AP Chemistry Unit 4 Notes: Quantifying Reactions with Stoichiometry and Titrations

Stoichiometry

What stoichiometry is (and why chemists rely on it)

Stoichiometry is the quantitative “accounting system” of chemistry. It lets you use a balanced chemical equation to predict how much product can form, how much reactant is required, which reactant runs out first, and how efficient a reaction is in practice.

The key idea is that a chemical equation is not just a story about substances changing—it is a numerical statement about mole ratios. When you balance an equation, you are saying that particles react in fixed whole-number ratios. Because individual atoms and molecules are too small to count directly, we count them in moles.

A balanced equation like

$2H_2 + O_2 \rightarrow 2H_2O$

means: for every 2 moles of $H_2$ that react, 1 mole of $O_2$ reacts, and 2 moles of $H_2O$ form. Those numbers (2:1:2) are the stoichiometric “exchange rate” that powers every calculation in this section.

Foundations: the mole, molar mass, and Avogadro’s number

A mole is a counting unit—like a dozen—so that we can relate macroscopic measurements (grams, liters) to microscopic counts (particles).

Avogadro’s number links moles to particles:

$1\text{ mol} = 6.022\times 10^{23}\text{ particles}$

Molar mass links grams to moles. The molar mass (in $g/mol$ ) is numerically equal to the atomic or molecular mass from the periodic table.

Common conversions you should be comfortable with:

You have	You want	Use
grams	moles	divide by molar mass
moles	grams	multiply by molar mass
particles	moles	divide by $6.022\times 10^{23}$
moles	particles	multiply by $6.022\times 10^{23}$

A frequent early mistake is trying to use stoichiometric coefficients directly with grams or liters. Coefficients apply to moles, not grams. Stoichiometry is usually a chain: grams $\rightarrow$ moles $\rightarrow$ mole ratio $\rightarrow$ moles $\rightarrow$ grams.

The central workflow: the “mole road map”

Stoichiometry problems look different on the surface, but most follow the same logic:

Balance the equation (coefficients determine mole ratios).
Convert given information to moles (often from mass or solution volume).
Use mole ratios from the balanced equation to find moles of the desired species.
Convert to requested units (grams, molecules, solution concentration, etc.).

The balanced equation is non-negotiable. If it is not balanced, every ratio you compute will be wrong.

Reaction stoichiometry with masses (basic mass-to-mass)

When you’re given a mass of one reactant and asked for a mass of product, you are doing mass-to-mass stoichiometry.

Worked example 1: mass of product from mass of reactant

Solid magnesium reacts with oxygen to form magnesium oxide.

$2Mg + O_2 \rightarrow 2MgO$

If you start with $6.00\text{ g}$ of $Mg$ and excess $O_2$ , what mass of $MgO$ can form?

Step 1: Convert grams $Mg$ to moles $Mg$ .

Molar mass of $Mg$ is about $24.31\text{ g/mol}$ .

$n(Mg) = \frac{6.00\text{ g}}{24.31\text{ g/mol}} = 0.2468\text{ mol}$

Step 2: Use mole ratio from the balanced equation.

From $2Mg \rightarrow 2MgO$ , the ratio $Mg:MgO$ is $1:1$ .

$n(MgO) = 0.2468\text{ mol}$

Step 3: Convert moles $MgO$ to grams $MgO$ .

Molar mass of $MgO$ is about $40.30\text{ g/mol}$ .

$m(MgO) = 0.2468\text{ mol}\times 40.30\text{ g/mol} = 9.95\text{ g}$

So, $9.95\text{ g}$ of $MgO$ can form.

What commonly goes wrong here: students sometimes use the coefficient “2” as a multiplier on grams. Coefficients relate moles to moles, not grams to grams.

Limiting reactant (when “excess” is not given)

In real reactions, you often mix two reactants and one runs out first. The reactant that is used up first is the limiting reactant. It limits how much product can form.

The other reactant is in excess—some will be left over.

Conceptually, think of making sandwiches: 2 slices of bread and 1 slice of cheese makes 1 sandwich. If you have 10 slices of bread and 20 slices of cheese, bread is limiting because you can only make 5 sandwiches.

How to find the limiting reactant (two reliable methods)

Method A: “Calculate product from each reactant.”

