A complete, printable study workbook covering Topic 1.4 (Composition of Mixtures & Gravimetric Analysis) with spiral review from Topic 1.1 (Moles & Atom Counting) and Topic 1.2 (Mass Spectrometry & Isotopes).
1. Review the Index & Fast Revision Diagrams: Use the quick-reference charts and decision trees for rapid recall before exams.
2. Study Part 1: Understand the core concepts, precipitation gravimetry protocol, and directional shift rules before calculating.
3. Follow Part 2: Step through all 12 walkthrough models. Each model directly prepares you for one of the 12 questions on your Week 3 FRQ homework.
4. Complete Part 4: Audit your skills on the Mastery Checklist, record lingering questions in your Mistake Vault, and build your study action plan.
Cannot be physically separated. Invariant chemical formulas.
Physically combined. Can be separated by physical techniques.
| Laboratory Error Source | Direct Physical Measurement Effect | Directional Impact on Final Calculated Result |
|---|---|---|
| Incomplete Drying (Wet Precipitate) | Water weight remains on filter paper; recorded precipitate mass is falsely high. | FALSELY HIGH (% of analyte or target salt is inflated). |
| Spattering During Heating | Solid particles burst out of crucible; final residue mass is falsely low. | FALSELY HIGH for calculated mass of lost gas/chlorine. |
| Washing with Warm Water | Precipitate partially dissolves and passes through filter paper into filtrate. | FALSELY LOW (% yield and mass of recovered precipitate). |
| Lighter Cation Contaminant (e.g. LiCl in NaCl) | Lower formula weight of cation means each gram of salt contains more Cl atoms. | HIGHER % Cl (83.6% in LiCl vs 60.7% in pure NaCl). |
| Heavier Cation Contaminant (e.g. KCl in NaCl) | Higher formula weight of cation dilutes the chlorine percentage. | LOWER % Cl (47.6% in KCl vs 60.7% in pure NaCl). |
| Moisture / Water in Organic Sample | Water contains 0% carbon; acts as an inert diluent. | LOWER % C (e.g. 38.2% vs pure 40.0% in glucose). |
In the laboratory, you cannot count invisible atoms directly; you can only measure grams on an analytical balance. Topic 1.4 is the mathematical and conceptual bridge between the macroscopic mass you weigh and the microscopic particles you cannot see. By combining molar masses and chemical formulas, you can verify if a substance is pure, measure the percent of active metal in an alloy, or determine how many water molecules are locked inside a hydrate crystal.
| Key Property | Pure Substance (Elements & Compounds) | Mixture (Homogeneous & Heterogeneous) |
|---|---|---|
| Constituent Particles | Exactly one type of particle (identical atoms or formula units). | Two or more distinct pure substances physically combined. |
| Composition by Mass | Fixed and Invariant. Dictated by chemical formula (Law of Definite Proportions). | Variable. Proportions change depending on how the mixture was made. |
| Separation Method | Only separable by chemical reactions (breaking chemical bonds). | Separable by physical methods (relying on property differences). |
| Properties | Uniform, characteristic chemical and physical constants (density, melting point). | Components retain their individual physical and chemical properties. |
| Examples | Pure H2O, elemental iron (Fe), pure NaCl, anhydrous Na2SO4, pure CaCO3. | Sterling silver alloy (Ag/Cu), air, salt water, brass (Cu/Zn), supplement tablets. |
Extensive (mass, volume) depends on sample size (10.0 g powder reveals nothing about purity). Intensive (elemental mass %, density) is independent of size and serves as the primary AP purity benchmark.
Homogeneous (Solution / Alloy): Uniform throughout at the particle level (sterling silver, salt water). Heterogeneous: Non-uniform with visible phases (sand and water, granite).
Because mixtures are physically mingled, components can be separated without breaking chemical bonds:
| Separation Technique | Physical Property Exploited | How It Works & AP Exam Application |
|---|---|---|
| Filtration | Particle size & solubility | Separates insoluble solid precipitate from solution. The liquid passing through is the filtrate; trapped solid is the residue. (Dissolved ions pass through filter paper!). |
| Distillation | Boiling point differences | Separates homogeneous liquids. The more volatile liquid boils first, condenses in a cooled condenser tube, and collects as pure distillate. |
| Evaporation to Dryness | Solvent volatility | Boiling off liquid solvent (evaporating H2O) to recover dry crystalline solute (NaCl). |
| Magnetism | Magnetic susceptibility | Separates ferromagnetic metals (iron, Fe) from non-magnetic powders without adding liquid solvents. |
| Chromatography (TLC) | Polarity & intermolecular forces | Separates dissolved components by their differential affinity for a mobile phase vs. a stationary phase. |
Principle: A liquid solvent (mobile phase) moves up a plate coated with a polar adsorbent (stationary phase, silica gel).
