Chemistry 2e · Stoichiometry of Chemical Reactions

Quantitative Chemical Analysis

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Qualitative analysis asks what is present; quantitative analysis asks how much. Stoichiometry is the engine: you measure something easy and reliable — a precipitate's mass, a standard solution's volume, a combustion gas's mass — then use mole ratios from a balanced equation to convert that measurement into the amount of the substance you care about (the ). This topic surveys the three classic approaches: (weighing a product), titrimetric analysis (measuring a solution's volume), and (weighing the gases a sample produces).

Every method shares the same logical chain:

$measured quantity \to moles of known species mole ratio ⟶ moles of analyte \to mass or concentration of analyte$

The balanced equation is the bridge at the middle of that chain. If the mole ratio is wrong, every downstream number is wrong no matter how precise the measurement.

Why this matters

Quantitative analysis is how chemists answer questions that affect health, safety, and money every day:

Clinical labs measure blood glucose, cholesterol, and electrolyte concentrations to guide medical decisions.
Environmental monitoring checks drinking water for nitrate, lead, or chloride levels against legal limits.
Food and beverage production verifies that products meet label claims, from vitamin content to alcohol percentage.

In all of these settings, a result drives a decision, so knowing how the numbers are derived — and where error creeps in — matters beyond the classroom.

The college version

Core Concepts

The analytical chain in practice

Quantitative analysis starts by choosing a reaction that is complete (goes essentially to completion), specific (tied to the analyte), and stoichiometrically known. The measured quantity can be a mass, a volume, or a gas mass; from there it is a stoichiometry problem in disguise.

Gravimetric analysis: weighing a precipitate

In gravimetric analysis the analyte is converted into a compound of known composition that precipitates from solution. The precipitate is collected, washed, dried, and weighed; its known formula and the mole ratio yield the mass of the analyte. A classic example is chloride determination: excess silver nitrate precipitates all chloride as AgCl, whose mass is measured accurately; the 1:1 Cl:AgCl ratio does the rest.

Titrimetric analysis: measuring a standard solution

In a titration, a solution of known concentration (the ) is added slowly to a measured portion of analyte until the reaction is just complete — the . Titrant volume and molarity give moles of titrant:

$n = M \times V$

where M is molarity (mol/L) and V is volume in liters. A mole ratio from the balanced equation converts titrant moles into analyte moles, and dividing by the analyte's volume gives its molarity. An indicator signals when the reaction is complete; the observed color change is the , which should sit close to the equivalence point.

Combustion analysis: weighing the gases

Combustion analysis determines the of a C–H–O compound. A weighed sample is burned; the CO₂ and H₂O produced are trapped and weighed. All carbon in the sample becomes CO₂, and all hydrogen becomes H₂O, so:

$n(C) = m(CO 2) 44.01 g/mol n(H) = 2 \times m(H 2 O) 18.02 g/mol$

Oxygen is found by difference: subtract the masses of C and H from the sample mass and convert to moles. Divide the three mole quantities by the smallest to get the empirical formula ratio.

Choosing a method

Gravimetry is slow but highly accurate, making it a reference standard; titrations are fast and versatile for routine acid–base and redox testing; combustion analysis characterizes new organic compounds. Choose by accuracy needs, sample size, and speed.

Common Confusions

Do Not Confuse	With	Difference
Endpoint	Equivalence point	Endpoint is the observed color change; equivalence point is the stoichiometric completion. A bad indicator makes them differ.
Volume in mL	Volume in L	Molarity needs liters: divide mL by 1000 first, or the mole count is off by 1000×.
Mass of precipitate	Mass of analyte	The precipitate contains the analyte plus the precipitating ion — convert via moles and the mole ratio.
Empirical formula	Molecular formula	Empirical is the simplest ratio; the molecular formula is a whole-number multiple (e.g., CH₂O vs C₆H₁₂O₆).
Gravimetric method	Titrimetric method	One weighs a solid product; the other measures a solution volume.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a detective who cannot see the thief but can see the thief's footprints. Quantitative analysis works the same way: instead of measuring the substance you want directly, you react it with something that makes an easy-to-see clue — a solid that falls out of the liquid, or a color change. From the size of the clue, you work backward to figure out exactly how much of the hidden substance was there.

Worked example

Example 1: Gravimetric determination of chloride

A 0.4550 g sample of an unknown salt is dissolved and treated with excess silver nitrate. The chloride precipitates completely:

$Ag + + Cl - \to AgCl (s)$

The dried precipitate has a mass of 0.6280 g. Molar masses: AgCl = 143.32 g/mol, Cl = 35.45 g/mol. What is the mass percent of chloride in the salt?

