Stoichiometry & Chemical Reactions

Stoichiometry & Chemical Reactions

Part 1 of 5 — Moles, Molar Mass & Avogadro's Number

Stoichiometry is the quantitative backbone of chemistry. Every MCAT general chemistry calculation—concentration, yield, titration, gas laws—relies on accurate mole conversions.

The Mole

One mole = $6.022 \times 10^{23}$ entities (Avogadro's number, $N_A$).

This is the bridge between atomic scale (amu) and lab scale (grams): $1\text{ amu} = 1\text{ g/mol}$

Key Conversions

$\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

$\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

$\text{moles (gas at STP)} = \frac{\text{volume (L)}}{22.4\text{ L/mol}}$

"STP" on the MCAT means 0°C and 1 atm. (Note: some modern definitions use 0°C / 100 kPa; the MCAT typically uses the traditional 22.4 L/mol value.)

Molar Mass Calculation

Add the atomic masses of all atoms in the formula.

Example — $\text{H}_2\text{SO}_4$ (sulfuric acid):

• $2 \times \text{H} = 2 \times 1.0 = 2.0$
• $1 \times \text{S} = 1 \times 32.1 = 32.1$
• $4 \times \text{O} = 4 \times 16.0 = 64.0$
• Molar mass $= 2.0 + 32.1 + 64.0 = \mathbf{98.1\text{ g/mol}}$

Example — $\text{Ca}_3(\text{PO}_4)_2$ (calcium phosphate):

• $3 \times \text{Ca} = 3 \times 40.1 = 120.3$
• $2 \times \text{P} = 2 \times 31.0 = 62.0$
• $8 \times \text{O} = 8 \times 16.0 = 128.0$
• Molar mass $= 310.3\text{ g/mol}$

Percent Composition

$\%\text{ by mass of element} = \frac{n \times M_{\text{element}}}{M_{\text{compound}}} \times 100$

Example — % O in $\text{H}_2\text{O}$: $\frac{16.0}{18.0} \times 100 = 88.9\%$

Empirical vs. Molecular Formula

• Empirical formula: Simplest whole-number ratio of atoms.
• Molecular formula: Actual number of atoms; always a whole-number multiple of the empirical formula.

Workflow to find empirical formula from % composition:

1. Assume 100 g sample → grams = percentages.
2. Divide each by atomic mass to get moles.
3. Divide all by the smallest mole value.
4. Round to integers (multiply if needed to clear fractions).

Moles & Molar Mass — Check Your Understanding 🎯

Worked Example: Multi-Step Mole Calculation

Problem: 9.80 g of $\text{H}_2\text{SO}_4$ reacts completely with excess NaOH. How many grams of $\text{Na}_2\text{SO}_4$ are produced?

$\text{H}_2\text{SO}_4 + 2\text{NaOH} \to \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

Step 1 — Moles of $\text{H}_2\text{SO}_4$: $n = \frac{9.80\text{ g}}{98.1\text{ g/mol}} = 0.100\text{ mol}$

Step 2 — Moles of $\text{Na}_2\text{SO}_4$ (1:1 ratio from balanced equation): $0.100\text{ mol } \text{H}_2\text{SO}_4 \times \frac{1\text{ mol } \text{Na}_2\text{SO}_4}{1\text{ mol } \text{H}_2\text{SO}_4} = 0.100\text{ mol}$

Step 3 — Grams of $\text{Na}_2\text{SO}_4$ (molar mass = $2(23) + 32 + 4(16) = 142\text{ g/mol}$): $0.100\text{ mol} \times 142\text{ g/mol} = \mathbf{14.2\text{ g}}$

MCAT Strategy

Always write the balanced equation first. Every stoichiometry problem requires it. Check that atoms and charges balance on both sides before calculating.