Intermolecular Attraction ("Like Dissolves Like"): Polar compounds adhere strongly to polar silica and move slower (lower Rf); less polar compounds dissolve better in the mobile solvent and travel farther up the plate (higher Rf).
Gravimetric analysis is an analytical technique where the amount of an analyte (the substance in a sample being determined) is calculated by measuring the mass of a pure solid substance containing that analyte. Unlike volumetric titrations that measure solution volumes, gravimetric methods rely on high-precision analytical balances (±0.0001 g).
A precipitate is an insoluble solid that forms when two aqueous solutions are mixed together. The reaction is typically a double-replacement reaction where cations and anions combine into an insoluble ionic lattice:
Weigh the dry mixture sample accurately on an analytical balance. Completely dissolve it in distilled water.
Add precipitating reagent in stoichiometric excess to guarantee 100% of the analyte ion is driven into the solid precipitate.
Filter the precipitate through quantitative ashless filter paper or a sintered glass crucible. The trapped solid is the residue.
Rinse the precipitate with small portions of cold distilled water to wash away adsorbed spectator ions (Na+, NO3−) without dissolving crystals.
Dry in an oven, cool in a desiccator, and weigh. Repeat heating and cooling until consecutive readings agree within ±0.002 g.
Convert mass of pure dry precipitate to moles, relate by mole ratio to analyte, and divide by initial sample mass to find % purity.
| Experimental Error Scenario | Physical Consequence | Calculated Impact on Result |
|---|---|---|
| Incomplete Drying (Wet Precipitate) | Trapped moisture adds extra mass to the filter paper. | Precipitate mass is falsely high → calculated analyte % is FALSELY HIGH. |
| Spattering During Heating | Solid particles eject out of crucible and are lost. | Residue mass is falsely low → apparent lost mass (H2O or CO2) is FALSELY HIGH. |
| Excess Warm Wash Water | Precipitate dissolves slightly and passes into filtrate. | Collected precipitate mass is falsely low → percent yield is FALSELY LOW. |
| Moisture Reabsorption on Cooling | Sample left uncovered in humid air rather than in a desiccator. | Final heated mass is falsely high → calculated mass lost is falsely low → hydrate value x is FALSELY LOW. |
| Contaminant with Lighter Cation (e.g. LiCl in NaCl) | Lighter cation means each formula unit has a higher mass % of anion. | Measured % Cl is shifted HIGHER than pure NaCl theoretical value. |
| Contaminant with Heavier Cation (e.g. KCl in NaCl) | Heavier cation means each formula unit has a lower mass % of anion. | Measured % Cl is shifted LOWER than pure NaCl theoretical value. |
| Term | Official AP Course Framework Definition |
|---|---|
| elemental analysis | An analytical laboratory technique used to determine the relative numbers of atoms of each element in a substance and to assess its purity. |
| elemental composition by mass | The percentage or proportion of each element present in a substance, expressed as a mass fraction or mass percentage. |
| mixture | Matter that contains atoms, molecules, or formula units of two or more types, whose relative proportions can vary. |
| pure substance | A material with a fixed, definite composition and invariant chemical/physical properties throughout. |
| purity | The degree to which a substance contains only one type of atom, molecule, or formula unit without contamination from other substances. |
| analyte | The specific substance or ion in an experimental sample whose presence or quantity is being determined. |
1. What is a mixture in AP Chemistry?
A mixture contains two or more distinct pure substances whose relative proportions can vary from sample to sample without forming new chemical bonds.
2. How is a pure substance different from a mixture?
A pure substance contains only one type of particle and has a fixed, invariant elemental mass percent. A mixture contains multiple types of particles with variable proportions.
3. What is elemental analysis used for?
Elemental analysis uses macroscopic mass measurements to determine empirical formulas and verify whether a sample is pure or contaminated.
