Stoichiometry

Stoichiometry is the quantitative backbone of chemistry — the tools that turn a balanced equation into masses, moles, and yields. The MCAT rewards knowing which conversion to apply and why the answer makes sense, not heavy arithmetic. Build each concept as a mental model first.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Molecular Weight and the Mole Concept

Atomic Mass and Molar Mass

Must know

Every element carries an atomic mass (in amu) — the weighted average of its naturally occurring isotopes. Scale to one mole (mol) and that number in amu becomes grams per mole; this equivalence is the definition of the amu.

$M = \sum_i n_i \cdot m_i$

where $M$ is the molar mass of a compound (g/mol), $n_i$ is the number of atoms of element $i$, and $m_i$ is that element's molar mass.

Avogadro's number, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$, is the number of particles in one mole. The mole is the bridge between the microscopic and macroscopic. Must know the central conversion chain (and its reverse):

$\text{mass (g)} \xrightarrow{\div M} \text{moles} \xrightarrow{\times N_A} \text{particles}$

Everything in stoichiometry reduces to this chain combined with mole ratios from a balanced equation.

Quick check: How many molecules are in 18.0 g of water?

Answer: $M_{\ce{H2O}} = 2(1.0) + 16.0 = 18.0\ \text{g/mol}$, so 18.0 g = 1.00 mol $= 1.00 \times 6.022 \times 10^{23} \approx 6.02 \times 10^{23}$ molecules — exactly Avogadro's number.

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Metric Units in Chemistry

SI Prefixes and Common Units

Must know

The MCAT uses SI units throughout, and unit errors are a common way to lose points. Know the prefixes and the units used most in stoichiometry:

Prefix	Symbol	Factor
mega	M	$10^6$
kilo	k	$10^3$
centi	c	$10^{-2}$
milli	m	$10^{-3}$
micro	μ	$10^{-6}$
nano	n	$10^{-9}$

• Mass: grams (g). Volume: L and mL; $1\ \text{L} = 1000\ \text{mL}$. Amount: moles (mol).
• Concentration: molarity $M = \text{mol/L}$; millimolar (mM) $= 10^{-3}\ \text{mol/L}$.
• Temperature: convert to Kelvin for gas-law/thermo work: $T(\text{K}) = T(°\text{C}) + 273$.

(Pressure units — atm, Pa — belong to the gas-law guides.)

Quick check: A solution is 50 mM. Express in mol/L and in g/L if the solute is glucose ($M = 180\ \text{g/mol}$).

Answer: $50\ \text{mM} = 0.050\ \text{mol/L}$; mass conc. $= 0.050 \times 180 = 9.0\ \text{g/L}$. Physiological glucose is roughly in this range — a useful sanity anchor.

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Percent Mass Composition

Calculating and Using Percent Composition

Must know

Percent mass composition is the fraction of a compound's total mass from each element:

$\% \text{ mass of element } i = \frac{n_i \cdot m_i}{M_{\text{compound}}} \times 100\%$

Know it in both directions: from a formula, and to derive a formula from given percents.

Example: Mass percent of N in urea, $\ce{(NH2)2CO}$. $M_{\text{urea}} = 2(14.0) + 4(1.0) + 12.0 + 16.0 = 60.0\ \text{g/mol}$; two N contribute 28.0 g/mol.

$\% \text{N} = \frac{28.0}{60.0} \times 100\% = 46.7\%$

Quick check: A compound is 28% N by mass; verify whether its empirical formula is $\ce{CH2N}$.

Answer: $M_{\ce{CH2N}} = 28\ \text{g/mol}$, so $\%\text{N} = 14/28 = 50\%$ — doesn't match 28%, so $\ce{CH2N}$ is wrong. Always check a derived formula against the data.

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Empirical vs. Molecular Formula

Deriving the Empirical Formula

Must know

The empirical formula is the simplest whole-number atom ratio; the molecular formula is the actual atom count — always a whole-number multiple of the empirical.

Must know the procedure from percent composition:

1. Assume 100 g (percent values become grams).
2. Divide each mass by the element's molar mass → moles.
3. Divide all by the smallest mole value.
4. If needed, multiply through by a small integer to reach whole numbers.