Applied Mole Calculations 🎯

Key Takeaways — Part 1

• 1 mole = $6.022 \times 10^{23}$ particles = molar mass in grams = 22.4 L at STP (gas)
• Molar mass = sum of (atomic mass × subscript) for all elements in formula
• % composition = (mass of element / molar mass of compound) × 100
• Empirical formula: divide % by atomic mass → divide by smallest → round to integers
• Molecular formula = empirical formula × (M_total / M_empirical)
• Balanced equation required before any stoichiometric calculation
• Mole ratio from balanced equation links moles of reactant to moles of product

Stoichiometry & Chemical Reactions

Part 2 of 5 — Limiting Reagent, Percent Yield & Reaction Types

Limiting Reagent

The limiting reagent is the reactant that is completely consumed first, limiting the amount of product formed. The excess reagent remains unreacted.

How to Find the Limiting Reagent

1. Convert all reactant masses to moles.
2. Divide each mole value by its stoichiometric coefficient.
3. The reactant with the smaller result is the limiting reagent.

Example:
$2\text{Al} + 3\text{Cl}_2 \to 2\text{AlCl}_3$

Given: 5.40 g Al and 21.3 g $\text{Cl}_2$.

• Moles Al $= 5.40/27.0 = 0.200$ mol → $0.200/2 = 0.100$
• Moles $\text{Cl}_2 = 21.3/70.9 = 0.300$ mol → $0.300/3 = 0.100$

Both give 0.100 — they are in exact stoichiometric ratio; neither is in excess. All reactants are consumed.

Second example:
$2\text{Al} + 3\text{Cl}_2 \to 2\text{AlCl}_3$

Given: 5.40 g Al and 14.2 g $\text{Cl}_2$.

• Moles Al $= 0.200$ mol → $0.200/2 = 0.100$
• Moles $\text{Cl}_2 = 14.2/70.9 = 0.200$ mol → $0.200/3 = 0.0667$

$\text{Cl}_2$ gives the smaller ratio → $\text{Cl}_2$ is the limiting reagent.

Moles of $\text{AlCl}_3$ produced (uses $\text{Cl}_2$): $0.200\text{ mol }\text{Cl}_2 \times \frac{2\text{ mol }\text{AlCl}_3}{3\text{ mol }\text{Cl}_2} = 0.133\text{ mol}$

Mass $= 0.133 \times 133.3 = 17.7\text{ g AlCl}_3$

Percent Yield

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

• Theoretical yield: calculated assuming 100% conversion of limiting reagent to product.
• Actual yield: experimentally measured.

Causes of < 100% yield: side reactions, incomplete reaction, product loss during isolation.

Types of Chemical Reactions

Combustion Stoichiometry Shortcut

For $\text{C}_x\text{H}_y$ + $\text{O}_2$: $\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \to x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}$

Limiting Reagent & Yield 🎯

Worked Example: Combustion Analysis

Propane ($\text{C}_3\text{H}_8$) burns completely in oxygen:

$\text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}$

How much $\text{CO}_2$ is produced by burning 44.1 g of propane?

Step 1 — Moles propane ($M = 44.1$ g/mol): $n = \frac{44.1}{44.1} = 1.00\text{ mol}$

Step 2 — Moles $\text{CO}_2$ (3:1 ratio): $1.00 \times 3 = 3.00\text{ mol CO}_2$

Step 3 — Grams: $3.00 \times 44.0 = \mathbf{132\text{ g CO}_2}$

MCAT Shortcut: Combustion $O_{2}$ Coefficient

For $\text{C}_3\text{H}_8$: $x=3$, $y=8$. $O_{2}$ needed $= 3 + 8/4 = 3 + 2 = 5$. ✓

Combustion & Reaction Types 🎯

Key Takeaways — Part 2

• Limiting reagent: divide moles by coefficient; smallest ratio = limiting reagent
• Theoretical yield is calculated from the limiting reagent using stoichiometric ratios
• % yield = (actual / theoretical) × 100; always ≤ 100% in practice
• Reaction types: combination, decomposition, single displacement, double displacement, combustion
• Combustion of $\text{C}_x\text{H}_y$: coefficient for $\text{O}_2 = x + y/4$; always produces $\text{CO}_2$ and $\text{H}_2\text{O}$
• On the MCAT, % yield questions often appear with multi-step syntheses — apply to each step separately

Stoichiometry & Chemical Reactions

Part 3 of 5 — Solutions: Molarity, Dilution & Solution Stoichiometry

Concentration Units

MCAT emphasis: Molarity is by far the most frequently tested. Molality appears in colligative property problems.

