Stoichiometry and Basic Concepts of Chemistry: Moles, Masses, and Formulae

Historical and Modern Standards for Atomic and Molecular Weight

• Evolution of Measurement Standards:     * Initially, atomic and molecular weights were measured on various different scales.     * The presently accepted standard was established by the International Union of Pure and Applied Chemistry (IUPAC) in 1961.     * The reference standard for comparison is the most stable isotope of carbon, known as Carbon-12 ($C^{12}$).

Definitions and Mathematical Expressions of Weight and Mass

• Atomic Weight:     * Definition: It is defined as the number of times an atom of an element is heavier than $\frac{1}{12}$th of the mass of one atom of $C^{12}$.     * Mathematical Formula:$\text{Atomic weight} = \frac{\text{Mass of one atom}}{\frac{1}{12} \times \text{Mass of one atom of } C^{12}}$

• Molecular Weight:     * Definition: It represents the number of times a molecule is heavier than $\frac{1}{12}$th of the mass of one atom of $C^{12}$.     * Mathematical Formula:$\text{Molecular weight} = \frac{\text{Mass of one molecule}}{\frac{1}{12} \times \text{Mass of one atom of } C^{12}}$     * Note on Units: Both atomic and molecular weights are unitless quantities. They are also termed relative atomic mass and relative molecular mass.

• Atomic Mass Unit (amu) or Unified Mass (u):     * With the development of mass spectroscopy, precise measurements of the mass of a single atom became possible.     * The value of $\frac{1}{12}$th of the mass of one atom of $C^{12}$ is defined as 1 amu.     * Exact Values:$1\,amu = 1.66 \times 10^{-24}\,g = 1.66 \times 10^{-27}\,kg$

• Atomic and Molecular Mass (Absolute Quantities):     * Representing atomic mass in amu depicts the actual mass of a single atom.     * $\text{Atomic mass} = \text{Atomic weight} \times amu$     * Representing molecular mass in amu depicts the actual mass of a single molecule.     * $\text{Molecular mass} = (\text{Molecular weight}) \times amu$     * Standard Examples:         * The atomic weight of F is 19; the atomic mass of F is $19\,amu$.         * The molecular weight of $F_2$ is 38; the molecular mass of $F_2$ is $38\,amu$.

Average Atomic and Molecular Masses of Mixtures

• Average Atomic Mass:     * Most elements are found in nature as mixtures of isotopes (atoms of the same element with different atomic masses).     * The atomic mass of any element is the average of all isotopic masses present in a sample.     * Formula:$\text{Average atomic mass} = \frac{\sum (\%\text{ abundance} \times \text{atomic mass})}{100}$     * Practice Problem (Try Yourself): Calculate the average atomic weight of $Fe$ if the abundance of isotopes $^{54}Fe$, $^{56}Fe$, and $^{57}Fe$ are $5\%$, $90\%$, and $5\%$ respectively.

• Average Molecular Weight of Gaseous Mixtures:     * Used for describing a mixture containing different gases.     * Formula:$\text{Average molecular weight} = \frac{\sum (\%\text{ abundance} \times \text{molecular wt. of constituent})}{100}$     * Practice Problem (Try Yourself): Calculate the average molecular weight of a mixture containing $25\%\,CH_4$ and $75\%\,SO_2$ by volume.

• Vapour Density of Gaseous Mixtures:     * Formula:$\text{Vapour density of gaseous mixture} = \frac{\text{Average molecular wt.}}{2}$

Avogadro's Hypothesis and Its Applications

• Hypothesis Statement:     * Equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of molecules.

• Applications (Knowledge Cloud):     1. It provides a method to determine the atomic weight of gaseous elements.     2. It establishes a relationship between vapour density and the molecular mass of substances.     3. It assists in determining the molecular formulae of gases and is crucial in gas analysis.

