Chemistry: Atoms First 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 answers "what is present?" Quantitative chemical analysis answers "how much is present?" This topic surveys the two classical laboratory strategies for measuring amounts of a substance in a sample: , which turns a mass measurement into an amount, and titrimetric (volumetric) analysis, which turns a carefully measured volume of reacting solution into an amount.

Both methods are stoichiometry in action. A measured quantity — the mass of a , or the volume and concentration of a — is converted into moles using molar mass or molarity, and the mole ratios from a balanced equation convert those moles into the amount of the , the substance being measured. The tools are the same conversion factors used in reaction stoichiometry, applied to real measurements.

Why this matters

Quality control: drug manufacturers, food producers, and water-treatment plants must certify how much of a key ingredient is present within tight tolerances.
Environmental and clinical labs: measuring pollutants in water, or blood glucose and drug levels, requires quantitative methods with known accuracy.
Legal weight: a reported concentration is only as trustworthy as the method behind it, which is why these analyses are held to strict standards of precision and accuracy.
Exams: expect titration and gravimetric calculation problems; fluency with mole conversions and balanced-equation ratios is the core skill being tested.

The college version

Core Concepts

Gravimetric analysis: mass as the measurement

In gravimetric analysis, the analyte is converted into a precipitate — an essentially insoluble solid — then filtered, dried, and weighed. The key insight: the mass of the precipitate is directly related to the mass of the original analyte through the mole ratios in a balanced equation.

The workflow: dissolve the sample; add a reagent that forms an insoluble precipitate of known, fixed composition (the reaction must go essentially to completion); filter, wash, and dry; weigh to constant mass; then convert precipitate mass → moles of precipitate → moles of analyte → mass of analyte.

For example, chloride ion can be precipitated as silver chloride:

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

Because the reaction is 1 mol Cl⁻ per 1 mol AgCl, the moles of chloride in the sample equal the moles of AgCl weighed.

Titrimetric (volumetric) analysis: volume as the measurement

In a titration, a solution of known concentration — the titrant — is added from a buret to a measured volume of analyte solution until the reaction is just complete. The point of chemical equivalence is the ; the point where an indicator visibly changes color is the . Ideally the two coincide within one drop.

The calculation chain is:

$mol titrant = M titrant \times V titrant$

then moles of titrant are converted to moles of analyte using the balanced equation's mole ratio, and finally to concentration or mass of analyte.

Standard solutions and standardization

A titration is only as good as the titrant's known concentration. A is a pure, stable, solid compound (e.g., potassium hydrogen phthalate, KHP, for base titrations) that can be weighed accurately and dissolved to make a solution of exactly known molarity. uses a primary standard to determine the exact concentration of a solution like NaOH, which absorbs CO₂ and water and cannot be weighed directly as a pure solid.

Common Confusions

Do Not Confuse	With	Difference
Equivalence point	Endpoint	Equivalence point is the stoichiometrically exact point; endpoint is where the indicator changes color. They should match within one drop
Gravimetric analysis	Titrimetric analysis	Gravimetric measures a mass of precipitate; titrimetric measures a volume of titrant solution
Mole ratio from coefficients	Subscripts in formulas	Coefficients (balanced equation) give mole ratios between substances; subscripts give atoms within one formula unit
Molarity M	Moles n	Molarity is mol L⁻¹; moles is the amount itself. n = M × V only when volume is in liters
Excess reagent	Limiting reagent	In gravimetry, excess precipitating agent forces complete reaction; the analyte is the limiting reagent
Standardization	Dilution	Standardization fixes a concentration by reacting with a primary standard; dilution just adds solvent

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine you want to know how much salt is in a jar of soup powder. One way: add a chemical that turns the salt into sand-like grains, catch the grains on a filter, dry them, and weigh them — heavier grains mean more salt. Another way: add drops of a "salt-detecting" liquid of known strength until the color changes, counting the drops to figure out the salt. Both ways use a recipe (the balanced equation) to turn a simple measurement — mass or drops — into an answer.

Worked example

Example 1: Gravimetric determination of chloride

A 0.4861 g sample of an ionic mixture is dissolved in water, and excess silver nitrate precipitates all chloride as AgCl. The dried precipitate has a mass of 0.4258 g. What is the mass percent of chloride in the sample?