Assume reactant 1 is limiting and calculate moles of product.
Assume reactant 2 is limiting and calculate moles of product.
The smaller amount of product is the real maximum; the reactant that produced that smaller amount is limiting.

Method B: “Compare needed ratio to available ratio.”
Convert to moles and compare $\frac{n(reactant\ A)}{n(reactant\ B)}$ to the coefficient ratio. This is faster but easier to mess up if you invert a ratio.

Worked example 2: identify limiting reactant and theoretical yield

Nitrogen reacts with hydrogen to form ammonia:

$N_2 + 3H_2 \rightarrow 2NH_3$

If you have $5.00\text{ mol}$ of $N_2$ and $12.0\text{ mol}$ of $H_2$ , which is limiting, and what is the theoretical yield of $NH_3$ in moles?

Step 1: Predict $NH_3$ from $N_2$ .

From $1\text{ mol }N_2$ you get $2\text{ mol }NH_3$ .

$n(NH_3) = 5.00\text{ mol }N_2\times \frac{2\text{ mol }NH_3}{1\text{ mol }N_2} = 10.0\text{ mol}$

Step 2: Predict $NH_3$ from $H_2$ .

From $3\text{ mol }H_2$ you get $2\text{ mol }NH_3$ .

$n(NH_3) = 12.0\text{ mol }H_2\times \frac{2\text{ mol }NH_3}{3\text{ mol }H_2} = 8.00\text{ mol}$

The smaller yield is $8.00\text{ mol}$ , so $H_2$ is the **limiting reactant** and the **theoretical yield** is $8.00\text{ mol }NH_3$ .

What commonly goes wrong: students pick the limiting reactant by choosing the smaller starting mole amount. That fails because the reaction may require different stoichiometric amounts (here you need 3 times as much $H_2$ as $N_2$ ).

Theoretical yield, actual yield, and percent yield

Even if stoichiometry predicts a maximum possible amount, real reactions may produce less product due to incomplete reaction, side reactions, loss during transfers, impurities, or equilibrium limitations.

Theoretical yield: the maximum amount predicted by stoichiometry (based on the limiting reactant).
Actual yield: what you actually isolate/measure in the lab.
Percent yield: how efficient the reaction was.

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

Percent yield is a powerful reality check: a calculated yield above 100% is a red flag. It usually means product was wet, impure, or your limiting-reactant/theoretical-yield calculation is wrong.

Stoichiometry with solutions: molarity as “moles per liter”

Many AP Chemistry reactions happen in aqueous solution. In that case, the most common way you’re given an amount is as a molarity.

Molarity is concentration in moles per liter:

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

Here $M$ is molarity, $n$ is moles of solute, and $V$ is volume of solution in liters.

The reason molarity matters for stoichiometry is that it gives you a direct path to moles:

$n = M\times V$

A very common error is forgetting that $V$ must be in liters, not milliliters.

Worked example 3: solution stoichiometry (precipitation)

Aqueous calcium chloride reacts with aqueous sodium carbonate to form solid calcium carbonate:

$CaCl_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaCl(aq)$

If you mix $25.0\text{ mL}$ of $0.150\text{ M }CaCl_2$ with excess $Na_2CO_3$ , what mass of $CaCO_3$ forms?

Step 1: Convert volume to liters and find moles of $CaCl_2$ .

$V = 0.0250\text{ L}$

$n(CaCl_2) = 0.150\text{ mol/L}\times 0.0250\text{ L} = 0.00375\text{ mol}$

Step 2: Use mole ratio to get moles of $CaCO_3$ .

The ratio $CaCl_2:CaCO_3$ is $1:1$ .

$n(CaCO_3) = 0.00375\text{ mol}$

Step 3: Convert moles $CaCO_3$ to mass.

Molar mass of $CaCO_3$ is about $100.09\text{ g/mol}$ .

$m(CaCO_3) = 0.00375\text{ mol}\times 100.09\text{ g/mol} = 0.375\text{ g}$

Dilution (often a “hidden” stoichiometry step)

Sometimes you are not reacting yet—you are preparing a solution by diluting a concentrated stock. The amount of solute (moles) stays the same; only volume changes.

The dilution relationship is:

$M_1V_1 = M_2V_2$

Here $M_1$ and $V_1$ refer to the initial (stock) solution, and $M_2$ and $V_2$ refer to the final (diluted) solution.

A misconception: students treat dilution like a reaction where moles change. In dilution, you are not consuming solute—just adding solvent.