4. Why can mixtures be separated physically, but compounds cannot?
Components in a mixture are held by intermolecular forces or physical mixing (no bonds broken during separation). Compounds consist of atoms joined by covalent or ionic bonds, which require chemical reactions to separate.
This comprehensive reference section provides the exact scientific laws, stoichiometry formulas, and diagnostic reasoning needed to solve every question in Week 3 Homework (Part 1) and Week 3 Homework (Part 2).
| Item | Core Principle & Topic | Formula & Quantitative Steps | Diagnostic Rule & AP Trap |
|---|---|---|---|
| Part 1 · Q1 | Liquid Solution as Homogeneous Mixture | Water + ethanol + dissolved NaCl form a single uniform phase. | Water and ethanol are completely miscible polar liquids; NaCl dissociates into hydrated Na+ and Cl- ions. It is a physical mixture, not a compound (no new chemical bonds). |
| Part 1 · Q2 | Molecular Arrangement of Heterogeneous Mixtures | Non-uniform spatial distribution with distinct microscopic or macroscopic boundaries. | Pure elements have identical particles; homogeneous mixtures have different particles evenly dispersed; compounds have chemically bonded atoms in fixed ratios. Heterogeneous mixtures form distinct, separated regions. |
| Part 1 · Q3 | Empirical Formula from Mass Percent |
% H = 100 - (39% S + 58.6% O) = 2.4% H. Moles S = 39 / 32.06 = 1.22 mol. Moles O = 58.6 / 16.00 = 3.66 mol. Moles H = 2.4 / 1.008 = 2.40 mol. |
Divide each mole value by 1.22: S = 1.0, O = 3.0, H = 1.97 ≈ 2. Empirical Formula: SO3H2 (or H2SO3). Trap: Never divide molar mass by %; always divide grams by atomic mass. |
| Part 1 · Q4 | Mass of Target Element in Compound Sample |
Molar mass NH4NO3 = 80.05 g/mol. Mass of N per mole = 2 × 14.01 = 28.02 g/mol. % N = (28.02 / 80.05) × 100% = 34.99%. Mass N = 23.15 g × 0.3499 = 8.10 g. |
Crucial: NH4NO3 has two nitrogen atoms per formula unit (total 28.02 g/mol N). Counting only one N yields 4.05 g (false distractor). |
| Part 1 · Q5 | Equivalent Mass Ratios Matching Formula Units | Butene (C4H8) simplifies to empirical formula CH2 (1:2 C:H mole ratio). |
For 72 g C and 12 g H: Moles C = 72 / 12.01 = 6.0 mol. Moles H = 12 / 1.008 = 11.9 ≈ 12 mol. Mole ratio = 6 : 12 = 1 : 2. Matches butene! |
| Part 1 · Q6 | Law of Definite Proportions | A pure compound (e.g. H2S) always contains its constituent elements in a fixed, constant mass ratio, regardless of source or preparation method. | The mass ratio of H to S in pure H2S is invariant. It does not vary by reaction conditions, geography, or batch size. |
| Part 1 · Q7 | Deducing Chemical Identity from Mass % |
In 100 g sample: Moles H = 5.00 / 1.008 = 4.96 mol. Moles N = 35.00 / 14.01 = 2.50 mol. Moles O = 60.00 / 16.00 = 3.75 mol. Divide by 2.50 → H = 2, N = 1, O = 1.5. Multiply by 2 → H4N2O3. |
The empirical formula H4N2O3 corresponds directly to ammonium nitrate, NH4NO3. |
| Item | Core Principle & Topic | Formula & Quantitative Steps | Diagnostic Rule & AP Trap |
|---|---|---|---|
| Part 2 · Q1 | Subatomic Particle Counts in Isotopes |
Chromium-53 (5324Cr): Protons = Z = 24. Neutral atom → Electrons = Protons = 24. Neutrons = A - Z = 53 - 24 = 29 neutrons. |
Atomic number (subscript) = protons. Mass number (superscript) = protons + neutrons. |
| Part 2 · Q2 | Coulomb's Law Equation & Principles | Fcoulombic = k (q1q2 / r2) | Attraction increases with greater nuclear charge (q1) and decreases with larger distance squared (r2). |
| Part 2 · Q3 | Subatomic Particles in Monatomic Ions |
