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Week 2 Coursework · Unit 1.3 Study Companion

Topic 1.3: Elemental Composition of Pure Substances

The Law of Definite Proportions, Mass Percent Composition, The 4-Step Empirical & Molecular Formula Pipeline, Hydrate Salt Dehydration, Combustion Analysis, and 12 Worked Problem Models for Week 2 Homework & FRQs.

📌 How to Use This Takeaway Guide:

Review the Law of Definite Proportions and empirical formula calculation pipelines in Part 1. Study the 12 worked problem models in Part 2 as step-by-step templates for your Week 2 homework and FRQ assignments. Audit your mastery with the checklist in Part 3 and note any errors in Part 4.

📑 Quick Navigation & Table of Contents

⚡ Fast Revision & Part 1 Notes:
✍️ Part 2: 12 Homework Walkthrough Models:
High-Yield AP Cheat Sheet

⚡ Fast-Revision Diagrams & Summary Charts

Visual Review

🛣️ Diagram 1: The 4-Step Empirical-to-Molecular Formula Pipeline

1. % to Mass (g)
→ [÷ Atomic Mass] →
2. Moles of Elements
→ [÷ Smallest Moles] →
3. Mole Ratio
→ [× Whole Multiplier] →
4. Empirical Formula
→ [× n = M/M_emp] →
Molecular Formula

🔢 Diagram 2: Decimal Multipliers for Whole-Number Subscripts

Ends in .50 Multiply all ratios by 2
(e.g., $1 : 1.5 \rightarrow 2 : 3$)
Ends in .33 or .67 Multiply all ratios by 3
(e.g., $1 : 1.33 \rightarrow 3 : 4$)
Ends in .25 or .75 Multiply all ratios by 4
(e.g., $1 : 1.25 \rightarrow 4 : 5$)
Ends in .20, .40, .60, .80 Multiply all ratios by 5
(e.g., $1 : 1.40 \rightarrow 5 : 7$)

🧭 Summary Table: Diagnostic Error Directional Shift Matrix

Experimental Error Direct Measurement Impact Impact on Calculated Result ($x$ in Salt · $x H_2O$)
Incomplete heating (sample not dried to constant mass) Water remains trapped in the crystals; measured final mass of anhydrous solid is falsely high. FALSELY LOW value of $x$. (Calculated mass of lost water is falsely low).
Spattering during heating Solid particles burst out of crucible and are lost; measured final residue mass is falsely low. FALSELY HIGH value of $x$. (Lost salt is incorrectly counted as lost water).
Sample weighed while still hot Upward buoyant thermal convection currents reduce the apparent weight on balance pan. FALSELY HIGH water mass loss, inflating calculated $x$. Always cool in desiccator!
Inert impurity in sample (e.g., sand in glucose) Impurity contains 0% carbon, increasing total sample mass denominator without adding carbon. LOWER % Carbon measured compared to pure theoretical standard (40.0%).

Part 1: Simple-Words Concept Notes & Big Picture Topic 1.3 Essentials

A. The Law of Definite Proportions (Proust's Law)

In 1799, French chemist Joseph Proust formulated the foundation of chemical stoichiometry:

A pure chemical compound always contains exactly the same elements in the exact same proportion by mass, regardless of the source or sample size.

Whether water is collected from an Antarctic glacier, generated in a combustion engine, or distilled from seawater, pure $H_2O$ always consists of exactly 11.19% Hydrogen and 88.81% Oxygen by mass.

If an unknown white powder has an elemental mass percent that deviates from the theoretical formula (for example, pure $NaCl$ is 60.66% Cl, but a test sample yields 54.2% Cl), the sample is definitively impure or is a completely different chemical compound!

B. Percent Composition by Mass

To calculate the mass percent of an element in a pure compound:

% Element = [ (Number of atoms of element × Atomic Mass) / (Molar Mass of Compound) ] × 100%

Mass percent is an intensive property. It does not change with sample size. A single grain of glucose ($C_6H_{12}O_6$) and a 50-pound sack of glucose both contain exactly 40.00% Carbon, 6.72% Hydrogen, and 53.28% Oxygen by mass.