Preparing a Solution

Problem: How many grams of $\text{NaOH}$ ($M = 40.0$ g/mol) are needed to make 250 mL of a 0.500 M solution?

$n = M \times V_{\text{L}} = 0.500\text{ M} \times 0.250\text{ L} = 0.125\text{ mol}$

$m = 0.125\text{ mol} \times 40.0\text{ g/mol} = \mathbf{5.00\text{ g}}$

Dilution

When a solution is diluted (solvent added), the amount of solute doesn't change:

$M_1 V_1 = M_2 V_2$

Problem: 25.0 mL of 12.0 M HCl is diluted to 300 mL. What is the final concentration?

$M_2 = \frac{M_1 V_1}{V_2} = \frac{12.0 \times 25.0}{300} = \mathbf{1.00\text{ M}}$

Lab safety note: Always add acid to water, never water to acid (heat released causes dangerous spattering).

Solution Stoichiometry

Combine molarity calculations with balanced-equation ratios.

Example: How many mL of 0.200 M $\text{H}_2\text{SO}_4$ are needed to neutralize 50.0 mL of 0.300 M $\text{NaOH}$?

$\text{H}_2\text{SO}_4 + 2\text{NaOH} \to \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

Moles NaOH $= 0.300 \times 0.0500 = 0.0150$ mol

Moles $\text{H}_2\text{SO}_4$ needed (2:1 NaOH:acid ratio): $0.0150/2 = 0.00750\text{ mol}$

Volume: $V = n/M = 0.00750/0.200 = 0.0375\text{ L} = \mathbf{37.5\text{ mL}}$

Colligative Properties (Molality-Based)

Properties that depend on the number (not identity) of solute particles:

$i$ = van 't Hoff factor (number of particles per formula unit after dissociation):

• Nonelectrolytes (glucose, urea): $i = 1$
• NaCl: $i = 2$; $\text{CaCl}_2$: $i = 3$; $\text{Al}_2(\text{SO}_4)_3$: $i = 5$

Example — Osmotic pressure:
$1.0\text{ M}$ NaCl solution at 37°C ($T = 310$ K):

$\Pi = iMRT = 2 \times 1.0 \times 0.0821 \times 310 \approx \mathbf{50.9\text{ atm}}$

Solution Concentration & Dilution 🎯

Net Ionic Equations

In solution, strong electrolytes dissociate completely. A net ionic equation shows only the species that actually participate in the reaction (spectator ions are omitted).

Example — Precipitation of $\text{AgCl}$:

Molecular: $\text{AgNO}_3(aq) + \text{NaCl}(aq) \to \text{AgCl}(s) + \text{NaNO}_3(aq)$

Complete ionic (split all soluble ionic compounds): $\text{Ag}^+(aq) + \text{NO}_3^-(aq) + \text{Na}^+(aq) + \text{Cl}^-(aq) \to \text{AgCl}(s) + \text{Na}^+(aq) + \text{NO}_3^-(aq)$

Cancel spectator ions ($\text{Na}^+$ and $\text{NO}_3^-$):

$\boxed{\text{Ag}^+(aq) + \text{Cl}^-(aq) \to \text{AgCl}(s)}$

Solubility Rules (MCAT-Relevant)

Always soluble: nitrates ($\text{NO}_3^-$), alkali metals ($\text{Na}^+$, $\text{K}^+$), ammonium ($\text{NH}_4^+$)