Phosphide ion (3115P3-): Protons = 15. Neutrons = 31 - 15 = 16. Electrons = 15 + 3 = 18 electrons. |
Anions gain electrons (15 + 3 = 18). Cations lose electrons. |
| Part 2 · Q4 | Ionic Dissociation & Total Ion Counts |
(NH4)3PO4 → 3 NH4+ + PO43-. Ions per unit = 3 + 1 = 4. Moles of ions = 0.02 mol × 4 = 0.08 mol. Total ions = 0.08 × (6.022 × 1023) = 4.8 × 1022 ions. |
Trap: Do not stop at moles of salt; multiply by ions per formula unit (4) and Avogadro's number. |
| Part 2 · Q5 | Atoms and Milligrams from Molecule Counts |
Maltose (C12H22O11, M = 342.3 g/mol): Atoms per molecule = 12 + 22 + 11 = 45. Total atoms = (1.5 × 1018) × 45 = 6.75 × 1019 atoms. Moles = (1.5 × 1018) / (6.022 × 1023) = 2.49 × 10-6 mol. Mass = (2.49 × 10-6) × 342.3 = 8.5 × 10-4 g = 0.85 mg. |
1 gram = 1,000 milligrams. Multiply grams by 1,000 to convert to mg. |
| Part 2 · Q6 | Weighted Average Atomic Mass |
Avg Mass = Σ (fractional abundance × isotopic mass) = (0.05845 × 53.94) + (0.91754 × 55.93) + (0.02119 × 56.94) + (0.00282 × 57.93) = 55.845 amu. |
The average atomic mass of iron (55.85 amu) is overwhelmingly dominated by 56Fe (91.8% abundance). |
| Part 2 · Q7 | Selective Precipitation in Ag/Cu Alloys |
Dissolve alloy in HNO3 → Ag+(aq) + Cu2+(aq). Add chloride (NaCl or HCl) → AgCl(s) precipitates selectively while CuCl2 remains soluble. |
Ksp of AgCl is 1.8 × 10-10 (insoluble). Copper(II) chloride is soluble and stays in the filtrate. Filter to separate silver. |
| Part 2 · Q8 | Stoichiometric Assay of Cu/Al via Acid Reaction |
Cu(s) + 4 HNO3(aq) → Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l) 0.010 mol Cu(NO3)2 formed = 0.010 mol Cu. Mass Cu = 0.010 mol × 63.55 g/mol = 0.636 g. % Cu = (0.636 g / 2.00 g) × 100% = 31.8% ≈ 32%. |
1:1 stoichiometric ratio between Cu consumed and Cu(NO3)2 produced. |
| Part 2 · Q9 | Essential Controls in Gravimetric Analysis |
Error evaluation: I. Subtract weighing paper: Essential. II. Temperature control: Not critical. III. Wash precipitate: Essential. IV. Heat to constant mass: Essential. |
Failure of I, III, and IV causes direct analytical error in gravimetric determination. |
| Part 2 · Q10 | Identifying Impurity Causing High Elemental % |
Pure NaCl contains 60.7% Cl. Impure sample contains 75% Cl. Impurity must have % Cl > 60.7%. LiCl: (35.45 / 42.39) × 100% = 83.6% Cl (Higher!). KCl: (35.45 / 74.55) × 100% = 47.6% Cl (Lower). NaI: 0% Cl (Lower). |
A mixture of NaCl (61% Cl) and LiCl (84% Cl) averages to 75% Cl. The impurity is LiCl. |
| Part 2 · Q11 | Identifying Impurity Causing Low Elemental % |
Pure glucose (C6H12O6) contains 40.0% C. Impure sample contains 38.2% C. Impurity must have % C < 40.0%. Water (H2O): 0% Carbon (Lowers % C!). Ribose (C5H10O5): 40.0% C (No change). Sucrose (C12H22O11): 42.1% C (Would raise % C). |
Trapped moisture / water (0% C) dilutes the carbon percentage below 40.0%. The impurity is water. |
| Part 2 · Q12 | Gravimetric Identification of Unknown Cation M+ |
Mass AgCl = 2.23 g - 0.80 g = 1.43 g. Moles AgCl = 1.43 / 143 g/mol = 0.0100 mol. Moles MCl = 0.0100 mol. Molar mass MCl = 0.74 g / 0.0100 mol = 74 g/mol. Atomic mass of M = 74 - 35.5 = 38.5 g/mol. |
Comparing with alkali metals (M+): Li = 6.94, Na = 23.0, K = 39.10 g/mol. The unknown chloride is KCl. |
| Part 2 · Q13 | Percent Compound from Elemental Assay |
Pure CaCO3 contains: 40.0 g Ca / 100.0 g CaCO3 = 40.0% Ca. Sample contains 30.0% Ca (no Ca in impurities). % CaCO3 = (30.0% / 40.0%) × 100% = 75%. |
Formula: % Compound = (% Element in sample / % Element in pure compound) × 100%.
