4.5 Stoichiometry · 4.6 Introduction to Titration

A chemical equation becomes experimentally useful only when its coefficients are treated as a quantitative recipe: they connect particles, moles, masses, gas volumes, and solution concentrations.

The stoichiometric map

A balanced equation such as

$$ 2H_2(g)+O_2(g)\rightarrow 2H_2O(g) $$

does not say that $2\ \mathrm{g}$ of hydrogen reacts with $1\ \mathrm{g}$ of oxygen. Its coefficients give mole relationships: $2$ mol $H_2$ react with $1$ mol $O_2$ to produce $2$ mol $H_2O$. Every advanced stoichiometry problem is a controlled conversion through that ratio.

The general pathway is:

$$ \text{given quantity} \rightarrow \text{moles of given substance} \rightarrow \text{moles of desired substance} \rightarrow \text{requested unit} $$

The middle conversion uses the coefficients of the balanced equation. The first and last conversions depend on the form of the data: molar mass for mass, $PV=nRT$ for gases, and molarity for aqueous solutions.

Gas stoichiometry: using $PV=nRT$

For a gas, the ideal gas law relates pressure, volume, amount, and temperature:

$$ PV=nRT $$

Here, $P$ is pressure, $V$ is volume, $n$ is amount in moles, $R$ is the gas constant, and $T$ is absolute temperature in kelvins. The temperature must be converted using $T(\mathrm{K})=T(^\circ\mathrm{C})+273.15$.

Worked example: identifying the limiting reactant

A reaction produces hydrogen gas according to

$$ Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g) $$

Suppose $0.120$ mol of magnesium reacts with $0.180$ mol of hydrochloric acid. Magnesium would require

$$ 0.120\ \mathrm{mol\ Mg}\times \frac{2\ \mathrm{mol\ HCl}}{1\ \mathrm{mol\ Mg}} =0.240\ \mathrm{mol\ HCl} $$

Only $0.180$ mol $HCl$ is available, so $HCl$ is the limiting reactant. The amount of hydrogen is therefore

$$ 0.180\ \mathrm{mol\ HCl}\times \frac{1\ \mathrm{mol\ H_2}}{2\ \mathrm{mol\ HCl}} =0.0900\ \mathrm{mol\ H_2} $$

At $298\ \mathrm{K}$ and $1.00\ \mathrm{atm}$,

$$ V=\frac{nRT}{P} =\frac{(0.0900)(0.08206)(298)}{1.00} =2.20\ \mathrm{L} $$

The limiting reactant controls the product amount; the excess reactant does not.

Misconception check — “The reactant with fewer grams is limiting.” Limiting status cannot be judged from mass alone. Convert each reactant to the amount of product it could form, or compare the available-to-required mole ratios. A large mass of a substance with a large molar mass may represent fewer moles than expected.

Aqueous stoichiometry and molarity

Molarity is the amount of solute per liter of solution:

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

where $M$ has units of $\mathrm{mol,L^{-1}}$ and $V$ must be measured in liters. Rearranging gives the especially useful relationship

$$ n=MV $$

Thus, a measured solution volume can enter a stoichiometric calculation directly after multiplication by its molarity.

For example, $25.00\ \mathrm{mL}$ of $0.1500\ \mathrm{M}$ $NaOH$ contains

$$ n=(0.1500\ \mathrm{mol,L^{-1}})(0.02500\ \mathrm{L}) =0.003750\ \mathrm{mol\ NaOH} $$

If the reaction requires a $1:1$ mole ratio between $NaOH$ and another reactant, then $0.003750$ mol of that reactant is consumed. If the ratio is not $1:1$, the balanced equation—not the volumes alone—determines the conversion.

CED traceability: Topic 4.5 Stoichiometry centers on Learning Objective 4.5.A and includes the quantitative integration identified in Essential Knowledge 4.5.A.3: stoichiometric reasoning may involve masses, gas quantities, and molarity in aqueous solutions.

Titration: measuring an unknown concentration

A titration determines an unknown concentration by reacting the unknown solution with a solution of known concentration. The known solution is the titrant; the solution being analyzed is the sample. A buret delivers the titrant while the volume required for the reaction is measured accurately.

At the reaction’s equivalence point, stoichiometrically sufficient titrant has been added. An indicator may produce a visible color change near this point, but the indicator’s color change is an experimental signal, not the definition of equivalence.

For a $1:1$ reaction, the calculation is often

$$ M_{\text{unknown}}V_{\text{unknown}}

M_{\text{titrant}}V_{\text{titrant}} $$

For any other ratio, first calculate titrant moles and then apply the balanced-equation mole ratio.

Titration example

If $18.60\ \mathrm{mL}$ of $0.1000\ \mathrm{M}$ $NaOH$ neutralizes $25.00\ \mathrm{mL}$ of a monoprotic acid $HA$,

$$ n_{NaOH}=(0.1000)(0.01860)=0.001860\ \mathrm{mol} $$

Because

$$ HA+NaOH\rightarrow NaA+H_2O $$

has a $1:1$ ratio,

$$ M_{HA}=\frac{0.001860}{0.02500} =0.07440\ \mathrm{M} $$

Misconception check — “Equivalence point means equal volumes.” Equal volumes matter only in special cases. Equivalence means equal stoichiometric amounts after accounting for the reaction coefficients. A $2:1$ reaction requires a different relationship between the measured volumes and concentrations.

Investigation 7: from measurement to purification

Investigation 7 provides a practical setting for connecting mathematical chemistry to a laboratory process such as determining composition, monitoring a reaction, or evaluating purification. A defensible investigation records initial and final buret readings, solution concentrations, sample volume, balanced-equation ratios, calculations, uncertainty, and a conclusion supported by measured evidence.

The laboratory sequence is:

1. Measure the sample or product carefully.
2. Add titrant until the endpoint is reached.
3. Convert titrant volume to moles using $n=MV$.
4. Use the balanced equation to determine moles of the unknown.
5. Convert to concentration, mass, purity, or yield.
6. Compare the result with a claimed or expected value.

This is where Science Practice 1: Models and Representations, especially 1.B, becomes visible: the balanced equation is a particulate-level model that predicts the macroscopic volume required in the buret. Science Practice 3: Representing Data and Phenomena supports recording and interpreting buret data; Science Practice 5: Mathematical Routines supports dimensional analysis and significant figures; and Science Practice 6: Argumentation supports a conclusion linking the calculated amount to evidence.

Retrieval check

A $0.250\ \mathrm{M}$ solution provides $0.00500\ \mathrm{mol}$ of titrant. If the balanced reaction requires $2$ mol of titrant per $1$ mol of analyte, the analyte amount is

$$ 0.00500\ \mathrm{mol}\times\frac{1\ \mathrm{mol\ analyte}}{2\ \mathrm{mol\ titrant}} =0.00250\ \mathrm{mol} $$

The essential question is not “What formula applies?” but “What measured quantity must be converted into moles, and what coefficient ratio connects it to the unknown?”