ AlCl_3}$

• Mass: $29.4\ \mathrm{mol\ AlCl_3} \times 133.5\ \mathrm{g/mol} = 3.93\ \mathrm{kg}$ (theoretical yield)

• If 2.85 kg is collected: $\frac{2.85}{3.93} \times 100 = 72.6\%$ yield

Combustion Reactions and Limiting Reactants

Combustion of Hydrocarbons

Combustion reactions involve a hydrocarbon reacting with oxygen to produce carbon dioxide and water. In these reactions, the hydrocarbon is almost always the limiting reactant, while oxygen is typically in excess.

• Example: $4\ \mathrm{mol\ C_4H_{10}}$ (butane) reacts with $8\ \mathrm{mol\ CO_2}$

• Stoichiometry: $4\ \mathrm{mol\ C_4H_{10}} \times \frac{8\ \mathrm{mol\ CO_2}}{2\ \mathrm{mol\ C_4H_{10}}} = 16\ \mathrm{mol\ CO_2}$

Summary Table: Key Stoichiometry Concepts

Concept	Definition	Key Formula
Limiting Reactant	Reactant that determines the maximum amount of product	Calculate product from each reactant; smallest is limiting
Theoretical Yield	Maximum possible amount of product from limiting reactant	Use stoichiometry and molar mass
Percent Yield	Efficiency of reaction	$\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$
Percent Composition	Percent by mass of each element in a compound	$\frac{\text{mass of element}}{\text{molar mass of compound}} \times 100$

Additional info: In combustion reactions, oxygen is usually in excess, and the hydrocarbon is the limiting reactant. For empirical formula calculations, always convert mass percentages to moles and simplify to the smallest whole-number ratio.