C. The 4-Step Empirical Formula Pipeline

The empirical formula is the lowest whole-number ratio of atoms in a chemical compound. Use the classic AP rhyme:

"Percent to mass, mass to mole, divide by small, multiply 'til whole."
  1. Percent to Mass: Assume a convenient $100.0\text{ g}$ sample. Thus, $40.0\%\text{ C}$ becomes $40.0\text{ g C}$.
  2. Mass to Mole: Divide each element's mass by its periodic table atomic mass ($n = m / M$). Do not round! Keep at least 4 significant figures.
  3. Divide by Smallest: Divide every mole value by the smallest mole value obtained in Step 2. This sets the smallest subscript to 1.
  4. Multiply 'Til Whole: If ratios are within $\pm 0.05$ of a whole number (e.g., $1.98 \approx 2$), round directly. If an obvious fraction appears ($.5, .33, .25$), multiply all subscripts by the appropriate integer.

D. Molecular Formula from Empirical Formula

The molecular formula represents the actual number of atoms of each element in a single molecule. It is always a whole-number multiple ($n$) of the empirical formula:

n = (Molar Mass of Compound) / (Empirical Formula Mass)
Molecular Formula = (Empirical Formula) × n

Example: Ethyne ($C_2H_2$) and benzene ($C_6H_6$) both have the exact same empirical formula ($CH$, mass $13.02\text{ g/mol}$) and the exact same mass percentages (92.3% C, 7.7% H). However, benzene has a molar mass of $78.11\text{ g/mol}$ ($n = 78.11 / 13.02 = 6$), so its molecular formula is $C_6H_6$.

E. Hydrate Salt Analysis ($Salt \cdot x H_2O$)

A hydrate is an ionic solid with a definite number of water molecules chemically bound inside its crystal lattice. Heating the hydrate drives off the water as steam, leaving behind the dry anhydrous salt:

Mass of Water Lost = Mass of Hydrate - Mass of Anhydrous Salt
x = (Moles of H₂O lost) / (Moles of Anhydrous Salt remaining)

F. Hydrocarbon Combustion Analysis Principles

When an organic compound containing $C, H,$ and possibly $O$ is combusted with excess $O_2$:

  • All Carbon converts quantitatively to $CO_2$ ($1\text{ mol } CO_2 = 1\text{ mol C}$).
  • All Hydrogen converts quantitatively to $H_2O$ ($1\text{ mol } H_2O = 2\text{ mol H}$).
  • Oxygen in the sample is determined by subtraction: $m_{\text{Oxygen}} = m_{\text{sample}} - m_{\text{C}} - m_{\text{H}}$.

G. AP Chemistry Core Vocabulary Glossary

Term AP Exam Operational Definition
Law of Definite Proportions A fundamental law stating that any pure chemical compound always contains elements in fixed, invariant ratios by mass.
Percent Composition The percentage by mass of each individual element in a pure compound.
Empirical Formula The simplest whole-number stoichiometric ratio of atoms of each element present in a compound.
Molecular Formula The true chemical formula indicating the actual number of atoms of each element in a molecule.
Hydrate An ionic compound containing water molecules structurally trapped in its crystalline matrix in stoichiometric proportions.
Anhydrous Salt The dry ionic crystal remaining after all water of hydration has been thermally expelled.

Part 2: 12 Fully Worked Walkthrough Problem Models Direct Alignment with Week 2 Coursework & FRQs

