Stoichiometry: The quantitative study of reactants and products in chemical reactions

From Greek:

• stoicheion = "element"
• metron = "measure"

What stoichiometry tells us:

• How much product forms from given reactants
• How much reactant needed to make desired product
• Which reactant limits the reaction
• Theoretical vs. actual yield

Foundation: Balanced chemical equations

Mole Ratios from Balanced Equations

Balanced equation provides mole ratios

Example:

$N2\text{(g)} + 3H2\text{(g)} \rightarrow 2NH3\text{(g)}$

Mole ratios:

• 1 mol N₂ : 3 mol H₂
• 1 mol N₂ : 2 mol NH₃
• 3 mol H₂ : 2 mol NH₃

These ratios are exact (not measured)

• Can have infinite sig figs
• Used as conversion factors

Interpretation:

• 1 mole of N₂ reacts with 3 moles of H₂
• 1 mole of N₂ produces 2 moles of NH₃
• 3 moles of H₂ produce 2 moles of NH₃

Key concept: Coefficients give mole-to-mole relationships

Basic Stoichiometry Calculations

Mass-to-Mass Calculations

Problem type: Given mass of reactant, find mass of product

General strategy:

$\text{Mass A} \xrightarrow{\text{÷ Molar mass}} \text{Moles A} \xrightarrow{\text{Mole ratio}} \text{Moles B} \xrightarrow{\text{× Molar mass}} \text{Mass B}$

Example:

How many grams of NH₃ can be produced from 28.0 g of N₂?

$N2 + 3H2 \rightarrow 2NH3$

Step 1: Convert mass N₂ to moles

$n_{N_2} = \frac{28.0 \text{ g}}{28.0 \text{ g/mol}} = 1.00 \text{ mol N}_2$

Step 2: Use mole ratio to find moles NH₃

$n_{NH_3} = 1.00 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 2.00 \text{ mol NH}_3$

Step 3: Convert moles NH₃ to mass

$\text{mass}_{NH_3} = 2.00 \text{ mol} \times 17.0 \text{ g/mol} = 34.0 \text{ g}$

Answer: 34.0 g NH₃

Mole-to-Mole Calculations

Simplest type: Already have moles

Example:

How many moles of O₂ needed to react with 5.0 mol glucose?

$C6H12O6 + 6O2 \rightarrow 6CO2 + 6H2O$

Solution:

$n_{O_2} = 5.0 \text{ mol C}_6\text{H}_{12}\text{O}_6 \times \frac{6 \text{ mol O}_2}{1 \text{ mol C}_6\text{H}_{12}\text{O}_6} = 30 \text{ mol O}_2$

Answer: 30 mol O₂

Mass-to-Particles Calculations

Use Avogadro's number: $N_A = 6.022 \times 10^{23}$ particles/mol

Example:

How many molecules of H₂O form from 9.0 g H₂?

$2H2 + O2 \rightarrow 2H2O$

Step 1: Moles H₂

$n_{H_2} = \frac{9.0 \text{ g}}{2.0 \text{ g/mol}} = 4.5 \text{ mol}$

Step 2: Moles H₂O

$n_{H_2O} = 4.5 \text{ mol H}_2 \times \frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2} = 4.5 \text{ mol}$

Step 3: Convert to molecules

$N = 4.5 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 2.7 \times 10^{24} \text{ molecules}$

Answer: $2.7 \times 10^{24}$ molecules H₂O

Limiting Reactants

Limiting reactant: Reactant that is completely consumed first, limiting the amount of product

Excess reactant: Reactant that remains after the reaction stops

Analogy: Making sandwiches

• 10 slices bread + 6 slices cheese
• 1 sandwich needs 2 bread + 1 cheese
• Can make only 5 sandwiches (limited by bread)
• Bread = limiting reactant
• Cheese = excess (1 slice left over)

Identifying Limiting Reactant

Method 1: Calculate product from each reactant

Compare: Which reactant produces less product?