Deriving the Molecular Formula

Must know

With the empirical formula and a measured molar mass:

$n = \frac{M_{\text{molecular}}}{M_{\text{empirical}}}$

Multiply the empirical formula through by $n$.

Combustion Analysis

Know the logic

Burning a C/H/O compound in excess $\ce{O2}$ sends all C to $\ce{CO2}$ and all H to $\ce{H2O}$:

• $n_{\ce{C}} = n_{\ce{CO2}} = \dfrac{m_{\ce{CO2}}}{44.0}$ ; $n_{\ce{H}} = 2\, n_{\ce{H2O}} = 2 \times \dfrac{m_{\ce{H2O}}}{18.0}$
• Oxygen is found by difference: $m_{\ce{O}} = m_{\text{sample}} - m_{\ce{C}} - m_{\ce{H}}$ (the $\ce{CO2}$/$\ce{H2O}$ oxygen comes mostly from the added $\ce{O2}$).

Fully Worked Example

Must know

Problem: A compound is 40.0% C, 6.7% H, 53.3% O by mass; molar mass 180 g/mol. Find both formulas.

Element	Mass (g)	÷ molar mass	Moles	÷ smallest
C	40.0	12.0	3.33	1
H	6.7	1.0	6.70	2
O	53.3	16.0	3.33	1

Empirical $\ce{CH2O}$, $M_{\text{emp}} = 30.0\ \text{g/mol}$. Then $n = 180/30.0 = 6$, so molecular formula $= \ce{C6H12O6}$ (glucose, or any hexose — molecular formula alone doesn't distinguish isomers).

Quick check: Benzene is $\ce{C6H6}$. What is its empirical formula?

Answer: $\ce{CH}$ (divide by 6). Acetylene ($\ce{C2H2}$) shares this empirical formula despite being structurally different.

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Density

Definition and Applications

Must know

Density is mass per unit volume:

$\rho = \frac{m}{V}$

Usually g/mL ($=$ g/cm³) for liquids/solids, g/L for gases. Density links volume to mass and therefore to moles:

$n = \frac{\rho \cdot V}{M}$

Passage-level anchors: water $\approx 1.00\ \text{g/mL}$; fat $\approx 0.9\ \text{g/mL}$ (so fat floats). At STP one mole of ideal gas occupies 22.4 L, so $\rho_{\text{gas, STP}} = M / 22.4\ \text{L/mol}$.

Quick check: Concentrated sulfuric acid is 1.84 g/mL and ~98% $\ce{H2SO4}$ by mass. How many moles of $\ce{H2SO4}$ in 10.0 mL?

Answer: Solution mass $= 18.4\ \text{g}$; $\ce{H2SO4} \approx 0.98 \times 18.4 = 18.0\ \text{g}$; moles $= 18.0/98.1 \approx 0.184\ \text{mol}$.

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Solution and Gas Routes to Moles

Most problems hinge on getting to moles. Beyond mass $\div$ molar mass, two routes recur.

Molarity and Dilution

Must know

Molarity $M = n/V$, so $n = M \times V$. Convert volume × concentration → moles, apply the mole ratio, convert back. (In-depth titration is in the Acid–Base guides.) On dilution (solute conserved):

$M_1 V_1 = M_2 V_2$

Quick check: What volume of $12.0\ \text{M}$ HCl makes $500\ \text{mL}$ of $0.10\ \text{M}$ HCl?

Answer: $V_1 = \dfrac{(0.10)(500)}{12.0} \approx 4.2\ \text{mL}$ (then dilute to 500 mL).

Ideal Gas Law

Must know

The 22.4 L/mol shortcut works only at STP. In general:

$n = \frac{PV}{RT}$

with $R = 0.0821\ \text{L·atm·mol}^{-1}\text{K}^{-1}$ and $T$ in kelvin. Use whenever a gas is at non-STP conditions before applying mole ratios.

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Oxidation Numbers

Assigning Oxidation States

Must know

An oxidation number is a bookkeeping device: it assigns each bond's electrons to the more electronegative atom. A formalism, but essential for identifying and balancing redox.

