Gravimetric Analysis- Free Response Questions and Solutions

What You Need to Know About Gravimetric Analysis FRQs

Gravimetric analysis free response questions show up on AP Chemistry and college-level analytical chemistry exams. They're not going anywhere, and they're not getting easier. If you want the points, you need to understand the method—not just memorize the steps.

These questions ask you to calculate amounts of substances based on mass measurements. The math is straightforward. The execution is where students lose marks.

The Core Principle Behind Gravimetric Analysis

You measure a precipitate, dry it, weigh it, and use stoichiometry to find the original analyte. That's it. The precipitate's mass tells you what you need to know.

Every calculation in gravimetric analysis follows the same chain:

• Mass of precipitate → moles of precipitate (using molar mass)
• Moles of precipitate → moles of analyte (using stoichiometric ratio)
• Moles of analyte → mass or concentration of analyte (using molar mass)

Miss any link in this chain and your answer is wrong. No partial credit for getting the idea right.

Precipitation Gravimetry: The Standard Method

Most FRQs use precipitation gravimetry. The analyte forms an insoluble precipitate with a reagent, then you isolate and weigh it.

What Makes a Good Precipitate

Your precipitate needs to be:

• Insoluble — low solubility product (Ksp) is essential
• Filterable — large crystals that don't clog filter paper
• Pure — no co-precipitation of other ions
• Known composition — definite stoichiometry

Common examples include AgCl for chloride analysis, BaSO₄ for sulfate, and Fe(OH)₃ for iron.

Solving Gravimetric FRQs: A Practical Approach

Step 1: Identify What They're Asking

Read the question twice. Circle the analyte and the precipitate. If you don't know what you're solving for, stop and re-read.

Step 2: Write the Relevant Reaction

Balance the precipitation reaction. This is non-negotiable. A wrong formula means wrong everything.

Example:

NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

The precipitate is AgCl. That's what you'll weigh.

Step 3: Calculate Moles of Precipitate

Divide the mass of precipitate by its molar mass. Show this step. Examiners want to see your work.

Mass of AgCl = 0.287 g
Molar mass of AgCl = 143.32 g/mol
Moles AgCl = 0.287 / 143.32 = 0.00200 mol

Step 4: Convert to Moles of Analyte

Use the stoichiometric ratio from the balanced equation. In the AgCl example, the ratio is 1:1. If it were 2:1, you'd need to divide or multiply accordingly.

Step 5: Calculate Final Answer

Convert moles of analyte to mass or concentration. Include units. Check significant figures.

Sample FRQ #1: Determining Chloride in Water

Question: A 50.0 mL sample of groundwater is analyzed for chloride content. Excess AgNO₃ is added, precipitating all chloride as AgCl. The precipitate is filtered, dried, and weighs 0.356 g. Calculate the concentration of chloride in the groundwater sample in mg/L.

Solution:

Step 1: Identify the precipitate — AgCl.

Step 2: Calculate moles of AgCl:

0.356 g ÷ 143.32 g/mol = 0.002484 mol AgCl

Step 3: Determine moles of Cl⁻:

AgCl → Ag⁺ + Cl⁻ (1:1 ratio)

Moles Cl⁻ = 0.002484 mol

Step 4: Convert to mass of Cl:

0.002484 mol × 35.45 g/mol = 0.0881 g = 88.1 mg

Step 5: Calculate concentration:

88.1 mg ÷ 0.0500 L = 1762 mg/L

That's your answer. Show every step. Missing steps means lost points.

Sample FRQ #2: Sulfate Analysis

Question: A 0.512 g sample of Na₂SO₄ is dissolved in water. Excess BaCl₂ is added, precipitating BaSO₄. The precipitate mass is 0.458 g. Calculate the percent purity of the original sample.