Step 1 — Balanced equation:

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

Step 2 — Convert precipitate mass to moles (formula first, then substitute):

$n AgCl = m AgCl M AgCl = 0.4258 g 143.32 g mol -1 = 2.971 \times 10 -3 mol$

Step 3 — Mole ratio (1 mol Cl⁻ : 1 mol AgCl):

$n Cl = n AgCl = 2.971 \times 10 -3 mol$

Step 4 — Convert moles of chloride to mass:

$m Cl = n Cl \times M Cl = (2.971 \times 10 -3 mol)(35.45 g mol -1) = 0.1053 g$

Step 5 — Mass percent:

$% Cl = m Cl m sample \times 100 % = 0.1053 g 0.4861 g \times 100 % = 21.66 %$

Dimensional check: grams cancel in the mole conversion, mol cancels in the mass conversion, and the percent is unitless. The sample is 21.66% chloride by mass.

Example 2: Titration of HCl with NaOH

A 25.00 mL sample of HCl requires 32.50 mL of 0.1500 M NaOH to reach the endpoint. What is the molarity of the HCl?

Step 1 — Balanced equation (1:1 mole ratio):

$HCl (aq) + NaOH (aq) \to NaCl (aq) + H 2 O (l)$

Step 2 — Moles of titrant:

$n NaOH = M NaOH \times V NaOH = (0.1500 mol L -1)(0.03250 L) = 4.875 \times 10 -3 mol$

Step 3 — Mole ratio 1:1, so n_HCl = n_NaOH.

Step 4 — Molarity of the acid:

$M HCl = n HCl V HCl = 4.875 \times 10 -3 mol 0.02500 L = 0.1950 M$

Units check: mol divided by L gives mol L⁻¹ (M), as expected for a concentration.

Example 3: Standardizing NaOH with KHP

A student dissolves 0.5012 g of potassium hydrogen phthalate (KHP, molar mass 204.22 g mol⁻¹) and titrates it with NaOH, requiring 24.55 mL to reach the endpoint. The reaction is 1:1 (one acidic proton per KHP). Find the molarity of the NaOH.

Step 1 — Moles of primary standard:

$n KHP = 0.5012 g 204.22 g mol -1 = 2.454 \times 10 -3 mol$

Step 2 — 1:1 mole ratio, so n_NaOH = n_KHP.

Step 3 — Molarity of NaOH:

$M NaOH = 2.454 \times 10 -3 mol 0.02455 L = 0.1000 M$

This standardized NaOH can now be used as a titrant for other acid samples.

Key takeaways

Quantitative analysis measures amount; qualitative analysis identifies what.
Gravimetric analysis: precipitate → filter → dry → weigh; convert precipitate mass to analyte mass via mole ratios.
Titrimetric analysis: moles of titrant = M × V; convert to analyte moles via the balanced equation.
The equivalence point (stoichiometric) and endpoint (indicator color change) should be within one drop of each other.
A primary standard is pure, stable, and of known molar mass; standardization fixes a titrant's exact concentration.
Percent composition by mass: % analyte = mass analytemass sample × 100%.
Always write and balance the equation first; the mole ratio is the bridge between substances.

Check yourself

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

What is the difference between qualitative and quantitative chemical analysis?
Show answer
Qualitative analysis identifies which substances are present; quantitative analysis determines how much of a substance is present.
A sample yields 0.1053 g of chloride from a 0.4861 g sample. What is the mass percent of chloride?
Show answer
% Cl = (0.1053 / 0.4861) × 100% = 21.66%.
In a titration, 32.50 mL of 0.1500 M NaOH neutralizes 25.00 mL of HCl. What is the molarity of the HCl?
Show answer
M_HCl = (0.1500 × 0.03250) / 0.02500 = 0.1950 M.
Why is a primary standard (like KHP) preferred over directly weighing NaOH for a standard solution?
Show answer
A primary standard is pure, stable, weighable, and of known molar mass, so its solution concentration is exactly known. NaOH absorbs water and CO₂ and cannot be weighed as a pure solid.
What role does the balanced chemical equation play in both gravimetric and titrimetric calculations?
Show answer
The balanced equation provides the mole ratio that converts moles of precipitate or titrant into moles of analyte — the essential bridge in every calculation.
A student stops the titration one drop past the color change. Is the reported analyte concentration too high or too low? Why?
Show answer
Too low: slightly more titrant was delivered than stoichiometrically required, so the extra volume inflates the moles of analyte and the calculated concentration is overstated.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Analyte: The substance whose amount is being measured
Gravimetric analysis: A method that measures the mass of a precipitate to find the amount of analyte
Titrant: The solution of known concentration added from a buret
Equivalence point: The point where added titrant is stoichiometrically equal to analyte
Endpoint: The point where the indicator changes color
Primary standard: A pure, stable solid of known molar mass used to make standard solutions
Standardization: Determining a solution's exact concentration using a primary standard
Precipitate: An insoluble solid that forms when two solutions react

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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