IMAT General Chemistry

Lesson 2: Stoichiometry, States of Matter, Gases & Redox

IMAT 100% 30 Questions

Introduction to Chemical Principles

Welcome to Lesson 2. In this extensive module, we rigorously bridge the conceptual gap between microscopic atoms and macroscopic laboratory measurements. We will comprehensively cover the mathematical language of chemistry (Stoichiometry & The Mole), how molecules interact physically in different phases (States of Matter & Gas Laws), and how the thermodynamic flow of electrons dictates energy production (Redox & Electrochemistry).

General Chemistry Overview Diagram

General Chemistry Overview: Mapping the core concepts of stoichiometry, matter, gases, and redox.

Core IMAT Objectives

  • Execute complex stoichiometric calculations: Empirical Formulas, Limiting Reagents, and Theoretical Yield.
  • Master concentration units: Molarity ($M$), Molality ($m$), Mass %, and Dilution equations.
  • Interpret Phase Diagrams (Triple Point, Critical Point) and distinguish Intra- vs Intermolecular forces.
  • Apply the Ideal Gas Law ($PV=nRT$) and simple gas laws (Boyle, Charles, Gay-Lussac).
  • Calculate exact Oxidation States and balance complex Redox half-reactions.
  • Differentiate Galvanic vs. Electrolytic cells and calculate Standard Cell Potential ($E^\circ_{cell}$).

Part 1: The Mole & Stoichiometry

Quantifying Chemical Reactions

Chemistry is fundamentally a quantitative science. Because individual atoms are unfathomably small, chemists utilize the Mole to mathematically group them into measurable, macroscopic quantities. Stoichiometry is the exact mathematical calculation of reactants and products in balanced chemical reactions based on the conservation of mass.

Stoichiometry and the Mole Diagram

Stoichiometry and the Mole: The Bridge to Macroscopic Measurement. Master visual map defining the relationships between particles, moles, mass, and molar mass.

1.1 Avogadro's Number & Molar Mass

The Mole Concept

One mole of strictly anything contains exactly $6.022 \times 10^{23}$ representative particles (atoms, molecules, or ions). This universal constant is known as Avogadro's Number ($N_A$). It is the bridge between the atomic mass unit (amu) and the gram.

  • Mass to Moles: $n = \frac{\text{mass (g)}}{\text{Molar Mass (g/mol)}}$
  • Moles to Particles: $N = n \times N_A$

Empirical vs Molecular Formula

The Empirical Formula represents the absolutely simplest whole-number ratio of atoms in a compound (e.g., $CH_2O$ for glucose). The Molecular Formula represents the exact, actual number of atoms in a single molecule (e.g., $C_6H_{12}O_6$).

To find the Molecular Formula:

Multiply the subscripts of the empirical formula by an integer $n$, where $n = \frac{\text{Actual Molar Mass}}{\text{Empirical Molar Mass}}$.

Diagram 1: The Master Mole Map

MOLES MASS (g) ÷ Molar Mass × Molar Mass PARTICLES × N_A ÷ N_A VOLUME (L) (Gas at STP) × 22.4 ÷ 22.4

Practice Problem 1.1: Mass to Moles

Question: How many moles of $CaCO_3$ are strictly present in 250 grams of pure calcium carbonate? (Atomic weights: Ca=40, C=12, O=16).

  1. Calculate Molar Mass (MM): $40 + 12 + (3 \times 16) = 100 \text{ g/mol}$.
  2. Apply Formula: $n = \frac{mass}{MM}$.
  3. Calculate: $n = \frac{250 \text{ g}}{100 \text{ g/mol}} = \mathbf{2.5 \text{ moles}}$.

Practice Problem 1.2: Empirical Formula

Question: A completely unknown compound consists of 40% Carbon, 6.7% Hydrogen, and 53.3% Oxygen by mass. What is its empirical formula?

  1. Assume a 100g sample: We have 40g C, 6.7g H, and 53.3g O.
  2. Convert to Moles:
    C = $40 / 12 \approx 3.33 \text{ mol}$
    H = $6.7 / 1 \approx 6.7 \text{ mol}$
    O = $53.3 / 16 \approx 3.33 \text{ mol}$
  3. Divide by Smallest (3.33):
    C = $3.33 / 3.33 = 1$
    H = $6.7 / 3.33 \approx 2$
    O = $3.33 / 3.33 = 1$
  4. Conclusion: The empirical formula is mathematically exactly $CH_2O$.