Usually insoluble: $\text{Ag}^+$ halides (except $\text{AgF}$), $\text{Ba}^{2+}$/$\text{Pb}^{2+}$/$\text{Ca}^{2+}$ sulfates, most carbonates/phosphates/hydroxides (except Group IA and $\text{Ba}^{2+}$)

Net Ionic Equations & Precipitation 🎯

Key Takeaways — Part 3

• Molarity: $M = n/V_{\text{L}}$. Memorize and apply in every concentration problem.
• Dilution: $M_1V_1 = M_2V_2$ — moles of solute don't change.
• Solution stoichiometry: convert to moles using $n = MV$, apply mole ratio, convert back.
• Colligative properties: depend on particle count ($i \times m$), not identity.
• Van 't Hoff factor $i$: 1 for nonelectrolytes; = ions per formula unit for strong electrolytes.
• Net ionic equations: cancel spectator ions; only reactive ions appear.
• Solubility rules: nitrates always soluble; $\text{Ag}^+$ halides (except AgF) insoluble; most carbonates/phosphates insoluble.
• $Q$ vs $K_{sp}$: if $Q > K_{sp}$, precipitation occurs.

Stoichiometry & Chemical Reactions

Part 4 of 5 — Gas Laws & Electrochemistry Stoichiometry

Ideal Gas Law

$PV = nRT$

$P$ = pressure (atm), $V$ = volume (L), $n$ = moles, $R = 0.0821$ L·atm/mol·K, $T$ = temperature (K)

Always convert temperature to Kelvin: $T(\text{K}) = T(°\text{C}) + 273$

Combined Gas Law (fixed $n$)

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Special Cases

Worked Example: Ideal Gas Law

What volume does 2.00 mol of gas occupy at 1.50 atm and 27°C?

$V = \frac{nRT}{P} = \frac{2.00 \times 0.0821 \times 300}{1.50} = \frac{49.26}{1.50} = \mathbf{32.8\text{ L}}$

Dalton's Law of Partial Pressures

$P_{\text{total}} = P_A + P_B + P_C + \cdots$

Each gas in a mixture exerts its partial pressure independently.

$P_A = \chi_A \times P_{\text{total}}$

where $\chi_A$ = mole fraction of gas A.

MCAT application — Gas collected over water:
When collecting a gas by water displacement: $P_{\text{gas}} = P_{\text{atm}} - P_{\text{H}_2\text{O (vapor)}}$

Faraday's Law: Electrochemical Stoichiometry

In electrolysis, electric current drives non-spontaneous redox reactions. The amount of product depends on the charge passed.

$\text{moles of electrons} = \frac{I \times t}{F}$

where $I$ = current (amperes), $t$ = time (seconds), $F = 96{,}485$ C/mol $\approx 96{,}500$ C/mol.

$\text{moles of product} = \frac{\text{moles of electrons}}{n}$

where $n$ = electrons required per atom/ion reduced or oxidized.

Example: How many grams of Cu are deposited by passing 2.00 A for 1930 s through $\text{CuSO}_4$ solution? ($\text{Cu}^{2+} + 2e^- \to \text{Cu}$, $M_{\text{Cu}} = 63.5$ g/mol)

$\text{mol } e^- = \frac{2.00 \times 1930}{96{,}500} = \frac{3860}{96{,}500} = 0.0400\text{ mol}$

$\text{mol Cu} = \frac{0.0400}{2} = 0.0200\text{ mol}$

$\text{mass Cu} = 0.0200 \times 63.5 = \mathbf{1.27\text{ g}}$

Gas Laws 🎯

Graham's Law of Effusion

The rate at which a gas escapes through a small orifice depends on its molar mass:

$\frac{\text{rate}_A}{\text{rate}_B} = \sqrt{\frac{M_B}{M_A}}$

Lighter gases effuse faster. MCAT application: Compare rates of diffusion/effusion for two gases at the same T and P.

Example: Which effuses faster — $\text{H}_2$ or $\text{O}_2$? By what factor?