Worked example 4: using dilution before reaction stoichiometry

You need $250.0\text{ mL}$ of $0.100\text{ M }NaCl$ from a $1.00\text{ M }NaCl$ stock. What volume of stock do you use?

$M_1V_1 = M_2V_2$

$1.00\text{ M}\times V_1 = 0.100\text{ M}\times 0.2500\text{ L}$

$V_1 = 0.0250\text{ L} = 25.0\text{ mL}$

Exam Focus

Typical question patterns:
- Convert mass to moles, use coefficients, convert to mass (often product mass or reactant required).
- Limiting reactant and theoretical yield from two given reactant quantities.
- Solution stoichiometry using $n = M\times V$ (sometimes paired with dilution using $M_1V_1 = M_2V_2$ ).
Common mistakes:
- Using an unbalanced equation (or using coefficients from a not-yet-balanced equation).
- Skipping the “moles step” and trying to apply mole ratios to grams or milliliters.
- Not converting $mL$ to $L$ when using molarity (leading to answers off by a factor of 1000).

Introduction to Titration

What a titration is (and what it’s for)

A titration is a lab method for determining an unknown concentration (or sometimes an unknown amount) by reacting it with a solution of known concentration in a carefully measured way.

In Unit 4, titration is mainly about stoichiometry in solution: you measure volumes, convert to moles using molarity, use the balanced equation to relate reactants, and then solve for the unknown concentration.

A useful analogy: titration is like using a “known-strength” cleaner (known concentration) dispensed in a measured amount to neutralize a spill of unknown size. If you know how strong the cleaner is and exactly how much it took to finish the job, you can work backward to how big the spill was.

Core titration vocabulary (each term has a specific job)

Titrant: the solution with known concentration, delivered from a buret.
Analyte: the solution with unknown concentration (the one you’re trying to determine), placed in a flask.
Standard solution: a solution whose concentration is accurately known (often the titrant).
Equivalence point: the moment when the reactants have been combined in exact stoichiometric proportions according to the balanced equation.
Endpoint: the observed signal (often an indicator color change) used to estimate that you’ve reached the equivalence point.
Indicator: a substance that changes color near a particular condition (commonly near a certain pH in acid–base titrations).

A common confusion is treating “endpoint” and “equivalence point” as the same. They are related but not identical: equivalence is the theoretical stoichiometric point; endpoint is what you see.

The key stoichiometric idea: moles at equivalence follow the balanced equation

At the equivalence point, the moles of titrant added and the moles of analyte originally present satisfy the mole ratio from the balanced equation.

For example, for a simple strong acid–strong base neutralization:

$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

The mole ratio is $1:1$ . At equivalence:

$n(HCl) = n(NaOH)$

But if the acid and base have different stoichiometric coefficients, you must use those coefficients. For instance:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)$

At equivalence, moles satisfy:

$n(NaOH) = 2\times n(H_2SO_4)$

This is the most important conceptual point in titration calculations: you are not assuming “moles acid equals moles base” unless the balanced equation supports that.

How titration data is collected (what you actually measure)

In a typical setup, you use a buret to deliver titrant. A buret reading is a volume mark; you determine volume delivered by subtracting the initial reading from the final reading.

So the delivered volume is:

$V_{delivered} = V_{final} - V_{initial}$

Then you convert that delivered volume into moles of titrant using:

$n = M\times V$

(again, $V$ must be in liters).

The standard titration calculation pattern

Most “intro titration” problems follow this chain:

Write and balance the relevant reaction.
Calculate moles of titrant added at equivalence using $n = M\times V$ .
Use mole ratios to determine moles of analyte originally present.
If asked for concentration, divide by analyte volume.

You can think of it as: measured volume $\rightarrow$ moles titrant $\rightarrow$ moles analyte $\rightarrow$ concentration analyte.

Worked example 1: find unknown molarity from a 1:1 neutralization

A $25.00\text{ mL}$ sample of $HCl(aq)$ of unknown concentration is titrated with $0.1000\text{ M }NaOH$ . The equivalence point is reached after $18.60\text{ mL}$ of $NaOH$ is delivered. Find the molarity of $HCl$ .

Step 1: Balanced equation and stoichiometric ratio.

$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

Ratio is $1:1$ .

Step 2: Convert delivered $NaOH$ volume to liters and find moles.