• That reactant is limiting

Example:

$2Al + 3Cl2 \rightarrow 2AlCl3$

Given: 5.0 mol Al and 6.0 mol Cl₂

From Al:

$n_{AlCl_3} = 5.0 \text{ mol Al} \times \frac{2 \text{ mol AlCl}_3}{2 \text{ mol Al}} = 5.0 \text{ mol AlCl}_3$

From Cl₂:

$n_{AlCl_3} = 6.0 \text{ mol Cl}_2 \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2} = 4.0 \text{ mol AlCl}_3$

Cl₂ produces less product → Cl₂ is limiting reactant

Al is in excess

Actual product formed: 4.0 mol AlCl₃ (limited by Cl₂)

Method 2: Compare mole ratios

Calculate: $\frac{\text{moles available}}{\text{coefficient}}$ for each reactant

Smallest ratio → limiting reactant

Example (same reaction):

$2Al + 3Cl2 \rightarrow 2AlCl3$

Given: 5.0 mol Al and 6.0 mol Cl₂

For Al:

$\frac{5.0 \text{ mol}}{2} = 2.5$

For Cl₂:

$\frac{6.0 \text{ mol}}{3} = 2.0$

Cl₂ has smaller ratio → Cl₂ is limiting ✓

Method 3: Calculate how much of one reactant is needed

Calculate: How much of reactant B is needed to react with all of reactant A?

Compare: Available B vs. needed B

• If available < needed → B is limiting
• If available > needed → A is limiting

Example (same reaction):

Given: 5.0 mol Al and 6.0 mol Cl₂

How much Cl₂ needed for all 5.0 mol Al?

$n_{Cl_2 \text{ needed}} = 5.0 \text{ mol Al} \times \frac{3 \text{ mol Cl}_2}{2 \text{ mol Al}} = 7.5 \text{ mol Cl}_2$

Available Cl₂: 6.0 mol Needed Cl₂: 7.5 mol

6.0 < 7.5 → Not enough Cl₂ → Cl₂ is limiting ✓

Calculating Excess Reactant Remaining

After reaction, how much excess reactant left?

Strategy:

1. Use limiting reactant to find how much excess reactant consumed
2. Subtract consumed from initial amount

Example:

$2Al + 3Cl2 \rightarrow 2AlCl3$

Given: 5.0 mol Al, 6.0 mol Cl₂ (Cl₂ is limiting)

How much Al remains?

Step 1: How much Al consumed by 6.0 mol Cl₂?

$n_{Al \text{ used}} = 6.0 \text{ mol Cl}_2 \times \frac{2 \text{ mol Al}}{3 \text{ mol Cl}_2} = 4.0 \text{ mol Al}$

Step 2: Calculate Al remaining

$n_{Al \text{ remaining}} = 5.0 \text{ mol initial} - 4.0 \text{ mol used} = 1.0 \text{ mol Al}$

Answer: 1.0 mol Al remains (excess)

Theoretical, Actual, and Percent Yield

Theoretical Yield

Theoretical yield: Maximum amount of product that can form from given reactants

• Calculated from stoichiometry
• Assumes 100% efficiency
• Based on limiting reactant

Always use limiting reactant for theoretical yield!

Actual Yield

Actual yield: Amount of product actually obtained in lab

• Measured experimentally
• Always less than theoretical yield
• Due to side reactions, losses, incomplete reactions

Percent Yield

Percent yield: Efficiency of reaction

$\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

Example:

Theoretical yield: 50.0 g product Actual yield: 42.5 g product

$\text{Percent yield} = \frac{42.5 \text{ g}}{50.0 \text{ g}} \times 100\% = 85.0\%$

Interpretation: Reaction is 85% efficient

Why actual < theoretical?