Model 1: Percent Composition of Pure Glucose ($C_6H_{12}O_6$) Core Topic Model
Problem Statement: Calculate the mass percent of Carbon, Hydrogen, and Oxygen in pure glucose, $C_6H_{12}O_6$. ($C = 12.01, H = 1.008, O = 16.00\text{ g/mol}$).
Step 1: Calculate Total Molar Mass
$$M = 6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.096 + 96.00 = 180.16\text{ g/mol}$$
Step 2: Calculate Individual Elemental Mass Percentages
• $\% \text{ C} = \frac{72.06}{180.16} \times 100\% = 40.00\%$
• $\% \text{ H} = \frac{12.096}{180.16} \times 100\% = 6.714\%$
• $\% \text{ O} = \frac{96.00}{180.16} \times 100\% = 53.29\%$
$$\text{Sum Check} = 40.00\% + 6.714\% + 53.29\% = 100.00\%$$
Final Answer: 40.00% C, 6.71% H, 53.29% O by mass.
Model 2: Verification of Definite Proportions Proust's Law Verification
Problem Statement: Sample A ($10.00\text{ g}$) of a copper chloride compound contains $4.73\text{ g}$ Cu and $5.27\text{ g}$ Cl. Sample B ($25.00\text{ g}$) from another supplier contains $11.83\text{ g}$ Cu and $13.17\text{ g}$ Cl. Do these samples represent the exact same pure compound?
Step 1: Calculate Mass % Cu in Sample A
$$\% \text{ Cu (A)} = \frac{4.73\text{ g}}{10.00\text{ g}} \times 100\% = 47.3\%$$
Step 2: Calculate Mass % Cu in Sample B
$$\% \text{ Cu (B)} = \frac{11.83\text{ g}}{25.00\text{ g}} \times 100\% = 47.32\% \approx 47.3\%$$
Step 3: Justify via Law of Definite Proportions
Both samples yield identical mass percentages of Copper (47.3%) and Chlorine (52.7%). According to the Law of Definite Proportions, both samples represent the exact same pure compound: Copper(II) chloride ($CuCl_2$).
Final Answer: Yes, both samples represent the identical pure compound (CuCl₂).
Model 3: Standard Empirical Formula from Mass % ($NO_2$) Standard AP Skill
Problem Statement: A gaseous oxide of nitrogen is analyzed and found to contain 30.45% Nitrogen and 69.55% Oxygen by mass. Determine its empirical formula.
Step 1: Assume 100.0 g Sample
$m_{\text{N}} = 30.45\text{ g}$,   $m_{\text{O}} = 69.55\text{ g}$
Step 2: Convert to Moles
$$n_{\text{N}} = \frac{30.45\text{ g}}{14.01\text{ g/mol}} = 2.173\text{ mol}$$ $$n_{\text{O}} = \frac{69.55\text{ g}}{16.00\text{ g/mol}} = 4.347\text{ mol}$$
Step 3: Divide by Smallest Mole Value
$$\text{N} = \frac{2.173}{2.173} = 1.000, \quad \text{O} = \frac{4.347}{2.173} = 2.000$$
Final Answer: Empirical formula is NO₂.
Model 4: Decimal Multipliers for Non-Integer Ratios ($Fe_2O_3$) Classic Decimal Trap
Problem Statement: An iron oxide mineral contains 69.94% Fe and 30.06% O by mass. Determine its empirical formula.
Step 1: Convert to Moles in 100 g
$$n_{\text{Fe}} = \frac{69.94\text{ g}}{55.85\text{ g/mol}} = 1.252\text{ mol}$$ $$n_{\text{O}} = \frac{30.06\text{ g}}{16.00\text{ g/mol}} = 1.879\text{ mol}$$
Step 2: Divide by Smallest
$$\text{Fe} = \frac{1.252}{1.252} = 1.000, \quad \text{O} = \frac{1.879}{1.252} = 1.501 \approx 1.50$$
Step 3: Multiply by 2 to Clear .50 Decimal
$$\text{Fe: } 1.00 \times 2 = 2, \quad \text{O: } 1.50 \times 2 = 3$$
Final Answer: Empirical formula is Fe₂O₃ (Rust / Hematite).
Model 5: Molecular Formula from Empirical Formula and Molar Mass Core AP Calculation
Problem Statement: A liquid hydrocarbon has an empirical formula of $CH$ and a measured molar mass of $78.11\text{ g/mol}$. Determine its molecular formula.
Step 1: Calculate Empirical Formula Mass
$$M_{\text{empirical}} = 1(12.01) + 1(1.008) = 13.02\text{ g/mol}$$
Step 2: Find Integer Ratio ($n$)
$$n = \frac{M_{\text{molecular}}}{M_{\text{empirical}}} = \frac{78.11\text{ g/mol}}{13.02\text{ g/mol}} = 5.999 \approx 6$$
Step 3: Multiply Subscripts
$$(CH)_6 = C_6H_6$$
Final Answer: Molecular formula is C₆H₆ (Benzene).
Model 6: Determining Water of Hydration in $CuSO_4 \cdot x H_2O$ Laboratory Gravimetry Model