Practice Problem 1.3: Limiting Reagent & Yield

Reaction: $N_2 + 3H_2 \rightarrow 2NH_3$

Question: You are given exactly 2.0 moles of $N_2$ and exactly 3.0 moles of $H_2$. What is the limiting reagent, and what is the theoretical yield of $NH_3$?

  • Check Stoichiometric Ratios: The balanced equation requires exactly 3 moles of $H_2$ for every 1 mole of $N_2$ (a 1:3 ratio).
  • Calculate Need: To react all 2.0 moles of $N_2$, you would mathematically need $2.0 \times 3 = 6.0$ moles of $H_2$.
  • Identify Limiting: You ONLY have 3.0 moles of $H_2$. Because you do not have enough, $H_2$ will run out entirely first. $H_2$ is the Limiting Reagent.
  • Calculate Yield: Use ONLY the limiting reagent to mathematically find the product.
    $3.0 \text{ mol } H_2 \times (\frac{2 \text{ mol } NH_3}{3 \text{ mol } H_2}) = \mathbf{2.0 \text{ moles of } NH_3}$.
IMAT Challenge

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Question 84 Official Paper: 2015 - Q42

In the reaction $\text{C}_{3}\text{H}_{7}\text{Br} + \text{KOH} \rightarrow \text{C}_{3}\text{H}_{7}\text{OH} + \text{KBr}$. $24.6\text{ g}$ of 1-bromopropane reacts with excess potassium hydroxide to produce $8.00\text{ g}$ of propan-1-ol. ($\text{M}_{r}: \text{C}_{3}\text{H}_{7}\text{Br}=123; \text{A}_{r}: \text{H}=1.0; \text{C}=12.0; \text{O}=16.0$.) What is the percentage yield of this reaction?
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Question 83 Official Paper: 2015 - Q41

$0.75\text{ g}$ of a hydrocarbon compound contains $0.60\text{ g}$ of carbon. ($\text{A}_{r}:\text{C}=12.0, \text{H}=1.0$) Which one of the following could be the molecular formula of the hydrocarbon compound?

Part 2: Solutions & Concentrations

Molarity, Molality, and Aqueous Mixtures

A solution is a perfectly homogeneous mixture of a solute (the substance dissolved, typically in lesser amount) and a solvent (the dissolving medium, typically water in biological systems). The concentration of a solution mathematically defines exactly how much solute is present.

2.1 Expressing Concentration

Molarity (M)

$$ M = \frac{\text{moles of Solute}}{\text{Liters of Total Solution}} $$

Most common unit. Highly temperature dependent because liquid volume expands upon heating.

Molality (m)

$$ m = \frac{\text{moles of Solute}}{\text{kg of Pure Solvent}} $$

Based purely on mass. Does not change with temperature. Crucial for colligative properties.

Mass Percent (%)

$$ \% = \frac{\text{mass of Solute}}{\text{Total mass of Solution}} \times 100 $$

Dilution Equation

$$ M_1V_1 = M_2V_2 $$

Used exclusively when adding pure solvent to a concentrated stock solution to lower its molarity.

Practice Problem 2.1: Molarity

Question: You dissolve exactly 58.5 g of $NaCl$ (Molar Mass = 58.5 g/mol) in enough water to make exactly 500 mL of solution. What is the Molarity?

  1. Find Moles: $58.5 \text{ g} / 58.5 \text{ g/mol} = 1.0 \text{ mol}$ of $NaCl$.
  2. Convert Volume to Liters: $500 \text{ mL} = 0.5 \text{ L}$.
  3. Calculate M: $M = 1.0 \text{ mol} / 0.5 \text{ L} = \mathbf{2.0 \text{ M}}$.

Practice Problem 2.2: Dilution

Question: You have 100 mL of a highly concentrated 12.0 M $HCl$ stock solution. You add pure water until the final total volume reaches exactly 400 mL. What is the new Molarity ($M_2$)?