1. Side reactions: Reactants form unwanted products
2. Incomplete reactions: Not all reactants convert to products
3. Losses during purification: Some product lost in transfer, filtering, etc.
4. Impure reactants: Starting materials not 100% pure

Percent yield range:

• 0-100% (cannot exceed 100% with pure products)
• Good lab technique: 80-95%
• Industrial processes often optimize for high percent yield

Using Percent Yield

If percent yield known, can calculate actual yield:

$\text{Actual yield} = \frac{\text{percent yield}}{100} \times \text{theoretical yield}$

Example:

If theoretical yield = 30.0 g and reaction is 75% efficient:

$\text{Actual yield} = \frac{75}{100} \times 30.0 \text{ g} = 22.5 \text{ g}$

Solution Stoichiometry

For reactions in solution, use molarity

Molarity (M):

$M = \frac{\text{moles solute}}{\text{liters solution}}$

Rearranged:

$\text{moles} = M \times V_{\text{liters}}$

Strategy for Solution Stoichiometry

Given volumes and molarities, find product:

1. Calculate moles of each reactant: $n = M \times V$
2. Identify limiting reactant
3. Use stoichiometry to find product moles
4. Convert to mass (if needed) or molarity (if in solution)

Example:

$2AgNO3\text{(aq)} + CaCl2\text{(aq)} \rightarrow 2AgCl\text{(s)} + Ca(NO3)2\text{(aq)}$

Mix 50.0 mL of 0.200 M AgNO₃ with 30.0 mL of 0.100 M CaCl₂

How many grams of AgCl precipitate?

Step 1: Calculate moles of reactants

$n_{AgNO_3} = 0.200 \text{ M} \times 0.0500 \text{ L} = 0.0100 \text{ mol}$

$n_{CaCl_2} = 0.100 \text{ M} \times 0.0300 \text{ L} = 0.00300 \text{ mol}$

Step 2: Identify limiting reactant

From AgNO₃:

$n_{AgCl} = 0.0100 \text{ mol AgNO}_3 \times \frac{2 \text{ mol AgCl}}{2 \text{ mol AgNO}_3} = 0.0100 \text{ mol}$

From CaCl₂:

$n_{AgCl} = 0.00300 \text{ mol CaCl}_2 \times \frac{2 \text{ mol AgCl}}{1 \text{ mol CaCl}_2} = 0.00600 \text{ mol}$

CaCl₂ produces less → CaCl₂ is limiting

Step 3: Calculate mass of AgCl

$\text{mass} = 0.00600 \text{ mol} \times 143.5 \text{ g/mol} = 0.861 \text{ g AgCl}$

Answer: 0.861 g AgCl precipitates

Consecutive Reactions

Sometimes reactions occur in sequence

Product of first reaction → reactant for second reaction

Example:

$S\text{(s)} + O2\text{(g)} \rightarrow SO2\text{(g)}$ (Reaction 1)

$2SO2\text{(g)} + O2\text{(g)} \rightarrow 2SO3\text{(g)}$ (Reaction 2)

If given amount of S, find SO₃:

Strategy: Chain the stoichiometry

$\text{S} \xrightarrow{\text{Reaction 1}} \text{SO}_2 \xrightarrow{\text{Reaction 2}} \text{SO}_3$

Use mole ratios from both equations

Summary of Stoichiometry Steps

General problem-solving strategy:

1. Write balanced equation

   • Double-check coefficients

2. Convert given to moles

   • Use molar mass (g → mol)
   • Use molarity × volume (M × L → mol)
   • Use Avogadro's number (particles → mol)

3. Identify limiting reactant (if multiple reactants given)

   • Calculate product from each reactant
   • Smallest product → that reactant is limiting
   • OR use ratio method

4. Use mole ratios (from balanced equation)

   • Convert moles of limiting reactant to moles of product

5. Convert to desired units

   • Mass (mol → g using molar mass)
   • Particles (mol → particles using $N_A$)
   • Volume (if gas at STP: 22.4 L/mol)
   • Molarity (if in solution)

6. Calculate percent yield (if actual yield given)

   • Compare actual to theoretical

Common Stoichiometry Errors

❌ Mistake 1: Using wrong mole ratio

• Check coefficients carefully
• Write ratio as fraction

❌ Mistake 2: Forgetting to identify limiting reactant

• If given multiple reactants, MUST determine limiting
• Calculate product from each reactant