Must know the rules (apply in order):

1. Pure element = 0 ($\ce{O2}$, $\ce{Fe}$).
2. Monatomic ion = its charge ($\ce{Na+}$ = +1, $\ce{Cl-}$ = −1).
3. Fluorine = −1 in compounds.
4. Oxygen = −2 usually; −1 in peroxides ($\ce{H2O2}$); +2 in $\ce{OF2}$.
5. Hydrogen = +1 with nonmetals; −1 in metal hydrides ($\ce{NaH}$, $\ce{LiAlH4}$).
6. Sum of oxidation numbers = charge of the species.

Example: Mn in $\ce{KMnO4}$: $+1 + \text{Mn} + 4(-2) = 0 \Rightarrow \text{Mn} = +7$ — highly oxidized, a strong oxidizer. Similarly Cr in $\ce{Cr2O7^{2-}}$ is +6.

Quick check: Oxidation state of sulfur in $\ce{H2SO4}$?

Answer: $2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$. Compare $\ce{H2S}$ (S = −2) and $\ce{SO2}$ (S = +4) — all fair MCAT targets.

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Oxidizing and Reducing Agents

Identifying OIL RIG

Must know

• Oxidation = loss of electrons; the oxidized species is the reducing agent.
• Reduction = gain of electrons; the reduced species is the oxidizing agent.

Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain. The agent does what its name says to the other species, so it itself undergoes the opposite.

Passage-level

Common oxidizing agents (themselves reduced): $\ce{MnO4-}$ (purple → colorless $\ce{Mn^{2+}}$ in acid, brown $\ce{MnO2}$ in neutral/base), $\ce{Cr2O7^{2-}}$ (orange → green $\ce{Cr^{3+}}$), $\ce{H2O2}$, $\ce{O2}$, the halogens, and $\ce{NAD+}$/$\ce{FAD}$ in metabolism. Common reducing agents (themselves oxidized): active metals, $\ce{H2}$, $\ce{NADH}$/$\ce{FADH2}$, and hydride reagents $\ce{LiAlH4}$/$\ce{NaBH4}$.

Quick check: In $\ce{Fe + CuSO4 -> FeSO4 + Cu}$, identify the oxidizing and reducing agents.

Answer: Fe goes 0 → +2 (oxidized) → reducing agent. Cu goes +2 → 0 (reduced) → $\ce{Cu^{2+}}$ is the oxidizing agent.

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Disproportionation Reactions

When One Species Is Both Oxidized and Reduced

Must know

In a disproportionation, a single element in one oxidation state is simultaneously oxidized and reduced, giving two products. Classic example:

$\ce{2 H2O2 -> 2 H2O + O2}$

O is −1 in $\ce{H2O2}$, goes to −2 in $\ce{H2O}$ (reduced) and 0 in $\ce{O2}$ (oxidized) — same element, same start, two fates. Passage-level further examples: $\ce{Cl2 + 2 OH- -> Cl- + ClO- + H2O}$ (Cl: 0 → −1 and +1) and $\ce{2 Cu+ -> Cu^0 + Cu^{2+}}$. The reverse (two states → one) is comproportionation.

Quick check: In $\ce{2 Cu+ -> Cu^0 + Cu^{2+}}$, which $\ce{Cu+}$ is oxidized and which reduced?

Answer: The one becoming $\ce{Cu^0}$ is reduced (+1→0); the one becoming $\ce{Cu^{2+}}$ is oxidized (+1→+2). Both occur simultaneously from the same pool.

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Writing and Balancing Chemical Equations

Conventions for Chemical Equations

Must know

• Reactants left, products right, separated by $\to$ (or $\rightleftharpoons$ for equilibrium).
• State symbols: (s), (l), (g), (aq).
• Coefficients are whole numbers in simplest ratio.
• A balanced equation conserves mass (atom counts) and charge (for ionic/redox).

