Unit 1: Some Basic Concepts of Chemistry

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The Big Picture: Why This Chapter Matters#

Welcome to chemistry. Before working with numerical formulas, let us understand the central challenge this entire chapter addresses:

We live in a macroscopic world of beakers, grams, litres, and analytical balances. However, chemical reactions occur in the microscopic world between individual atoms, ions, and molecules that are far too small to observe directly or manipulate individually.

Consider the orders of magnitude involved:

• In just $18\text{ grams}$ of water (approximately one tablespoon), there are roughly $602,200,000,000,000,000,000,000$ molecules.
• A single hydrogen atom has a mass of only about $0.00000000000000000000000166\text{ grams}$.

How can a chemist in a laboratory measure out the exact number of molecules required for a stoichiometric reaction using a standard mass balance?

This chapter provides the universal translator between the microscopic atomic world and the macroscopic laboratory world. Once you master the mental models, unit relationships, and stoichiometric laws established here, subsequent branches of chemistry (Thermodynamics, Chemical Equilibrium, Electrochemistry, and Organic synthesis) become clear and systematic.

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1. Development and Scope of Chemistry#

Chemistry is the central science because it links fundamental physics (atomic structure, energy states, and forces) with biology (biochemical processes in living cells) and materials science.

Historical Roots: Alchemy to Modern Science#

Historically, chemistry emerged from two long-standing practical pursuits:

1. The Philosopher's Stone (Paras): A substance sought by alchemists believed capable of transmuting base metals such as lead and copper into gold.
2. The Elixir of Life: A medicinal preparation sought for longevity and the cure of diseases.

Ancient Indian Chemical Traditions (Rasayan Shastra)#

Long before modern systematic laboratory nomenclature was standardized, ancient Indian traditions developed empirical chemistry:

• Acharya Kanada (Kashyap, ~600 BCE): In the Vaiseshika Sutras, he conceptualized atomic theory centuries before John Dalton, postulating that all matter consists of eternal, indivisible, spherical particles termed Paramãnu (atoms) that combine into Dvinuka (diatomic) and Trinuka (triatomic) aggregates.
• Applied Metallurgy and Medicine: Charaka Samhita and Sushruta Samhita documented preparations of mineral acids, caustic alkalies (Kshara), and metallic bhasmas (which modern analytical electron microscopy confirms consist of biocompatible nanoparticles).
• Analytical Methods: The text Rasarnavam (~800 CE) described laboratory furnaces, crucibles, and the systematic identification of specific metals by their characteristic flame emissions.

Modern Practical Applications#

Modern society relies on synthetic and physical chemistry across numerous sectors:

• Pharmaceuticals: Cisplatin and Taxol (chemotherapy agents for cancer treatment); AZT (Azidothymidine) for HIV/AIDS treatment regimens.
• Environmental Management: Developing halogen-free alternatives to ozone-depleting chlorofluorocarbons (CFCs).
• Materials Engineering: Superconductors, conducting polymers, carbon nanotubes, and solid-state battery electrolytes.

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2. Nature and Classification of Matter#

Matter is defined as anything that possesses mass and occupies physical space.

The Three Physical States of Matter#

Matter exists in three primary states: Solid, Liquid, and Gas. At the particulate level, the state is governed by the equilibrium between two competing influences:

1. Intermolecular Attractive Forces: Draw particles together and keep them in ordered positions.
2. Thermal Kinetic Energy ($T$): Thermal agitation that causes particles to separate and move.

These states interconvert reversibly upon modifying temperature ($T$) and pressure ($P$): $\text{Solid} \xrightleftharpoons[\text{cool}]{\text{heat}} \text{Liquid} \xrightleftharpoons[\text{cool}]{\text{heat}} \text{Gas}$

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Macroscopic Classification of Matter#

All matter can be categorized chemically into Mixtures or Pure Substances:

Key Insight: The Emergent Nature of Chemical Compounds

A chemical compound exhibits distinct, emergent properties that differ fundamentally from the elements that formed it. For example:

• Dihydrogen gas ($\text{H}_2$) is flammable and combustible.
• Dioxygen gas ($\text{O}_2$) strongly supports combustion.
• When chemically combined in a $2:1$ atomic stoichiometry, they yield Water ($\text{H}_2\text{O}$), which is an incombustible liquid used to extinguish combustion.