❌ Mistake 3: Using excess reactant for calculations

• Always use limiting reactant for product calculations
• Excess reactant doesn't limit product

❌ Mistake 4: Rounding too early

• Keep extra sig figs during calculation
• Round at final answer

❌ Mistake 5: Unit errors

• Track units throughout
• Cancel units properly
• Convert volumes to liters for molarity

Tips for Success

✓ Always start with balanced equation

✓ Write out conversion factors explicitly

✓ Track units through calculation

✓ For limiting reactant: Calculate product from EACH reactant

✓ Check reasonableness of answer

• Product mass shouldn't exceed reactant mass (usually)
• Percent yield can't exceed 100%

✓ Use dimensional analysis

• Set up so units cancel

✓ For solution problems: V must be in liters for molarity

(c) Excess Al remaining: Al consumed = 0.113 mol Cl₂ × (2 Al/3 Cl₂) = 0.0753 mol Al remaining = 0.200 - 0.0753 = 0.125 mol = 3.37 g

Find: (a) Theoretical yield of H₂, (b) Percent yield

Part (a): Calculate theoretical yield

Step 1: Convert mass Al to moles

$n_{Al} = \frac{\text{mass}}{\text{molar mass}}$

$n_{Al} = \frac{15.0 \text{ g}}{27.0 \text{ g/mol}}$

$n_{Al} = 0.556 \text{ mol Al}$

Step 2: Use stoichiometry to find moles H₂

From balanced equation:

$2Al + 6HCl \rightarrow 2AlCl3 + 3H2$

Mole ratio: $\frac{3 \text{ mol H}_2}{2 \text{ mol Al}}$

$n_{H_2} = 0.556 \text{ mol Al} \times \frac{3 \text{ mol H}_2}{2 \text{ mol Al}}$

$n_{H_2} = 0.556 \times 1.5$

$n_{H_2} = 0.834 \text{ mol H}_2$

Step 3: Convert moles H₂ to volume at STP

At STP: 1 mol gas = 22.4 L (molar volume)

$V_{H_2} = n_{H_2} \times 22.4 \text{ L/mol}$

$V_{H_2} = 0.834 \text{ mol} \times 22.4 \text{ L/mol}$

$V_{H_2} = 18.7 \text{ L}$

Answer (a):

$\boxed{\text{Theoretical yield} = 18.7 \text{ L H}_2}$

This is the maximum H₂ that could be produced from 15.0 g Al

Part (b): Calculate percent yield

Step 4: Use percent yield formula

$\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

Given:

• Actual yield = 16.8 L (measured in lab)
• Theoretical yield = 18.7 L (calculated above)

Calculate:

$\text{Percent yield} = \frac{16.8 \text{ L}}{18.7 \text{ L}} \times 100\%$

$\text{Percent yield} = 0.898 \times 100\%$

$\text{Percent yield} = 89.8\%$

Answer (b):

$\boxed{\text{Percent yield} = 89.8\%}$

Or: $\boxed{90\%}$ (2 sig figs)

Interpretation:

What does 89.8% yield mean?

• Reaction produced 89.8% of theoretical maximum
• This is actually quite good efficiency for lab reaction
• 10.2% of potential H₂ not collected

Why is actual < theoretical?

Possible reasons:

1. Gas escaped during collection

   • Not all H₂ captured in collection apparatus
   • Small leaks in setup

2. Incomplete reaction

   • Not all Al dissolved
   • Oxide coating on Al surface prevents complete reaction

3. Impure aluminum

   • Al sample may contain impurities
   • Less than 15.0 g pure Al actually present

4. Measurement errors

   • Volume measurement not exact
   • Temperature or pressure not exactly STP

5. Side reactions

   • Some Al reacts with O₂ in air instead of HCl

In industrial processes:

• 89.8% would be excellent yield
• Industries optimize conditions for high percent yield
• Economic motivation (more product per reactant)

Summary of calculations:

Stoichiometry path:

$15.0 \text{ g Al} \xrightarrow{÷27.0} 0.556 \text{ mol Al} \xrightarrow{×\frac{3}{2}} 0.834 \text{ mol H}_2 \xrightarrow{×22.4} 18.7 \text{ L H}_2$

Percent yield:

$\frac{16.8 \text{ L (actual)}}{18.7 \text{ L (theoretical)}} \times 100\% = 89.8\%$

Comparison table:

Additional practice:

What if we wanted mass of H₂ instead of volume?