Describing Reactions by Type

Know the logic

Recognize the categories and predict products:

• Combination: $\ce{A + B -> AB}$
• Decomposition: $\ce{AB -> A + B}$
• Single displacement: $\ce{A + BC -> AC + B}$ (more active element displaces another)
• Double displacement (metathesis): $\ce{AB + CD -> AD + CB}$ — includes precipitation and acid–base neutralization
• Combustion: a hydrocarbon + $\ce{O2} \to \ce{CO2 + H2O}$

Net Ionic Equations and Spectator Ions

Must know

Three forms, using $\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}$:

1. Molecular — full neutral formulas (above).
2. Complete ionic — strong electrolytes as separated ions.
3. Net ionic — cancel the spectator ions ($\ce{Na+}$, $\ce{NO3-}$): $\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$.

Optional

Solubility rules: Group 1, $\ce{NH4+}$, and nitrates are always soluble; most halides soluble except with $\ce{Ag+}$, $\ce{Pb^{2+}}$; most carbonates, phosphates, sulfides, hydroxides insoluble except with Group 1 and $\ce{NH4+}$.

Balancing Non-Redox Equations by Inspection

Must know

Balance atoms appearing in one reactant/product first; treat unchanged polyatomic ions as units; do H and O last; clear fractions at the end. E.g. $\ce{4 Al + 3 O2 -> 2 Al2O3}$.

Balancing Redox Equations: The Half-Reaction Method

Must know

Use the half-reaction method. In acidic solution:

1. Split into oxidation and reduction half-reactions.
2. Balance atoms except H and O.
3. Balance O with $\ce{H2O}$; balance H with $\ce{H+}$.
4. Balance charge with $e^-$.
5. Scale the half-reactions to equal electrons, then add and cancel.

In basic solution: same, then add $\ce{OH-}$ to both sides to neutralize each $\ce{H+}$ ($\ce{H+ + OH- -> H2O}$) and simplify.

Worked Example: Acidic Solution

Must know

Problem: Balance $\ce{MnO4- + Fe^{2+} -> Mn^{2+} + Fe^{3+}}$ (acidic).

Reduction (Mn +7 → +2, gains 5$e^-$):

$\ce{MnO4- + 8 H+ + 5 e- -> Mn^{2+} + 4 H2O}$

Oxidation (Fe +2 → +3) ×5:

$\ce{5 Fe^{2+} -> 5 Fe^{3+} + 5 e-}$

Add:

$\ce{MnO4- + 8 H+ + 5 Fe^{2+} -> Mn^{2+} + 5 Fe^{3+} + 4 H2O}$

Check charge: left $-1+8+10 = +17$; right $+2+15 = +17$ ✓.

Quick check: What changes if this occurs in basic solution?

Answer: Add 8 $\ce{OH-}$ to both sides to neutralize the 8 $\ce{H+}$; the $\ce{H2O}$ partially cancel, giving $\ce{MnO4- + 5 Fe^{2+} + 4 H2O -> Mn^{2+} + 5 Fe^{3+} + 8 OH-}$.

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Limiting Reactants and Theoretical Yield

Identifying the Limiting Reactant

Must know

The limiting reactant is consumed first and caps product formed; the rest are in excess. Must know the method: convert each reactant to moles, divide by its stoichiometric coefficient — the smallest ratio is limiting.

$\frac{\text{moles of reactant}}{\text{stoichiometric coefficient}} \quad (\text{smallest} \to \text{limiting})$

Theoretical, Actual, and Percent Yield

Must know

• Theoretical yield: max product from the limiting reactant (100% completion).
• Actual yield: amount actually obtained.
• Percent yield: $\% \text{ yield} = \dfrac{\text{actual}}{\text{theoretical}} \times 100\%$

Percent yield is almost always < 100% (side reactions, losses); > 100% signals an error (e.g., wet product).

Fully Worked Example

Must know

Problem: $\ce{N2(g) + 3 H2(g) -> 2 NH3(g)}$. Mix 28 g $\ce{N2}$ and 4.0 g $\ce{H2}$; find theoretical yield of $\ce{NH3}$ and percent yield if actual = 14 g.

Moles: $n_{\ce{N2}} = 1.00$, $n_{\ce{H2}} = 2.0$. Ratios: $\ce{N2}: 1.00/1 = 1.00$; $\ce{H2}: 2.0/3 = 0.667$ → $\ce{H2}$ limits.