  1. Identify variables: $M_1 = 12.0 \text{ M}$, $V_1 = 100 \text{ mL}$, $V_2 = 400 \text{ mL}$, $M_2 = ?$
  2. Setup Equation: $(12.0 \text{ M})(100 \text{ mL}) = (M_2)(400 \text{ mL})$
  3. Solve: $1200 = M_2 \times 400 \rightarrow M_2 = 1200 / 400 = \mathbf{3.0 \text{ M}}$.
IMAT Challenge

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Question 94 Official Paper: 2024 - Q37

How many $\text{mL}$ of water must be added to $15\text{ mL}$ of a $0.25\text{ M}$ solution of $\text{H}_{2}\text{SO}_{4}$ to obtain a $0.05\text{ M}$ solution?

Part 3: States of Matter

Phase Changes, Heating Curves, and Intermolecular Forces

The macroscopic state of a substance is determined strictly by the microscopic thermodynamic battle between the Kinetic Energy of the particles (driven by temperature) violently pushing them apart, and the Intermolecular Forces (IMFs) pulling them firmly together.

3.1 Intermolecular Forces (IMFs)

While covalent bonds hold atoms together within a molecule, IMFs hold entirely separate molecules together. They dictate boiling points, melting points, and physical states.

Force Type Physical Basis Strength
Ion-Dipole Attraction strictly between a fully charged ion and the partial charge of a polar molecule. (e.g., $Na^+$ in water). Strongest
Hydrogen Bonding Occurs ONLY when H is covalently bonded to N, O, or F. The bare $\delta^+ H$ is strongly attracted to a lone pair. Strong
Dipole-Dipole Attraction between the permanent positive ($\delta^+$) end of one polar molecule and the negative ($\delta^-$) end of another. Moderate
London Dispersion Temporary induced dipoles caused by completely random fluctuations in electron clouds. Exists in ALL molecules. Weakest
Properties of States of Matter and Visualizing IMFs Diagram

Properties of States of Matter and Visualizing IMFs: Analyzing the anatomy of solid, liquid, and gas phases, and the visual hierarchy of intermolecular force strengths.

Diagram 2: Standard Pressure-Temperature Phase Diagram

Temperature (T) Pressure (P) SOLID LIQUID GAS Triple Point Critical Point Supercritical Fluid

Specific Heat vs. Latent Heat

  • Specific Heat Capacity ($q = mc\Delta T$): The energy required to change the temperature of a substance without changing its phase. Energy goes into increasing the kinetic motion of particles.
  • Latent Heat ($q = mL$): The energy completely absorbed or released exactly during a phase change (e.g., Latent Heat of Fusion/Vaporization). The temperature remains absolutely constant during a phase change because 100% of the energy is strictly used to physically break intermolecular forces, not to increase kinetic motion.

Practice Problem 3.1: Specific Heat

Question: How much heat energy ($q$) is required to precisely raise the temperature of 100 grams of liquid water from 20°C to 50°C? (Specific heat of water $c = 4.18 \text{ J/g}\cdot^\circ\text{C}$).

  1. Formula: $q = mc\Delta T$.
  2. Identify variables: $m = 100 \text{ g}$, $c = 4.18$, $\Delta T = (50 - 20) = 30^\circ\text{C}$.
  3. Calculate: $q = 100 \times 4.18 \times 30 = \mathbf{12,540 \text{ Joules (or 12.54 kJ)}}$.

Practice Problem 3.2: Latent Heat

Question: How much heat is required to completely melt 50 grams of ice at exactly 0°C? (Latent heat of fusion for water $L_f = 334 \text{ J/g}$).

  1. Formula: Because there is NO temperature change during melting, use $q = mL$.
  2. Identify variables: $m = 50 \text{ g}$, $L_f = 334 \text{ J/g}$.
  3. Calculate: $q = 50 \times 334 = \mathbf{16,700 \text{ Joules (or 16.7 kJ)}}$.