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3. Properties and the International System of Units (SI)#

Every scientific measurement must record two components: a numerical magnitude and a standardized unit (for example, $5.0\text{ kg}$).

The Seven Fundamental SI Base Units#

In 2019, the General Conference on Weights and Measures (CGPM) updated the SI system so that every base unit is defined by fixing the exact numerical values of fundamental physical constants of nature:

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Important Derived Quantities and Practical Conversions#

1. Mass versus Weight:

   • Mass is the invariant measure of the quantity of matter contained within a body. It is measured on an analytical balance and remains identical regardless of gravitational field strength.
   • Weight is the downward gravitational force acting on that mass ($W = mg$), which varies with local gravitational acceleration $g$.

2. Volume ($V$):

   • SI derived unit: $\text{m}^3$ (cubic metre).
   • Practical chemical laboratory relationships: $1\text{ L} = 1\text{ dm}^3 = 1000\text{ mL} = 1000\text{ cm}^3 = 10^{-3}\text{ m}^3$

3. Density ($\rho$): $\rho = \frac{\text{Mass}}{\text{Volume}}$

   • SI unit: $\text{kg m}^{-3}$.
   • Practical laboratory unit: $\text{g cm}^{-3}$ or $\text{g mL}^{-1}$ (for example, density of pure water at $4^{\circ}\text{C} \approx 1.00\text{ g mL}^{-1}$).

4. Temperature Scales ($^{\circ}\text{C}$, $^{\circ}\text{F}$, $\text{K}$):

   • Celsius scale ($^{\circ}\text{C}$): Based on the freezing point ($0^{\circ}\text{C}$) and boiling point ($100^{\circ}\text{C}$) of pure water at $1\text{ atm}$.
   • Kelvin scale ($\text{K}$): The thermodynamic SI absolute temperature scale.

   $\mathbf{^{\circ}\text{F} = \frac{9}{5}\left(^{\circ}\text{C}\right) + 32} \qquad \mathbf{\text{K} = {^{\circ}\text{C}} + 273.15}$

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4. Uncertainty in Measurement and Significant Figures#

Experimental laboratory equipment has finite precision, meaning no empirical measurement has infinite certainty.

Scientific Notation ($N \times 10^n$)#

Scientific notation expresses very large or very small magnitudes in standardized exponential form: $N \times 10^n \quad (1.000\dots \le N < 10.000\dots, \quad n \in \mathbb{Z})$

• Speed of light: $299,792,458\text{ m/s} = 2.99792458 \times 10^8\text{ m/s}$
• Mass of a single hydrogen atom: $0.00000000000000000000000166\text{ g} = 1.66 \times 10^{-24}\text{ g}$

Significant Figures#

Five Rules for Counting Significant Figures:

1. All non-zero digits are significant: $285\text{ cm}$ has 3 significant figures; $0.25\text{ mL}$ has 2 significant figures.
2. Leading zeros are never significant: Zeros appearing before the first non-zero digit serve only to indicate decimal position: $0.03\text{ g}$ has 1 significant figure; $0.0052\text{ L}$ has 2 significant figures.
3. Trapped zeros are always significant: Zeros between non-zero digits are significant: $2.005\text{ g}$ has 4 significant figures; $5.0001\text{ s}$ has 5 significant figures.
4. Trailing zeros after a decimal point are significant: $0.200\text{ g}$ has 3 significant figures; $100.0\text{ mL}$ has 4 significant figures.
   (If written without a decimal point like $100$, the precision is ambiguous; write in scientific notation: $1.00 \times 10^2$ denotes 3 significant figures, whereas $1 \times 10^2$ denotes 1).
5. Exact counting numbers possess an infinite ($\infty$) number of significant figures: $2\text{ beakers} = 2.00000\dots$ has an infinite number of significant figures.