Moles of precipitate:

$n(AgCl) = 0.6280 g 143.32 g/mol = 4.382 \times 10 -3 mol$

Moles of chloride (1:1 ratio):

$n(Cl -) = 4.382 \times 10 -3 mol$

Mass of chloride:

$m(Cl) = 4.382 \times 10 -3 mol \times 35.45 g/mol = 0.1553 g$

Mass percent:

$% Cl = 0.1553 g 0.4550 g \times 100 % = 34.1 %$

Example 2: Titration of a diluted vinegar sample

A 25.00 mL sample of diluted vinegar is titrated with 0.1000 M NaOH. The endpoint is reached after 32.50 mL of NaOH. The reaction is:

$HC 2 H 3 O 2 + NaOH \to NaC 2 H 3 O 2 + H 2 O$

Moles of NaOH (titrant):

$n(NaOH) = 0.1000 mol L \times 0.03250 L = 3.250 \times 10 -3 mol$

Moles of acetic acid (1:1 ratio):

$n(HC 2 H 3 O 2) = 3.250 \times 10 -3 mol$

Molarity of the vinegar sample:

$M = 3.250 \times 10 -3 mol 0.02500 L = 0.1300 M$

As a mass–volume percentage: 0.1300 mol/L × 60.05 g/mol = 7.81 g/L, about 0.78 g per 100 mL.

Example 3: Combustion analysis of an unknown compound

A 0.2402 g sample of an unknown compound containing only carbon, hydrogen, and oxygen is burned completely, producing 0.3521 g of CO₂ and 0.1442 g of H₂O. Find the empirical formula.

Moles of C:

$n(C) = 0.3521 g 44.01 g/mol = 8.00 \times 10 -3 mol$

Moles of H (two per water molecule):

$n(H) = 2 \times 0.1442 g 18.02 g/mol = 1.60 \times 10 -2 mol$

Masses of C and H, then O by difference:

$m(C) = 8.00 \times 10 -3 mol \times 12.01 g/mol = 0.0961 g$

$m(H) = 1.60 \times 10 -2 mol \times 1.008 g/mol = 0.0161 g$

$m(O) = 0.2402 - 0.0961 - 0.0161 = 0.1280 g n(O) = 0.1280 g 16.00 g/mol = 8.00 \times 10 -3 mol$

Ratio C : H : O = 8.00×10⁻³ : 1.60×10⁻² : 8.00×10⁻³ = 1 : 2 : 1, so the empirical formula is CH₂O.

Key takeaways

Every method follows the same chain: measurement → moles of a known species → mole ratio → moles of analyte.
Gravimetry converts analyte to a weighable precipitate of known formula; the mole ratio is the key conversion.
Titration uses n = M × V for the titrant, then a mole ratio to the analyte.
The equivalence point is the ideal completion point; the endpoint is the observed signal — they should nearly coincide.
In combustion analysis, all C ends up in CO₂ and all H in H₂O; O is found by mass difference.
Units must cancel at every step: molarity in mol/L requires volume in liters, not milliliters.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

Write the general chain of conversions used in any quantitative analysis.
Show answer
Measured quantity → moles of known species → mole ratio from the balanced equation → moles of analyte → mass or concentration of analyte.
In a titration, 20.00 mL of 0.1500 M HCl neutralizes 25.00 mL of NaOH. What is the NaOH molarity?
Show answer
Moles HCl = 0.1500 × 0.02000 = 3.000×10⁻³ mol = moles NaOH; M = 3.000×10⁻³ ÷ 0.02500 = 0.1200 M.
Why is it important that a gravimetric precipitation be complete?
Show answer
If precipitation is incomplete, some analyte never becomes precipitate and the result is too low — the measured mass no longer represents the full sample.
In combustion analysis, how do you find the oxygen content of a C–H–O sample?
Show answer
Oxygen is found by difference: subtract the masses of carbon and hydrogen from the original sample mass, then convert the remainder to moles using 16.00 g/mol.
What is the difference between the equivalence point and the endpoint of a titration?
Show answer
The equivalence point is where stoichiometrically equivalent amounts have reacted; the endpoint is the observable signal (color change) that approximates it. Indicator choice keeps them close.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Analyte: The substance whose amount or concentration is being determined.
Gravimetric analysis: Method that measures the mass of a precipitate to find analyte amount.
Titrant: Solution of known concentration added during a titration.
Equivalence point: Point where titrant and analyte react in exact stoichiometric ratio.
Endpoint: Observed signal (color change) indicating the reaction is complete.
Combustion analysis: Burning a sample and weighing the CO₂ and H₂O produced.
Empirical formula: Simplest whole-number ratio of atoms in a compound.

Sources & references

openstax.org — Chemistry 2e

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