$\text{mass}_{H_2} = 0.834 \text{ mol} \times 2.0 \text{ g/mol} = 1.67 \text{ g H}_2$ (theoretical)

$\text{mass}_{H_2 \text{ actual}} = 1.67 \text{ g} \times 0.898 = 1.50 \text{ g}$ (actual)

Or from volume:

At STP: 22.4 L H₂ = 2.0 g H₂

$18.7 \text{ L} \times \frac{2.0 \text{ g}}{22.4 \text{ L}} = 1.67 \text{ g}$ ✓

Key concepts demonstrated:

1. Stoichiometry with limiting reactant
2. Gas volume at STP (molar volume = 22.4 L/mol)
3. Theoretical yield calculation
4. Percent yield formula and interpretation
5. Understanding why actual < theoretical

Theoretical mass = 0.145 mol × 180.16 g/mol = 26.1 g

Step 3: Percent yield Actual yield = 18.5 g Percent yield = (actual/theoretical) × 100% Percent yield = (18.5 g / 26.1 g) × 100% = 70.9%

Find: (a) Limiting reactant, (b) Mass PbI₂, (c) [K⁺] in final solution

Part (a): Identify limiting reactant

Step 1: Calculate moles of each reactant

Moles of Pb(NO₃)₂:

$n_{Pb(NO_3)_2} = M \times V$

$n_{Pb(NO_3)_2} = 0.200 \text{ M} \times 0.0500 \text{ L}$

$n_{Pb(NO_3)_2} = 0.0100 \text{ mol}$

Moles of KI:

$n_{KI} = M \times V$

$n_{KI} = 0.300 \text{ M} \times 0.0300 \text{ L}$

$n_{KI} = 0.00900 \text{ mol}$

Step 2: Identify limiting reactant

Method 1: Calculate product from each reactant

From balanced equation:

$Pb(NO3)2 + 2KI \rightarrow PbI2 + 2KNO3$

Mole ratio: 1 Pb(NO₃)₂ : 2 KI : 1 PbI₂

From Pb(NO₃)₂:

$n_{PbI_2} = 0.0100 \text{ mol Pb(NO}_3)_2 \times \frac{1 \text{ mol PbI}_2}{1 \text{ mol Pb(NO}_3)_2}$

$n_{PbI_2} = 0.0100 \text{ mol}$

From KI:

$n_{PbI_2} = 0.00900 \text{ mol KI} \times \frac{1 \text{ mol PbI}_2}{2 \text{ mol KI}}$

$n_{PbI_2} = 0.00450 \text{ mol}$

Compare:

• From Pb(NO₃)₂: 0.0100 mol PbI₂
• From KI: 0.00450 mol PbI₂

KI produces less product → KI is limiting reactant ✓

Method 2: Ratio method (verification)

$\frac{n_{Pb(NO_3)_2}}{\text{coefficient}} = \frac{0.0100}{1} = 0.0100$

$\frac{n_{KI}}{\text{coefficient}} = \frac{0.00900}{2} = 0.00450$

KI has smaller ratio → KI is limiting ✓

Answer (a):

$\boxed{\text{KI is the limiting reactant}}$

Pb(NO₃)₂ is in excess

Part (b): Calculate mass of PbI₂

Step 3: Calculate moles of PbI₂ formed

Use limiting reactant (KI):

$n_{PbI_2} = 0.00900 \text{ mol KI} \times \frac{1 \text{ mol PbI}_2}{2 \text{ mol KI}}$