Significant Figures in Calculations:

• Addition and Subtraction: The calculated answer can have no more decimal places than the measurement having the fewest decimal places: $12.11 + 18.0 \text{ (least precise: 1 decimal place)} + 1.012 = 31.122 \xrightarrow{\text{round}} \mathbf{31.1}$
• Multiplication and Division: The calculated answer can retain no more significant figures than the input value possessing the fewest significant figures: $2.5 \text{ (2 significant figures)} \times 1.25 = 3.125 \xrightarrow{\text{round}} \mathbf{3.1}$

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Precision versus Accuracy#

Experimental Demonstration: True Standard Mass = 2.000 g

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5. The Five Foundational Laws of Chemical Combinations#

Stoichiometric calculations rest upon five empirical laws established between 1789 and 1811:

1. Law of Conservation of Mass (Lavoisier, 1789)#

The total mass of reactants consumed in a chemical reaction strictly equals the total mass of products generated: $\mathbf{\sum m_{\text{reactants}} = \sum m_{\text{products}}}$

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2. Law of Definite Proportions (Proust, 1799)#

Pure water obtained from natural ice, rainwater, or synthesized in a laboratory will always contain hydrogen and oxygen combined in a fixed mass ratio of $1 : 8$ ($11.11\%\ \text{H}$ and $88.89\%\ \text{O}$).

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3. Law of Multiple Proportions (Dalton, 1803)#

Consider Carbon and Oxygen forming two distinct gases:

1. Carbon Monoxide ($\text{CO}$): $12\text{ g Carbon}$ combines with $16\text{ g Oxygen}$.
2. Carbon Dioxide ($\text{CO}_2$): $12\text{ g Carbon}$ combines with $32\text{ g Oxygen}$.

Holding the mass of carbon fixed at $12\text{ g}$, the ratio of oxygen masses is: $\frac{16\text{ g}}{32\text{ g}} = \frac{1}{2} \implies \mathbf{1 : 2} \quad \text{(a small whole-number ratio)}$

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4. Gay-Lussac's Law of Gaseous Volumes (1808)#

$2\text{H}_2(g) + 1\text{O}_2(g) \longrightarrow 2\text{H}_2\text{O}(g)$ $100\text{ mL} + 50\text{ mL} \longrightarrow 100\text{ mL}$ The combining volume ratio is $2 : 1 : 2$. Notice that $100\text{ mL} + 50\text{ mL}$ yields $100\text{ mL}$ of water vapor, not $150\text{ mL}$. This observation demonstrated that gas volumes reflect particle counts rather than particle sizes.

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5. Avogadro's Law (1811)#

Amedeo Avogadro clarified the distinction between individual atoms and bonded molecules, establishing that common gaseous elements exist as diatomic molecules ($\text{H}_2, \text{O}_2, \text{N}_2$).

Consequently: $2\text{ volumes of } \text{H}_2 + 1\text{ volume of } \text{O}_2 \longrightarrow 2\text{ volumes of } \text{H}_2\text{O}$ $2n\text{ molecules of } \text{H}_2 + 1n\text{ molecule of } \text{O}_2 \longrightarrow 2n\text{ molecules of } \text{H}_2\text{O}$ Avogadro's hypothesis directly established the relationship between macroscopic gas volumes and microscopic mole quantities ($V \propto n$).

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6. Dalton's Atomic Theory#

In 1808, John Dalton published 'A New System of Chemical Philosophy', formalizing the first systematic atomic model:

1. All matter is composed of indivisible particles called atoms.
2. All atoms of a given element have identical mass and chemical properties; atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in fixed, simple whole-number ratios.
4. Chemical reactions involve only the combination, separation, or rearrangement of atoms; atoms are neither created nor destroyed.