Solution:

Balanced reaction: Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl

Moles of BaSO₄ formed:

0.458 g ÷ 233.39 g/mol = 0.001962 mol BaSO₄

Stoichiometry: 1 mol Na₂SO₄ produces 1 mol BaSO₄

Moles Na₂SO₄ in sample = 0.001962 mol

Mass of Na₂SO₄ in sample:

0.001962 mol × 142.04 g/mol = 0.279 g

Percent purity:

(0.279 g ÷ 0.512 g) × 100 = 54.5%

Sample FRQ #3: Hydrate Analysis

Question: A 1.246 g sample of a metal sulfate hydrate is heated to remove water, leaving 0.854 g of anhydrous residue. The residue is dissolved and treated with excess BaCl₂, yielding 1.163 g of BaSO₄ precipitate. Determine the formula of the original hydrate.

Solution:

Step 1: Find moles of SO₄²⁻ from BaSO₄:

1.163 g ÷ 233.39 g/mol = 0.004983 mol BaSO₄ = 0.004983 mol SO₄²⁻

Step 2: Find mass of SO₄²⁻:

0.004983 mol × 96.06 g/mol = 0.479 g SO₄²⁻

Step 3: Find mass of metal (mass of residue minus mass of SO₄):

0.854 g - 0.479 g = 0.375 g metal

Step 4: Find moles of metal (1:1 ratio with SO₄):

0.004983 mol metal

Step 5: Calculate molar mass of metal:

0.375 g ÷ 0.004983 mol = 75.2 g/mol

The metal is Zn (65.38) or something close. Let's check for Cu (63.55). Not exact. Try Mg (24.31) — no. The closest atomic mass suggests Fe (55.85) if we recalculate with better precision or account for experimental error.

Water mass: 1.246 - 0.854 = 0.392 g

Moles water: 0.392 g ÷ 18.02 g/mol = 0.0218 mol

Divide by moles of metal sulfate:

0.0218 mol ÷ 0.004983 mol = 4.37 ≈ 4

Formula: MSiO₄·4H₂O (where M is the metal)

Common Mistakes That Cost You Points

• Forgetting to dry the precipitate — moisture adds mass, inflating your results
• Using wrong molar mass — check your periodic table values twice
• Dropping the 1:1 assumption — always verify the stoichiometric ratio
• Rounding too early — keep extra digits until the final answer
• Ignoring significant figures — match them to the given data
• Not showing work — partial credit exists for a reason

Volatilization Gravimetry

Some questions involve heating a sample until one component decomposes and escapes as gas. You weigh what's left. The difference in mass gives you the volatile component.

Classic example: CaCO₃ heated → CaO + CO₂

Mass lost = mass of CO₂ = mass of carbonate in original sample

These problems require the same stoichiometric thinking. Calculate moles of product or residue, then work backward to the original analyte.

Quick Reference: Gravimetric Analysis Problem Types

Getting Started: Your Study Plan

Don't just read solutions. Practice solving problems from scratch.

• Work through 5 precipitation problems without looking at answers first
• Time yourself — you have roughly 12-15 minutes per FRQ in the AP exam
• Write every step. Examiners grade what they can see.
• Check your answers against the solutions. Find where you went wrong.
• Repeat until you can solve without hesitation

Use past AP Chemistry FRQs. The College Board releases them. They're free. Use them.

What to Memorize

You need these on exam day without hesitation:

• Common precipitate formulas and their molar masses
• The general formula: mass ÷ molar mass = moles
• How to extract stoichiometric ratios from balanced equations
• Percent purity formula: (actual ÷ theoretical) × 100
• Concentration formula: mass (mg) ÷ volume (L) = mg/L

Everything else you can derive. These you need instantly.

The Bottom Line

Gravimetric analysis FRQs are calculation problems. You solve them by following a method, showing your work, and getting the right number with correct units. There's no trick. There's no hidden complexity.

Students fail these questions because they rush, skip steps, or don't practice enough. That's the whole list of reasons.

Get the method down. Practice the calculations. Write every step on exam day.