Part 4: The Gas Laws

Kinetic Theory and Macroscopic Behavior

Gases behave highly predictably under varying conditions of Pressure (P), Volume (V), and Temperature (T). The Kinetic Molecular Theory strictly assumes that Ideal Gas particles possess negligible physical volume, exert absolutely no intermolecular forces of attraction on each other, and undergo perfectly elastic collisions (conserving total kinetic energy).

The Gas Laws: Mathematical Connections and Graphs Diagram

The Gas Laws: Mathematical Connections and Graphs. Visual breakdown of PV=nRT and graphical comparisons of Boyle's, Charles's, and Gay-Lussac's Laws.

The Ideal Gas Law: $PV = nRT$

  • P: Pressure (atm)
  • V: Volume (Liters)
  • n: Moles of gas
  • R: Ideal Gas Constant ($0.0821 \text{ L}\cdot\text{atm/mol}\cdot\text{K}$)
  • T: Temperature MUST be mathematically in Kelvin! ($K = ^\circ C + 273$)

Diagram 3: Graphical Representation of Simple Gas Laws

Boyle's Law P₁V₁ = P₂V₂ (Const T, n) Volume (V) Pressure (P) P ∝ 1/V Charles's Law V₁/T₁ = V₂/T₂ (Const P) Temp (K) Volume (V) Direct Gay-Lussac's Law P₁/T₁ = P₂/T₂ (Const V) Temp (K) Pressure (P) Direct

Practice Problem 4.1: Boyle's Law

Question: A gas occupies 4.0 L at a pressure of 2.0 atm. If the volume is expanded to 8.0 L at a constant temperature, what is the new theoretical pressure?

  1. Formula: $P_1 V_1 = P_2 V_2$.
  2. Variables: $P_1 = 2.0$, $V_1 = 4.0$, $V_2 = 8.0$, $P_2 = ?$
  3. Calculate: $(2.0 \text{ atm}) \times (4.0 \text{ L}) = (P_2) \times (8.0 \text{ L}) \rightarrow 8.0 = P_2 \times 8.0 \rightarrow \mathbf{P_2 = 1.0 \text{ atm}}$.
  4. Logic check: Volume doubled, so pressure halved (Inverse relationship).

Practice Problem 4.2: Ideal Gas Law

Question: What exact volume will 3.0 moles of pure Oxygen gas ($O_2$) occupy at a temperature of 27°C and a pressure of 1.5 atm? (R = 0.0821)

  1. Variables: $n = 3.0 \text{ mol}$, $P = 1.5 \text{ atm}$, $V = ?$
  2. Convert Temperature to Kelvin! $T = 27 + 273 = 300 \text{ K}$.
  3. Set up equation: $V = \frac{nRT}{P}$
  4. Calculate: $V = \frac{3.0 \times 0.0821 \times 300}{1.5} = \frac{73.89}{1.5} = \mathbf{49.26 \text{ Liters}}$.

Practice Problem 4.3: Dalton's Law of Partial Pressures

Question: A gas mixture contains 2.0 moles of $N_2$, 3.0 moles of $O_2$, and 5.0 moles of $He$. If the total pressure of the tank is 10.0 atm, what is the partial pressure of $O_2$?

  1. Concept: Partial pressure strictly equals the Total Pressure multiplied by the Mole Fraction ($X$).
  2. Find Total Moles: $2.0 + 3.0 + 5.0 = 10.0 \text{ total moles}$.
  3. Mole Fraction of $O_2$ ($X_{O_2}$): $3.0 \text{ mol } O_2 / 10.0 \text{ total mol} = 0.30$.
  4. Calculate Partial Pressure: $P_{O_2} = X_{O_2} \times P_{total} = 0.30 \times 10.0 \text{ atm} = \mathbf{3.0 \text{ atm}}$.
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Question 13 Official Paper: 2023 - Q34

A sample of $n\text{ moles}$ of an ideal gas is contained in a closed system of fixed volume $V\text{ m}^{3}$ at pressure $P\text{ Pa}$ and temperature $27.0^{\circ}\text{C}$. If the gas is heated to $327.0^{\circ}\text{C}$, what will be the pressure, in $\text{Pa}$, at this new temperature? Assume absolute zero is at $-273.0^{\circ}\text{C}$

Part 5: Redox Reactions

Electron Transfer and Oxidation States

Oxidation-Reduction (Redox) reactions involve the absolute, fundamental transfer of electrons from one chemical species to another. These reactions drive cellular respiration, photosynthesis, and all batteries.