Problem Statement: A student heats a $2.500\text{ g}$ sample of blue hydrated copper(II) sulfate. After heating to constant mass, $1.598\text{ g}$ of white anhydrous $CuSO_4$ remains ($M = 159.61\text{ g/mol}$). Determine the value of $x$.
Step 1: Determine Mass of Water Lost
$$m_{\text{water}} = 2.500\text{ g} - 1.598\text{ g} = 0.902\text{ g } H_2O$$
Step 2: Convert Masses to Moles
$$n_{\text{anhydrous}} = \frac{1.598\text{ g}}{159.61\text{ g/mol}} = 0.01001\text{ mol } CuSO_4$$ $$n_{\text{water}} = \frac{0.902\text{ g}}{18.02\text{ g/mol}} = 0.05006\text{ mol } H_2O$$
Step 3: Calculate Mole Ratio ($x$)
$$x = \frac{n_{\text{water}}}{n_{\text{anhydrous}}} = \frac{0.05006\text{ mol}}{0.01001\text{ mol}} = 5.00 \approx 5$$
Final Answer: x = 5 (Formula: CuSO₄ · 5H₂O).
Model 7: Incomplete Dehydration Error Directional Shift Laboratory Audit Trap
Problem Statement: During the hydrate heating experiment in Model 6, the student removed the crucible before all water had evaporated. Explain the directional effect of this error on the calculated value of $x$.
Step 1: Physical Consequence on Measurements
Trapped water remains inside the solid residue, causing the recorded mass of the "anhydrous" salt to be falsely high.
Step 2: Impact on Water Mass and Calculated $x$
Since $m_{\text{water}} = m_{\text{hydrate}} - m_{\text{residue}}$, an inflated residue mass results in a falsely low calculated mass of water lost. $$x = \frac{\text{moles } H_2O (\text{falsely low})}{\text{moles anhydrous} (\text{falsely high})} \quad \Longrightarrow \quad x \text{ is FALSELY LOW}$$
Final Answer: The calculated value of x will be falsely low.
Model 8: Impurity Raising or Lowering Percent Composition AP Diagnostic MC & FRQ
Problem Statement: A bottle labeled "Pure Sodium Chloride, $NaCl$" is contaminated with potassium chloride, $KCl$. How does this impurity affect the measured mass percent of Chlorine compared to pure $NaCl$?
Step 1: Calculate Theoretical % Cl in Pure Compounds
• Pure $NaCl$ ($M = 58.44\text{ g/mol}$): $\% \text{ Cl} = \frac{35.45}{58.44} \times 100\% = 60.66\%$
• Pure $KCl$ ($M = 74.55\text{ g/mol}$): $\% \text{ Cl} = \frac{35.45}{74.55} \times 100\% = 47.55\%$
Step 2: Assess Directional Impact of Mixture
Because Potassium has a higher atomic mass ($39.10$) than Sodium ($22.99$), $KCl$ has a lower mass percentage of Chlorine (47.55%) than pure $NaCl$ (60.66%). Any presence of $KCl$ dilutes the chlorine concentration, making the measured % Cl lower than theoretical pure NaCl.
Final Answer: The measured mass percent of chlorine will be lower than pure NaCl.
Model 9: Organic Combustion Analysis ($C_xH_yO_z$) Advanced AP Problem
Problem Statement: A $0.5438\text{ g}$ sample of an organic liquid containing only C, H, and O is combusted, yielding $1.039\text{ g } CO_2$ and $0.6369\text{ g } H_2O$. Find its empirical formula.
Step 1: Calculate Grams of Carbon and Hydrogen
$$m_{\text{C}} = 1.039\text{ g } CO_2 \times \frac{12.01\text{ g C}}{44.01\text{ g } CO_2} = 0.2835\text{ g C}$$ $$m_{\text{H}} = 0.6369\text{ g } H_2O \times \frac{2.016\text{ g H}}{18.02\text{ g } H_2O} = 0.07125\text{ g H}$$
Step 2: Calculate Oxygen by Subtraction
$$m_{\text{O}} = 0.5438\text{ g} - (0.2835\text{ g} + 0.07125\text{ g}) = 0.18905\text{ g O}$$
Step 3: Moles and Whole Number Ratio
• $n_{\text{C}} = 0.2835 / 12.01 = 0.02361\text{ mol}$ → $0.02361 / 0.01182 = 2.00$
• $n_{\text{H}} = 0.07125 / 1.008 = 0.07068\text{ mol}$ → $0.07068 / 0.01182 = 5.98 \approx 6.00$
• $n_{\text{O}} = 0.18905 / 16.00 = 0.01182\text{ mol}$ → $0.01182 / 0.01182 = 1.00$
Final Answer: Empirical formula is C₂H₆O (Ethanol / Dimethyl ether).
Model 10: Differentiating Isomers with Identical % Composition Conceptual Mastery