$V(NaOH) = 0.01860\text{ L}$

$n(NaOH) = 0.1000\text{ mol/L}\times 0.01860\text{ L} = 0.001860\text{ mol}$

Step 3: Use stoichiometry to find moles of $HCl$ .

At equivalence, $n(HCl) = n(NaOH)$ .

$n(HCl) = 0.001860\text{ mol}$

Step 4: Divide by analyte volume to get molarity.

$V(HCl) = 0.02500\text{ L}$

$M(HCl) = \frac{0.001860\text{ mol}}{0.02500\text{ L}} = 0.07440\text{ M}$

So the $HCl$ concentration is $0.07440\text{ M}$ .

What commonly goes wrong: using $25.00$ as liters instead of milliliters; that would make your molarity 1000 times smaller.

Worked example 2: titration with a 2:1 stoichiometric relationship

A $10.00\text{ mL}$ sample of $H_2SO_4(aq)$ is titrated to equivalence with $0.2500\text{ M }NaOH$ . It takes $32.40\text{ mL}$ of base to reach equivalence. Find the molarity of $H_2SO_4$ .

Step 1: Balanced equation.

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l)$

This tells you $2$ moles of $NaOH$ react per $1$ mole of $H_2SO_4$ .

Step 2: Moles of $NaOH$ used.

$V(NaOH) = 0.03240\text{ L}$

$n(NaOH) = 0.2500\text{ mol/L}\times 0.03240\text{ L} = 0.008100\text{ mol}$

Step 3: Convert to moles of $H_2SO_4$ using the mole ratio.

$n(H_2SO_4) = 0.008100\text{ mol}\times \frac{1\text{ mol }H_2SO_4}{2\text{ mol }NaOH} = 0.004050\text{ mol}$

Step 4: Divide by analyte volume to get molarity.

$V(H_2SO_4) = 0.01000\text{ L}$

$M(H_2SO_4) = \frac{0.004050\text{ mol}}{0.01000\text{ L}} = 0.4050\text{ M}$

What commonly goes wrong: assuming a 1:1 relationship for all acid–base titrations. The coefficients decide the mole relationship, not the fact that it’s “acid vs base.”

Choosing the reaction: molecular vs net ionic (why it matters)

In aqueous reactions, especially acid–base and precipitation, you may be asked to write the net ionic equation. For titration calculations, you can often use a correct molecular equation (as long as it’s balanced and represents the reacting species), but net ionic can clarify what actually reacts.

For example, for a strong acid–strong base neutralization, the net ionic form is:

$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

This highlights that spectator ions (like $Na^+$ and $Cl^-$ ) do not affect the stoichiometric “reactive core.”

A common trap: writing an incorrect net ionic equation by canceling species that aren’t actually identical (for example, canceling weak acids as if they fully dissociate). In this Unit 4 “intro” context, the main emphasis is usually correct balancing and mole ratios, not equilibrium treatment—still, your equation must represent the chemistry you claim is occurring.

Practical lab details that affect calculations

Even without doing a full lab report, AP-style questions sometimes include realistic measurement details.

Buret readings: You calculate delivered volume by subtraction. Using the final reading directly (instead of final minus initial) is a frequent mistake.
Significant figures: Volumetric glassware is precise; report concentrations consistent with given data.
Overshooting the endpoint: Adding extra titrant past the endpoint makes the calculated analyte concentration appear too high (because you think more titrant was required than truly necessary).

Why titration connects strongly to stoichiometry

Titration is essentially stoichiometry with a convenient measurement tool. Mass measurements can be inconvenient for solutions, but volume is easy to measure accurately. Molarity converts that volume right back into moles, and once you’re in moles, the balanced equation does the rest.

Exam Focus

Typical question patterns:
- Given titrant molarity and volume at equivalence, find analyte molarity (often a neutralization).
- Use a non-1:1 balanced equation (diprotic acid, carbonate reacting with acid, etc.) so you must apply coefficients.
- Interpret buret data (initial and final readings) to find volume delivered before doing stoichiometry.
Common mistakes:
- Confusing equivalence point with endpoint, or assuming the indicator changes exactly at equivalence without being told.
- Forgetting to convert $mL$ to $L$ in $n = M\times V$ .
- Assuming $n(acid) = n(base)$ even when the balanced equation shows a different ratio (for example, $1:2$ ).