$n_{\ce{NH3}} = 2.0 \times \tfrac{2}{3} = 1.33\ \text{mol} \Rightarrow m = 1.33 \times 17.0 = 22.7\ \text{g}$

$\% \text{ yield} = \frac{14}{22.7} \times 100\% \approx 62\%$

Sanity: we had 2.0 mol $\ce{H2}$ but needed 3.0 to consume all $\ce{N2}$, so $\ce{H2}$ runs out first.

Quick check: With 6.0 g $\ce{H2}$ instead, does the limiting reactant change?

Answer: $n_{\ce{H2}} = 3.0\ \text{mol}$; ratio $3.0/3 = 1.00$ = the $\ce{N2}$ ratio. Exact stoichiometric proportion — neither limits; yield $= 2.0\ \text{mol}\ \ce{NH3} = 34.0\ \text{g}$.

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Common Confusions & Tricks

1. Empirical ≠ molecular formula (but might be). Glucose, ribose, and formaldehyde all share $\ce{CH2O}$. The molecular formula needs molar mass data.

2. The limiting reactant is NOT just the one with fewer moles. Divide by the stoichiometric coefficient — only that ratio decides.

3. Oxidizing agent is reduced; reducing agent is oxidized. The agent does what its name says to the other substance, so it undergoes the opposite.

4. Hydrogen oxidation state: +1 vs. −1. +1 with nonmetals; −1 in metal hydrides ($\ce{NaH}$, $\ce{LiAlH4}$) — which is why hydride reagents reduce.

5. Oxygen exceptions. Peroxides ($\ce{H2O2}$): O = −1. In $\ce{OF2}$: O = +2 (F always −1).

6. Percent yield > 100% is always wrong. Look for an error — usually undried product or a miscalculated theoretical yield.

7. Molar mass vs. molecular weight. Numerically equal; don't waste time on the distinction.

8. Balancing redox — don't use $\ce{H2O}$ to balance charge. Only $e^-$ balance charge; $\ce{H2O}$ balances O, $\ce{H+}$/$\ce{OH-}$ balance H.

9. Disproportionation: both half-reactions start with the same species (e.g., both from $\ce{H2O2}$).

10. See $\ce{KMnO4}$ or $\ce{K2Cr2O7}$ → think redox. Iconic strong oxidizers with passage-ready color changes (purple→colorless/brown; orange→green).

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Key Equations

Equation	Variables	When to Use
$M = \sum_i n_i \cdot m_i$	$M$ = molar mass, $n_i$ = atom count, $m_i$ = elemental molar mass	Molar mass of any compound
$n = \dfrac{m}{M}$	$n$ = moles, $m$ = mass (g), $M$ = molar mass	Mass ↔ moles
$N = n \cdot N_A$	$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$	Moles ↔ particles
$\% \text{ mass}_i = \dfrac{n_i \cdot m_i}{M_{\text{cpd}}} \times 100\%$	$n_i$ = atoms of element $i$	Percent composition from formula
$n = \dfrac{M_{\text{mol}}}{M_{\text{emp}}}$	$n$ = integer multiplier	Molecular from empirical formula
$\rho = \dfrac{m}{V}$	$\rho$ = density, $m$ = mass, $V$ = volume	Mass ↔ volume
$\sum(\text{ox. numbers}) = \text{charge}$	0 for neutral, ion charge for ions	Assigning oxidation numbers
$\% \text{ yield} = \dfrac{\text{actual}}{\text{theoretical}} \times 100\%$	Same units	Reaction efficiency
$\dfrac{n_A}{\text{coeff}_A}$ smallest → limiting	—	Identifying the limiting reactant
$\rho_{\text{gas, STP}} = \dfrac{M}{22.4\ \text{L/mol}}$	valid at 0°C, 1 atm	Gas density at STP
$n = M \cdot V$	$M$ = molarity, $V$ = volume (L)	Moles of solute from concentration
$M_1 V_1 = M_2 V_2$	1/2 = before/after dilution	Dilution
$PV = nRT$	$R = 0.0821\ \text{L·atm·mol}^{-1}\text{K}^{-1}$; $T$ in K	Moles of gas at non-STP