The Mole Concept and Molar Quantities

• Definition of One Mole:     * A mole is the amount of substance that contains as many entities (atoms, molecules, etc.) as there are atoms in exactly $12.00\,g$ of $C^{12}$.     * Avogadro's Number ($N_A$): The number of carbon atoms in $12\,g$ of $C^{12}$ is $6.022 \times 10^{23}$.         $1\,mol = 6.022 \times 10^{23}\text{ microscopic entities}$

• Molar Volume:     * $1\,mol = 22.4\,litres$ of gas at STP (Standard Temperature and Pressure).

• Calculated Properties of a Mole (Example $CO_2$):     * $1\text{ mole of } CO_2 = 44\,g$     * $1\text{ mole of } CO_2 = 22.4\,litres\text{ of gas at STP}$     * Contains $1\text{ mole of carbon atoms}$     * Contains $2\text{ moles of oxygen atoms}$     * Contains $6.022 \times 10^{23}\text{ molecules (entities)}$

• Fundamental Formulas for Calculating Moles ($n$):     1. Based on Mass: $n = \frac{\text{Weight (in gram)}}{\text{Molar mass}}$     2. Based on Particles: $n = \frac{\text{Number of molecules or entities}}{\text{Avogadro number}}$     3. Based on Volume (Gases): $n = \frac{\text{Volume of gas at STP (in litre)}}{22.4\text{ litre}}$

Molar Mass and Percentage Composition

• Molar Mass:     * The mass of one mole of a substance in grams is its molar mass.     * Numerically, it is equal to the atomic or molecular mass expressed in atomic mass units (u).     * Example: Molar mass of ammonia ($NH_3$) is $17.0\,g$.

• Gram Atomic Mass (Gram Atom):     * The atomic weight of an element expressed in grams.     * It represents the weight of $6.022 \times 10^{23}\text{ atoms}$.     * $1\,g\text{ atom of carbon} = 1\text{ mole atom of carbon} = 12\,g$.

• Gram Molecular Mass:     * The molecular mass of a compound expressed in grams.

• Percentage of an Element in a Compound:     * $\% \text{ by weight} = \frac{\text{wt. of element}}{\text{mol. wt. of compound}} \times 100$     * Example Case Studies:         1. Acetic Acid ($CH_3COOH$):             * $\%C = \frac{2 \times 12}{60} \times 100 = 40\%$             * $\%O = \frac{32}{60} \times 100 = 53.3\%$             * $\%H = \frac{4}{60} \times 100 = 6.66\%$         2. Calcium Carbonate ($CaCO_3$):             * Molecular mass = $40 + 12 + (16 \times 3) = 100\,u$             * $\%Ca = \frac{40}{100} \times 100 = 40\%$     * Practice Problem (Try Yourself): Calculate the percentage composition of carbon in methane ($CH_4$). Options: (1) 80% (2) 75% (3) 20% (4) 25%.

Chemical Formulae: Empirical and Molecular

• Empirical Formula (EF):     * Represents the smallest whole number ratio of the constituent atoms in a molecule.     * Different compounds may share the same empirical formula (e.g., $C_6H_{12}O_6$, $CH_3COOH$, and $HCHO$ all have the EF of $CH_2O$).

• Molecular Formula (MF):     * Represents the actual number of each individual atom present in a molecule.     * Relationship formulas:         * $\text{Molecular formula} = (\text{Empirical formula})_n$         * $\text{Molecular formula weight} = \text{Empirical formula weight} \times n$         * $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}}$     * Vapour Density Relationship:$2 \times \text{Vapour density} = \text{Molecular weight}$

• Mathematical Example for Empirical Formula (Example 3):     * Composition: Carbon 10.06%, Hydrogen 0.84%, Chlorine 89.10%.     * Conversion to EF:         1. $C: \frac{10.06}{12} = 0.84 \implies \text{Ratio } \frac{0.84}{0.84} = 1$         2. $H: \frac{0.84}{1} = 0.84 \implies \text{Ratio } \frac{0.84}{0.84} = 1$         3. $Cl: \frac{89.10}{35.5} = 2.50 \implies \text{Ratio } \frac{2.50}{0.84} = 2.97 \approx 3$     * Empirical Formula: $CHCl_3$

• Calculation of Molecular Formula (Example 4):     * Given: Empirical formula is $CH_2O$, Molecular weight is 180.     * Steps:         1. Empirical formula weight = $(12 + 2 + 16) = 30$.         2. $n = \frac{180}{30} = 6$.         3. Molecular formula = $(CH_2O)_6 = C_6H_{12}O_6$.