$\frac{\text{rate}_{\text{H}_2}}{\text{rate}_{\text{O}_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$

$\text{H}_2$ effuses 4 times faster than $\text{O}_2$.

Real Gases vs. Ideal Gases

Ideal gas assumptions (particles have no volume, no intermolecular forces) deviate at:

• High pressure (volume of particles becomes significant)
• Low temperature (attractive forces between molecules become significant)

van der Waals equation: $\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$

• $a$ accounts for intermolecular attractions (reduces effective pressure)
• $b$ accounts for molecular volume (reduces available volume)

On the MCAT, you don't calculate with van der Waals, but you must identify when gases deviate from ideal behavior and in which direction.

Faraday's Law & Graham's Law 🎯

Key Takeaways — Part 4

• Ideal gas law: $PV = nRT$; $R = 0.0821$ L·atm/mol·K; always use Kelvin.
• Combined gas law: $P_1V_1/T_1 = P_2V_2/T_2$ (fixed $n$).
• Dalton's law: $P_{\text{total}} = \sum P_i$; $P_i = \chi_i P_{\text{total}}$.
• Graham's law: lighter gas effuses faster; rate ratio $= \sqrt{M_B/M_A}$.
• Real gases deviate most from ideal at high $P$ and low $T$.
• Faraday's law: mol $e^- = It/F$; mol product = mol $e^-$ / $n$ electrons transferred per ion.
• Gas collected over water: subtract water vapor pressure to get gas pressure.

Stoichiometry & Chemical Reactions

Part 5 of 5 — Mixed MCAT Review

This section integrates all stoichiometry concepts you'll encounter on the MCAT. Each question may draw on moles, limiting reagents, gas laws, solutions, colligative properties, electrochemistry, or multiple topics at once.

High-Yield Checklist Before You Practice

• ✅ Molar mass from periodic table; $n = m/M$
• ✅ Limiting reagent = divide moles by stoichiometric coefficient; choose smaller result
• ✅ Percent yield = (actual / theoretical) × 100%
• ✅ Molarity $M = n/V(\text{L})$; dilution: $M_1V_1 = M_2V_2$
• ✅ Ideal gas law $PV = nRT$; $R = 0.0821$ L·atm/mol·K; $T$ in Kelvin
• ✅ Dalton's law $P_{\text{total}} = \sum P_i$
• ✅ Graham's law rate $\propto 1/\sqrt{M}$
• ✅ Colligative: $\Delta T_f = K_f m i$; $\Delta T_b = K_b m i$; $\Pi = iMRT$
• ✅ Faraday: mol $e^- = It/F$; $F \approx 96{,}500$ C/mol
• ✅ Net ionic equations: cancel spectator ions; know solubility rules

Mixed Stoichiometry — MCAT-Style Questions 🎯

Stoichiometry — Complete Topic Summary

Part 1: Moles, molar mass, Avogadro's number, empirical/molecular formula, combustion analysis.

Part 2: Limiting reagent, percent yield, reaction classification (synthesis, decomposition, single/double displacement, combustion).

Part 3: Molarity, dilution ($M_1V_1 = M_2V_2$), solution stoichiometry, colligative properties, van 't Hoff factor, net ionic equations, solubility rules, $Q$ vs $K_{sp}$.

Part 4: Ideal gas law ($PV=nRT$), combined gas law, Dalton's partial pressures, Graham's law of effusion, real gas deviations, Faraday's law of electrolysis.

Part 5: Integrated MCAT practice combining all above topics.

Most Common MCAT Pitfalls

1. Forgetting to convert °C → K before using gas laws
2. Dividing by stoichiometric coefficient to find limiting reagent (not just comparing moles)
3. Forgetting van 't Hoff $i$ in colligative property calculations
4. Subtracting water vapor pressure from total pressure for gases collected over water
5. Using $n =$ electrons per ion (not per formula unit) in Faraday's law