The OIL RIG Mnemonic
  • Oxidation Is Loss of electrons. (The oxidation state mathematically increases/becomes more positive).
  • Reduction Is Gain of electrons. (The oxidation state mathematically decreases/is reduced).

The Agents: The substance that is explicitly oxidized loses electrons, forcing them onto another substance; therefore, it is the Reducing Agent. Conversely, the substance that is reduced forcefully takes electrons, acting as the Oxidizing Agent.

Redox Reactions and Oxidation States Diagram

Redox Reactions and Oxidation States: Mastery and Logic map. Visualizing OIL RIG, step-by-step oxidation state assignment, and agent identification.

5.1 Master Rules for Assigning Oxidation States

To identify if redox has occurred, we must assign an oxidation state (a bookkeeping number of electrons) to every single atom in a reaction.

  1. Any element in its pure, uncombined standard state has an oxidation state of exactly 0 (e.g., $O_2$, $Na(s)$, $Cl_2$, $Fe$).
  2. For a simple monoatomic ion, the oxidation state strictly equals the physical charge of the ion (e.g., $Na^+$ is +1, $S^{2-}$ is -2).
  3. Oxygen is almost universally exactly -2 in its compounds. (Exception: In peroxides like $H_2O_2$, it is -1).
  4. Hydrogen is almost universally exactly +1 when bonded to nonmetals. (Exception: in metal hydrides like $NaH$, it is -1).
  5. The Master Sum Rule: The total mathematical sum of all oxidation states in a neutral molecule must exactly equal 0. In a polyatomic ion, the sum must exactly equal the net charge of the ion.

Practice Problem 5.1: Assigning Oxidation States

Question: Determine the exact oxidation state of Chromium (Cr) in the highly toxic Dichromate ion ($Cr_2O_7^{2-}$).

  1. We know Oxygen is reliably -2. There are 7 Oxygen atoms. Total Oxygen contribution = $7 \times (-2) = -14$.
  2. The overall net charge of the entire ion is -2.
  3. Set up the mathematical equation: $2(Cr) + (-14) = -2$.
  4. Solve for Cr: $2(Cr) = +12$. Therefore, $Cr = +6$.
  5. Answer: The oxidation state of Chromium is exactly +6.

Practice Problem 5.2: Identifying Agents

Question: In the reaction $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$, identify the Oxidizing Agent and the Reducing Agent.

  1. Assign Oxidation States:
    $Zn(s)$ = 0. $Cu^{2+}$ = +2. $Zn^{2+}$ = +2. $Cu(s)$ = 0.
  2. Analyze Changes:
    Zinc goes from 0 to +2. It lost electrons. Zinc is Oxidized.
    Copper goes from +2 to 0. It gained electrons. Copper is Reduced.
  3. Assign Agents: Because Zinc is oxidized, it forces reduction on Copper. Therefore, $Zn(s)$ is the Reducing Agent. Because Copper is reduced, it forces oxidation on Zinc. Therefore, $Cu^{2+}$ is the Oxidizing Agent.
IMAT Challenge

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Question 113 Official Paper: 2024 - Q40

Zinc nitrate, nitrogen dioxide, and water are obtained from the reaction of metallic zinc and nitric acid in an aqueous solution. What is the reducing species?

Part 6: Electrochemistry

Galvanic Cells, Electrolysis, and Standard Potentials

Electrochemistry physically harnesses the energy of redox reactions. If the reaction is completely spontaneous ($\Delta G < 0$), it can generate a massive voltage (Galvanic Cell). If the reaction is totally non-spontaneous ($\Delta G > 0$), we must explicitly force it to happen by applying an external voltage from a battery (Electrolytic Cell).

Galvanic (Voltaic) Cells

A standard battery. A spontaneous redox reaction physically generates an electrical current.