Modern Scientific Revisions:#

• Dalton considered the atom indivisible; modern atomic physics demonstrated subatomic structure consisting of protons, neutrons, and electrons.
• Dalton postulated that all atoms of an element possess identical mass; the discovery of isotopes (such as $^{12}\text{C}, {}^{13}\text{C}, {}^{14}\text{C}$) showed that atoms of the same element can possess different masses.
• Dalton postulated that atoms of different elements always possess different masses; the discovery of isobars (such as $^{40}\text{Ar}$ and $^{40}\text{Ca}$) demonstrated that distinct elements can share the same atomic mass number.

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7. Atomic Mass and the Carbon-12 Standard#

Because an individual atom possesses an imperceptible mass (a hydrogen atom weighs $1.67 \times 10^{-24}\text{ g}$), absolute mass is inconvenient for routine work. In 1961, IUPAC internationally standardized atomic masses relative to the Carbon-12 ($^{12}\text{C}$) isotope.

The Unified Mass Unit ($u$)#

$\mathbf{1\text{ u} = \frac{\text{Mass of one }{}^{12}\text{C atom}}{12} = 1.66056 \times 10^{-24}\text{ g} = 1.66056 \times 10^{-27}\text{ kg}}$

• Mass of one Hydrogen atom $= 1.008\text{ u}$
• Mass of one Oxygen atom $= 16.00\text{ u}$

Average Atomic Mass#

Most elements occur in nature as a combination of two or more stable isotopes. The atomic weight listed on the periodic table represents the weighted average based on natural isotopic abundance: $\mathbf{\bar{A} = \sum (f_i \times A_i)}$

Example: Average Atomic Mass of Carbon

• $^{12}\text{C}$: $98.892\%$ abundance ($f_1 = 0.98892$), mass $= 12\text{ u}$
• $^{13}\text{C}$: $1.108\%$ abundance ($f_2 = 0.01108$), mass $= 13.00335\text{ u}$ $\bar{A} = (0.98892 \times 12) + (0.01108 \times 13.00335) = \mathbf{12.011\text{ u}}$

Molecular Mass versus Formula Mass#

• Molecular Mass: The sum of atomic masses of all atoms present in a covalent molecule (for example, $\text{H}_2\text{O} = 2(1.008) + 16.00 = 18.016\text{ u}$).
• Formula Mass: Used for ionic network solids like Sodium Chloride ($\text{NaCl}$), which do not exist as isolated discrete single molecules, but as an alternating three-dimensional lattice: $\text{Formula Mass of NaCl} = 23.0\text{ u} + 35.5\text{ u} = \mathbf{58.5\text{ u}}$

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8. The Mole Concept: The Avogadro Bridge#

The mole concept provides the quantitative bridge between atomic scale units and macroscopic laboratory measurements:

The Origin of the Avogadro Constant#

The constant $6.02214076 \times 10^{23}$ represents the reciprocal of the unified atomic mass unit expressed in grams:

$\frac{1}{1.66056 \times 10^{-24}\text{ g}} = \mathbf{6.02214 \times 10^{23}}$

Consequently: $\mathbf{1\text{ u} \times (6.02214 \times 10^{23}) = 1\text{ gram}}$

This specific quantity ($6.02214076 \times 10^{23}$ particles) is defined as $1\text{ Mole}$ (symbol: $\text{mol}$).

Molar Mass ($M$)#

Molar mass is defined as the mass of one mole of a substance expressed in grams per mole ($\text{g mol}^{-1}$). It is numerically equal to the atomic, molecular, or formula mass in $u$.

Master Mole Interconversion Formula#

$\mathbf{n = \frac{w}{M} = \frac{N}{N_A} = \frac{V_{\text{STP}}}{22.7\text{ L}}}$

Where:

• $n$ is the number of moles.
• $w$ is the sample mass in grams.
• $M$ is the molar mass ($\text{g mol}^{-1}$).
• $N$ is the number of elementary entities (atoms, molecules, or formula units).
• $N_A$ is the Avogadro constant ($6.022 \times 10^{23}\text{ mol}^{-1}$).
• $V_{\text{STP}}$ is the gas volume at standard temperature and pressure ($273.15\text{ K}, 1\text{ bar}$), where molar volume $V_m = 22.7\text{ L mol}^{-1}$ (or $22.4\text{ L}$ under the historical $1\text{ atm}$ standard).