Problem Statement: Formaldehyde ($CH_2O$, $M = 30.03\text{ g/mol}$), acetic acid ($C_2H_4O_2$, $M = 60.05\text{ g/mol}$), and glucose ($C_6H_{12}O_6$, $M = 180.16\text{ g/mol}$) all have the exact same empirical formula ($CH_2O$). Explain why elemental mass percent analysis alone cannot distinguish between them.
Chemical Explanation
Because all three compounds share the exact same empirical formula ($CH_2O$), they have identical mass ratios of Carbon (40.0%), Hydrogen (6.7%), and Oxygen (53.3%). Mass percentage depends strictly on the empirical ratio. To distinguish them, an analytical technique that measures molar mass (such as mass spectrometry or freezing-point depression) is required.
Final Answer: Mass % depends only on empirical ratio; molar mass is required to differentiate.
Model 11: Quantitative Metal Oxide Synthesis ($MgO$) Lab Crucible Model
Problem Statement: A strip of magnesium metal ($0.486\text{ g}$) is strongly heated in a crucible until it reacts completely with atmospheric oxygen to produce $0.806\text{ g}$ of white magnesium oxide powder. Determine the empirical formula of the oxide.
Step 1: Determine Mass of Bound Oxygen
$$m_{\text{O}} = 0.806\text{ g} - 0.486\text{ g} = 0.320\text{ g O}$$
Step 2: Convert to Moles
$$n_{\text{Mg}} = \frac{0.486\text{ g}}{24.31\text{ g/mol}} = 0.0200\text{ mol}$$ $$n_{\text{O}} = \frac{0.320\text{ g}}{16.00\text{ g/mol}} = 0.0200\text{ mol}$$
Step 3: Determine Ratio
$$\frac{n_{\text{Mg}}}{n_{\text{O}}} = \frac{0.0200}{0.0200} = 1.00 \quad \Longrightarrow \quad 1 : 1$$
Final Answer: Empirical formula is MgO.
Model 12: Comprehensive Laboratory Audit Protocol AP FRQ Gold Standard
Problem Statement: A sample of an unknown white crystalline substance is suspected of being calcium carbonate, $CaCO_3$ ($M = 100.09\text{ g/mol}$). A student performs thermal decomposition ($CaCO_3 \rightarrow CaO + CO_2$). What mass loss percentage should be observed if the sample is 100% pure?
Step 1: Identify Evolving Gas
The lost mass corresponds quantitatively to carbon dioxide gas ($CO_2$, $M = 44.01\text{ g/mol}$).
Step 2: Theoretical Percent Mass Loss
$$\% \text{ Mass Loss} = \frac{M(CO_2)}{M(CaCO_3)} \times 100\% = \frac{44.01\text{ g/mol}}{100.09\text{ g/mol}} \times 100\% = 43.97\%$$
Step 3: Verification Standard
If the experimental mass loss is substantially less than 43.97%, the sample contains thermally stable non-carbonate impurities (such as $SiO_2$ sand).
Final Answer: Theoretical mass loss is 43.97% CO₂.

✅ Part 3: Student Subtopic Mastery Checklist (Topic 1.3)

Core Competency / Skill Confidence (1–5) Homework Problem Verification
I can state and apply the Law of Definite Proportions to verify purity. [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] Model 2, Part 1-A
I can calculate elemental mass percent composition from a chemical formula. [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] Model 1, Model 8
I can determine an empirical formula from mass percentages using the 4-step pipeline. [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] Model 3, Model 4
I can determine the molecular formula using empirical formula and molar mass. [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] Model 5
I can determine water of hydration ($x$) from thermal gravimetric lab data. [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] Model 6, Model 11
I can evaluate laboratory errors (e.g. incomplete drying) and explain directional shifts in calculated formulas. [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] Model 7, Model 8

🔒 Part 4: Personal Misconception Vault & Post-Assignment Reflection

Error Log 1 (What went wrong?):
Chemical Remedy (What is the true principle?):
Error Log 2 (What went wrong?):
Chemical Remedy (What is the true principle?):