$n_{PbI_2} = 0.00450 \text{ mol}$

Step 4: Calculate molar mass of PbI₂

$M_{PbI_2} = 207.2 + 2(126.9)$

$M_{PbI_2} = 207.2 + 253.8$

$M_{PbI_2} = 461.0 \text{ g/mol}$

Step 5: Convert moles to mass

$\text{mass}_{PbI_2} = n_{PbI_2} \times M_{PbI_2}$

$\text{mass}_{PbI_2} = 0.00450 \text{ mol} \times 461.0 \text{ g/mol}$

$\text{mass}_{PbI_2} = 2.07 \text{ g}$

Answer (b):

$\boxed{2.07 \text{ g PbI}_2 \text{ precipitates}}$

This is bright yellow precipitate

Part (c): Calculate [K⁺] in final solution

Step 6: Find total moles of K⁺

K⁺ comes from KI:

• Each KI provides 1 K⁺
• Started with 0.00900 mol KI
• Therefore: 0.00900 mol K⁺ total

Note: All KI dissolves initially (before precipitation)

• KI → K⁺ + I⁻ (complete dissociation)
• Then: Pb²⁺ + 2I⁻ → PbI₂(s)
• K⁺ is spectator ion (stays dissolved)

All K⁺ remains in solution

Step 7: Calculate total volume

Assume volumes are additive:

$V_{total} = V_{Pb(NO_3)_2} + V_{KI}$

$V_{total} = 50.0 \text{ mL} + 30.0 \text{ mL} = 80.0 \text{ mL}$

$V_{total} = 0.0800 \text{ L}$

Step 8: Calculate molarity of K⁺

$[K^+] = \frac{n_{K^+}}{V_{total}}$

$[K^+] = \frac{0.00900 \text{ mol}}{0.0800 \text{ L}}$

$[K^+] = 0.1125 \text{ M}$

$[K^+] = 0.113 \text{ M}$

Answer (c):

$\boxed{[K^+] = 0.113 \text{ M}}$

Summary Table:

Additional analysis:

How much Pb(NO₃)₂ remains?

Pb(NO₃)₂ consumed:

$n_{Pb(NO_3)_2 \text{ used}} = 0.00900 \text{ mol KI} \times \frac{1 \text{ mol Pb(NO}_3)_2}{2 \text{ mol KI}}$

$n_{Pb(NO_3)_2 \text{ used}} = 0.00450 \text{ mol}$

Pb(NO₃)₂ remaining:

$n_{Pb(NO_3)_2 \text{ remaining}} = 0.0100 - 0.00450 = 0.00550 \text{ mol}$

Concentration of Pb²⁺ in final solution:

$[Pb^{2+}] = \frac{0.00550 \text{ mol}}{0.0800 \text{ L}} = 0.0688 \text{ M}$

Wait! This ignores PbI₂ solubility equilibrium

• PbI₂ is "insoluble" but has tiny solubility
• Some Pb²⁺ and I⁻ in equilibrium with solid
• For this problem, assume PbI₂ completely insoluble

What about NO₃⁻ concentration?

NO₃⁻ from Pb(NO₃)₂:

• Each Pb(NO₃)₂ provides 2 NO₃⁻
• Started with 0.0100 mol Pb(NO₃)₂
• Total NO₃⁻ = 0.0100 × 2 = 0.0200 mol

NO₃⁻ is spectator ion (stays dissolved)

$[NO_3^-] = \frac{0.0200 \text{ mol}}{0.0800 \text{ L}} = 0.250 \text{ M}$

Complete ion inventory in final solution:

Plus: Solid PbI₂ precipitate (2.07 g)

Key concepts demonstrated:

1. Solution stoichiometry using M × V = n
2. Limiting reactant determination with multiple reactants
3. Precipitate formation (PbI₂ is insoluble)
4. Spectator ions (K⁺ and NO₃⁻ don't react)
5. Final concentration calculation with mixed volumes
6. Excess reactant remaining after reaction

Observable evidence:

• Solutions mixed → immediately bright yellow precipitate forms
• Yellow solid settles to bottom
• Clear solution above contains K⁺, NO₃⁻, excess Pb²⁺