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9. Empirical and Molecular Formulas#

• Empirical Formula: Expresses the simplest whole-number ratio of atoms of each element present in a compound.
• Molecular Formula: Expresses the actual number of atoms of each element in a single molecule of the compound.

$\mathbf{\text{Molecular Formula} = (\text{Empirical Formula})_n} \qquad \mathbf{n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}}$

Example: The empirical formula of glucose is $\text{CH}_2\text{O}$ (mass $= 30.03\text{ g/mol}$).
With an actual molar mass of $180.16\text{ g/mol}$, the scaling factor is: $n = \frac{180.16}{30.03} = 6 \implies \text{Molecular Formula} = (\text{CH}_2\text{O})_6 = \mathbf{\text{C}_6\text{H}_{12}\text{O}_6}$

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Step-by-Step Method for Empirical Formula Calculations#

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10. Stoichiometry and the Limiting Reagent#

Stoichiometry deals with quantitative calculations of masses, moles, and volumes of reactants and products in chemical reactions.

Interpreting a Balanced Chemical Equation#

$\text{CH}_4(g) + 2\text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$

The stoichiometric coefficients ($1, 2, 1, 2$) represent:

• Molecules: $1\text{ molecule of } \text{CH}_4 + 2\text{ molecules of } \text{O}_2 \to 1\text{ molecule of } \text{CO}_2 + 2\text{ molecules of } \text{H}_2\text{O}$
• Moles: $1\text{ mol of } \text{CH}_4 + 2\text{ mol of } \text{O}_2 \to 1\text{ mol of } \text{CO}_2 + 2\text{ mol of } \text{H}_2\text{O}$
• Volumes at STP: $22.7\text{ L of } \text{CH}_4 + 45.4\text{ L of } \text{O}_2 \to 22.7\text{ L of } \text{CO}_2 + 45.4\text{ L of } \text{H}_2\text{O}$
• Masses: $16\text{ g of } \text{CH}_4 + 64\text{ g of } \text{O}_2 \to 44\text{ g of } \text{CO}_2 + 36\text{ g of } \text{H}_2\text{O}$
  (Total reactant mass equals $80\text{ g}$; total product mass equals $80\text{ g}$, adhering to Conservation of Mass).

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The Limiting Reagent Principle#

Systematic Test for the Limiting Reagent:

For a general reaction: $a\text{A} + b\text{B} \longrightarrow \text{Products}$

1. Compute the available moles for each reactant: $n_A$ and $n_B$.
2. Divide the moles by the respective stoichiometric coefficient: $\mathbf{\text{Compare: } \frac{n_A}{a} \quad \text{with} \quad \frac{n_B}{b}}$
3. The reactant with the smaller ratio is strictly the Limiting Reagent.
4. All subsequent product yield calculations must be based upon this limiting reactant.

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11. Reactions in Solutions and Concentration Expressions#

The relative proportion of solute dissolved in a solution is expressed in four standard ways:

1. Mass Percent ($w/w\ \%$)#

$\mathbf{\text{Mass \%} = \frac{\text{Mass of Solute}}{\text{Total Mass of Solution}} \times 100 = \frac{w_{\text{solute}}}{w_{\text{solute}} + w_{\text{solvent}}} \times 100}$

• Independent of temperature because it is based strictly on mass.

2. Mole Fraction ($x$)#

The ratio of moles of a designated component to the total moles of all components present: $\mathbf{x_A = \frac{n_A}{n_A + n_B}} \qquad \mathbf{x_B = \frac{n_B}{n_A + n_B}} \qquad \mathbf{\sum x_i = x_A + x_B = 1}$

• Dimensionless and independent of temperature.