|
| Part 2 · Q14 | Gas Pressure & Calibration Curve Analysis |
Gas produced: ΔP = 0.870 atm - 0.800 atm = 0.070 atm. From calibration line: 0.070 atm corresponds to 0.15 g CaCO3. % CaCO3 = (0.15 g / 0.200 g eggshell) × 100% = 75%. |
Pressure difference ΔP isolates reaction-generated CO2 gas from initial air in the closed vessel. |
| Part 2 · Q15 | Redox Displacement & Mass Recovery Discrepancies |
Reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). If mass of Cu produced exceeds theoretical yield: |
Solid copper collected was contaminated with excess unreacted zinc metal granules or was incompletely dried (retaining moisture). |
Study each model below. The principles, unit conversions, and justifications match your 12 Week 3 FRQ homework questions in exact sequence.
| Concept | Mathematical Formula | AP Exam Key Strategy |
|---|---|---|
| Percent by Mass | % X = (Mass of X ÷ Total Mass) × 100% | Intensive property: independent of sample mass. |
| Precipitate to Analyte | Mass = Massppt × (1 ÷ Mppt) × (Mole Ratio) × Manalyte | Check formula subscripts for mole ratios (1 BaSO4 : 1 BaCl2; 1 Ag2S : 2 Ag). |
| Gas Evolved by Mass | Massgas = Initial Mass − Final Residue Mass | Assumes the gas was the only species escaping the container. |
| Hydrate Water Ratio (x) | x = Moles of H2O lost ÷ Moles of anhydrous salt residue | Round to nearest whole integer (4.98 → 5). |
| Atom Counting | Total Atoms = Moles × (6.022 × 1023) × (subscript) | Multiply by subscript to count constituent atoms inside a molecule! |
| Average Atomic Mass | Average Mass = Σ (% × mass) ÷ 100 | Result must lie between the lightest and heaviest isotope masses. |
Complete this section after finishing your Week 3 FRQ assignment to diagnose your conceptual mastery and guide your exam review.
| Subtopic & Problem Type | Confidence (1–5) | Status |
|---|---|---|
| 1. Gravimetric Precipitation: Mass difference (mtotal − mpaper) & moles of precipitate | [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] | [ ] Mastered [ ] Review |
| 2. Error Directionality: Explaining why splattering or wet paper makes calculated % too high or low | [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] | [ ] Mastered [ ] Review |
| 3. Washing Technique: Crystal solubility loss with excess or room-temperature solvent | [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] | [ ] Mastered [ ] Review |
| 4. Alloy & Tarnish: Sterling silver composition and calculating moles of Ag from Ag2S | [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] | [ ] Mastered [ ] Review |
| 5. Hydrate Analysis: Finding water of crystallization x from dehydration mass loss | [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] | [ ] Mastered [ ] Review |
| 6. Carbonate Decomposition: Finding molar mass of MCO3 from evolved CO2 to identify metal | [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] | [ ] Mastered [ ] Review |
| 7. Moles to Atom Counting: Converting grams to molecules and multiplying by formula subscripts | [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] | [ ] Mastered [ ] Review |
| 8. Mass Spectrometry: Identifying peak abundance, finding protons, neutrons, and average atomic mass | [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] | [ ] Mastered [ ] Review |
| FRQ # | The Mistake I Made | Chemistry Reason (Why It Happened) | Correct Strategy for Next Time |
|---|---|---|---|
| Ex. | Forgot Ag2S has 2 Ag atoms | Used 1:1 ratio instead of formula subscript | Always multiply moles of compound by atom subscript |