  • Anode: Where Oxidation occurs ("An Ox"). It is the Negative (-) electrode because it literally generates and pushes out electrons.
  • Cathode: Where Reduction occurs ("Red Cat"). It is the Positive (+) electrode pulling electrons in.
  • Electrons physically flow exclusively from Anode $\rightarrow$ Cathode through the external wire.
  • The Salt Bridge is absolutely crucial; it supplies inert ions to balance the charge build-up in both half-cells, perfectly completing the circuit.

Electrolytic Cells

Requires an external power source to forcefully drive a non-spontaneous reaction (e.g., splitting water, electroplating).

  • Oxidation still strictly occurs at the Anode ("An Ox"), and Reduction still occurs at the Cathode ("Red Cat").
  • The polarities are mathematically REVERSED! Because the battery physically forces electrons into the Cathode, the Cathode becomes Negative (-). The battery pulls electrons from the Anode, making it Positive (+).
Electrochemistry: Galvanic vs. Electrolytic Comparative Anatomy Diagram

Electrochemistry: Galvanic vs. Electrolytic Comparative Anatomy. Side-by-side visualization of spontaneous energy generation vs. non-spontaneous electroplating.

Diagram 4: A Standard Zn-Cu Galvanic Cell

Zn ANODE (-) Oxidation: Zn $\rightarrow$ Zn²⁺ + 2e⁻ Cu CATHODE (+) Reduction: Cu²⁺ + 2e⁻ $\rightarrow$ Cu V e⁻ flow e⁻ flow Salt Bridge

Standard Cell Potential ($E^\circ_{cell}$)

The measurable voltage generated by a galvanic cell under standard conditions. A strictly positive voltage ($E^\circ_{cell} > 0$) explicitly indicates a spontaneous reaction (a working battery).

$$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $$

(IMAT Trick: You use the standard reduction potentials exactly as written on the table without manually changing their signs. The minus sign in the mathematical equation flips the sign for the anode automatically!)

Practice Problem 6.1: Calculating Cell Potential

Question: A galvanic cell is constructed using Copper ($Cu^{2+} + 2e^- \rightarrow Cu, E^\circ = +0.34\text{V}$) and Zinc ($Zn^{2+} + 2e^- \rightarrow Zn, E^\circ = -0.76\text{V}$). What is the standard cell potential ($E^\circ_{cell}$), and which metal is the Anode?

  1. Identify Cathode and Anode: In a spontaneous Galvanic cell, the metal with the higher (more positive) reduction potential wants electrons more, so it is the Cathode (Reduction). The lower potential is the Anode (Oxidation).
    Therefore, Cu is the Cathode and Zn is the Anode.
  2. Apply Formula: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$.
  3. Calculate: $E^\circ_{cell} = (+0.34\text{V}) - (-0.76\text{V}) = 0.34 + 0.76 = \mathbf{+1.10\text{ Volts}}$.

Practice Problem 6.2: Faraday's Laws of Electrolysis

Question: An electrolytic cell passes a current of 10 Amperes for exactly 9650 seconds through a solution of $Cu^{2+}$ ions. How many moles of solid Copper are deposited at the cathode? (Faraday's Constant $F \approx 96500 \text{ C/mol } e^-$).

  1. Calculate Total Charge (Q): $Q = Current (I) \times Time (t)$.
    $Q = 10 \text{ A} \times 9650 \text{ s} = 96,500 \text{ Coulombs}$.
  2. Calculate Moles of Electrons: $n_{e-} = Q / F$.
    $n_{e-} = 96,500 \text{ C} / 96,500 \text{ C/mol} = \mathbf{1.0 \text{ mole of electrons}}$.
  3. Use Stoichiometry of Reduction: The reaction is $Cu^{2+} + 2e^- \rightarrow Cu(s)$. This requires exactly 2 moles of electrons for every 1 mole of Cu.
  4. Calculate Moles of Cu: $1.0 \text{ mol } e^- \times (\frac{1 \text{ mol Cu}}{2 \text{ mol } e^-}) = \mathbf{0.5 \text{ moles of Cu}}$.
IMAT Challenge

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Question 128 Official Paper: 2024 - Q47

According to the Brønsted-Lowry theory:
PRACTICE EXAM

General Chemistry Final Exam

30 High-Yield Questions (Stoichiometry, Gases, Redox)