3. Molarity ($M$)#

The number of moles of solute dissolved per $1\text{ litre}$ ($1\text{ dm}^3$) of total solution: $\mathbf{M = \frac{\text{Moles of Solute}}{\text{Volume of Solution in Litres}} = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times V_{\text{mL}}}}$

• Units: $\text{mol L}^{-1}$ or $\text{M}$ (molar).
• Dilution Relationship: Dilution with pure solvent maintains constant solute moles ($n_1 = n_2$): $\mathbf{M_1 V_1 = M_2 V_2}$

4. Molality ($m$)#

The number of moles of solute dissolved in $1\text{ kilogram}$ of pure solvent: $\mathbf{m = \frac{\text{Moles of Solute}}{\text{Mass of Solvent in kg}} = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times w_{\text{solvent in g}}}}$

• Units: $\text{mol kg}^{-1}$ or $\text{m}$ (molal).

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Core Concept: Molarity vs. Molality Under Temperature Variations#

• Molarity ($M$) depends directly upon the volume of the solution: $M = \frac{n}{V}$ Liquids undergo thermal expansion upon heating and contraction upon cooling. When temperature increases, solution volume $V$ expands, causing Molarity to decrease, even though the quantity of solute remains unchanged.
• Molality ($m$) depends upon the mass of the solvent: $m = \frac{n}{w_{\text{solvent}}}$ Mass is invariant to temperature variations. One kilogram of water remains exactly one kilogram regardless of heating. Therefore, Molality remains strictly constant across all temperatures, making it scientifically preferred in thermodynamic studies.

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12. Walkthrough of Key NCERT Solved Problems#

Problem 1.1: Molecular Mass Calculation#

Question: Calculate the molecular mass of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$).
Solution: Glucose contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms: $\text{Mass} = 6(12.011\text{ u}) + 12(1.008\text{ u}) + 6(16.00\text{ u}) = 72.066 + 12.096 + 96.00 = \mathbf{180.162\text{ u}}$

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Problem 1.2: Empirical and Molecular Formula#

Question: A compound contains $4.07\%$ hydrogen, $24.27\%$ carbon, and $71.65\%$ chlorine. Its molar mass is $98.96\text{ g}$. Determine its empirical and molecular formulas.
Solution:

1. In $100\text{ g}$ of sample: $\text{H} = 4.07\text{ g}$, $\text{C} = 24.27\text{ g}$, $\text{Cl} = 71.65\text{ g}$.
2. Convert to mole quantities:
   • $n_{\text{H}} = 4.07 / 1.008 = 4.04\text{ mol}$
   • $n_{\text{C}} = 24.27 / 12.01 = 2.021\text{ mol}$
   • $n_{\text{Cl}} = 71.65 / 35.453 = 2.021\text{ mol}$
3. Divide by the smallest value ($2.021$): $\text{H} : \text{C} : \text{Cl} = \frac{4.04}{2.021} : \frac{2.021}{2.021} : \frac{2.021}{2.021} = 2 : 1 : 1$ Empirical Formula = $\mathbf{\text{CH}_2\text{Cl}}$
4. Empirical formula mass: $12.01 + 2(1.008) + 35.45 = 49.48\text{ g/mol}$.
5. Ratio factor $n = \frac{98.96}{49.48} = 2$.
   Molecular Formula = $(\text{CH}_2\text{Cl})_2 = \mathbf{\text{C}_2\text{H}_4\text{Cl}_2}$ (1,2-dichloroethane).