Conversions and the Gas Laws

• Memory Map for Molar Conversions:     * Mole to Mass (g): Multiply by molar mass.     * Mass (g) to Mole: Divide by molar mass.     * Mole to Number of Entities: Multiply by $6.022 \times 10^{23}$.     * Number of Entities to Mole: Divide by $6.022 \times 10^{23}$.     * Mole to Volume (Litres at STP): Multiply by $22.4\,litre$.     * Volume (Litres at STP) to Mole: Divide by $22.4\,litre$.

• Non-STP Volume Conversion:     * If gas conditions are not at STP, use the Combined Gas Law formula to convert volume:         $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$     * Conversion to STP ($V_2$):         $V_2 = \frac{P_1 V_1}{T_1} \times \frac{273}{1}$         (Where $P_2 = 1\,atm$ and $T_2 = 273\,K$).

Problems and Exercises

• Illustration 1:     * Question: 5 moles of $AB_2$ weigh $125 \times 10^3\,kg$ and 10 moles of $A_2B_2$ weigh $300 \times 10^3\,kg$. Determine the molar mass of A ($M_A$) and B ($M_B$) in $kg\,mol^{-1}$.     * Solution Logic:         1. $AB_2 = \frac{125}{5} = 25\,kg/mol$         2. $A_2B_2 = \frac{300}{10} = 30\,kg/mol$         3. Equation 1: $A + 2B = 25$         4. Equation 2: $2A + 2B = 30$         5. Subtracting Eq 1 from Eq 2: $A = 5$.         6. Solving for B: $5 + 2B = 25 \implies 2B = 20 \implies B = 10$.     * Answer: (3) $M_A = 5 \times 10^3$ and $M_B = 10 \times 10^3$.

• Calculation Examples (Example 1):     * (a) Weight of 1000 atoms of N:$n = \frac{1000}{6.022 \times 10^{23}}$$\text{Weight} = n \times 14\,g = 2.32 \times 10^{-20}\,g$     * (b) Weight of 500 ml $CH_4$ at STP:$n = \frac{500}{1000 \times 22.4} = \frac{1}{44.8}$$\text{Weight} = \frac{1}{44.8} \times 16\,g = 0.36\,g$     * (c) Weight of one molecule of $CH_4$:$\text{Weight} = \frac{16}{6.022 \times 10^{23}}\,g = 2.66 \times 10^{-23}\,g$

• Self-Test Exercise:     * Question 5: Number of carbon atoms present in $22\,g\,CO_2$ is:         (1) $6.02 \times 10^{23}$ (2) $3.01 \times 10^{23}$ (3) $3.01 \times 10^{-23}$ (4) $6.02 \times 10^{-23}$.         (Calculation Note: 22g is 0.5 mole, thus $0.5 \times N_A = 3.01 \times 10^{23}$).     * Question 6: Which of the following possess highest mass?         (1) 0.2 mol of $CO_2$ gas (2) 10.8 g of Ag (108) (3) $3.01 \times 10^{23}$ molecules of water (4) 20 g calcium.     * Question 7: Which has maximum number of atoms?         (1) 10.8 g of Ag (108) (2) 2.4 g of C (12) (3) 5.6 g of Fe (56) (4) 54 g of Al (27).     * Question 8: How many moles of magnesium phosphate, $Mg_3(PO_4)_2$, will contain 0.25 mole of oxygen atoms?