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Problem 1.5: Limiting Reagent in Haber Ammonia Synthesis#

Question: $50.0\text{ kg of } \text{N}_2(g)$ and $10.0\text{ kg of } \text{H}_2(g)$ are mixed to produce $\text{NH}_3(g)$. Identify the limiting reagent and calculate the mass of $\text{NH}_3(g)$ formed.
Solution: $\text{N}_2(g) + 3\text{H}_2(g) \longrightarrow 2\text{NH}_3(g)$

1. Convert masses to moles:
   • $n(\text{N}_2) = \frac{50.0 \times 10^3\text{ g}}{28.02\text{ g/mol}} = 1786\text{ mol}$
   • $n(\text{H}_2) = \frac{10.0 \times 10^3\text{ g}}{2.016\text{ g/mol}} = 4960\text{ mol}$
2. Compare stoichiometric requirements:
   • $1786\text{ mol of } \text{N}_2$ requires $3 \times 1786 = 5358\text{ mol of } \text{H}_2$.
   • The available quantity of $\text{H}_2$ is only $4960\text{ mol}$.
   • Therefore, $\text{H}_2$ is the Limiting Reagent.
3. Calculate theoretical yield based on limiting $\text{H}_2$: $n(\text{NH}_3) = \frac{2}{3} \times 4960\text{ mol} = 3307\text{ mol}$ $\text{Mass of } \text{NH}_3 = 3307\text{ mol} \times 17.03\text{ g/mol} = 56.3 \times 10^3\text{ g} = \mathbf{56.1\text{ kg of } \text{NH}_3}$

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Problem 1.7: Molarity Calculation#

Question: Calculate the molarity of $\text{NaOH}$ in a solution prepared by dissolving $4\text{ g}$ in enough water to form $250\text{ mL}$ of solution.
Solution:

1. Molar mass of $\text{NaOH} = 23 + 16 + 1 = 40.0\text{ g/mol}$.
2. Moles of $\text{NaOH}$: $n = \frac{4.0\text{ g}}{40.0\text{ g/mol}} = 0.10\text{ mol}$
3. Volume in litres: $V = \frac{250\text{ mL}}{1000\text{ mL/L}} = 0.250\text{ L}$
4. Molarity: $M = \frac{0.10\text{ mol}}{0.250\text{ L}} = \mathbf{0.40\text{ mol L}^{-1}} = \mathbf{0.40\text{ M}}$

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Problem 1.8: Molality from Molarity and Density#

Question: The density of a $3\text{ M}$ solution of $\text{NaCl}$ is $1.25\text{ g mL}^{-1}$. Calculate the molality of the solution.
Solution:

1. A $3\text{ M}$ solution contains $3\text{ moles of } \text{NaCl}$ in $1\text{ litre}$ of solution.
2. Mass of solute: $w_{\text{solute}} = 3\text{ mol} \times 58.5\text{ g/mol} = 175.5\text{ g}$
3. Mass of $1\text{ L}$ ($1000\text{ mL}$) of solution using density: $w_{\text{solution}} = 1000\text{ mL} \times 1.25\text{ g/mL} = 1250\text{ g}$
4. Mass of solvent (water): $w_{\text{solvent}} = w_{\text{solution}} - w_{\text{solute}} = 1250\text{ g} - 175.5\text{ g} = 1074.5\text{ g} = 1.0745\text{ kg}$
5. Molality: $m = \frac{\text{Moles of Solute}}{\text{kg of Solvent}} = \frac{3\text{ mol}}{1.0745\text{ kg}} = \mathbf{2.79\text{ mol kg}^{-1}} = \mathbf{2.79\text{ m}}$

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13. Master Summary and Reference Formula Sheet#

Universal Formula Card#

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Important Examination Considerations#

1. Always compute mole ratios rather than mass ratios: Chemical reactions proceed by mole stoichiometry, not by raw gram quantities.
2. Distinguish solvent mass from total solution mass in Molality: Molality uses the mass of pure solvent in kilograms, not the mass of the overall solution.
3. Account for thermal expansion with Molarity: Because liquid volumes expand with increasing temperature, molarity decreases under heating.
4. Identify non-significant leading zeros: Values such as $0.0025$ possess only two significant figures.
5. Exact counted quantities have infinite significant figures: Pure counting numbers do not restrict the precision of a calculated result.