1766515026.pdf

# Mole concept and stoichiometry This topic introduces the fundamental concept of the mole, its relationship to molar mass, and the quantitative analysis of chemical reactions through stoichiometry, including determining empirical and molecular formulas [2](#page=2). ### 1.1 Fundamental Concepts and Definitions #### 1.1.1 Dalton's atomic theory Dalton's atomic theory, proposed in the early 19th century, laid the groundwork for modern chemistry with several key postulates: * Matter is composed of indivisible particles called atoms [3](#page=3). * All atoms of a specific element are identical in properties, including mass. Atoms of different elements possess different masses [3](#page=3) [5](#page=5). * Compounds are formed when atoms of different elements combine in fixed, whole-number ratios [3](#page=3) [4](#page=4). * Chemical reactions involve the rearrangement, creation, or destruction of atoms [3](#page=3) [4](#page=4). * **Incorrect Postulate:** The theory incorrectly suggested that atoms are indivisible and that all atoms of a given element have identical properties, including mass, which has been disproven by isotopes [5](#page=5). #### 1.1.2 The mole concept The mole is a fundamental unit in chemistry representing a specific quantity of a substance [6](#page=6). * **Definition:** One mole of any substance contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of the carbon-12 isotope [6](#page=6). * **Avogadro's Number ($N_A$):** This number, approximately $6.022 \times 10^{23}$ represents the number of entities in one mole [6](#page=6). * 1 mole of atoms = $6.022 \times 10^{23}$ atoms [6](#page=6). * 1 mole of molecules = $6.022 \times 10^{23}$ molecules [6](#page=6). * 1 mole of ions (e.g., Na$^+$) = $6.022 \times 10^{23}$ Na$^+$ ions [6](#page=6). #### 1.1.3 Atomic mass and molar mass * **Atomic Mass (or Relative Atomic Mass):** The atomic mass of an element is defined as the ratio of the mass of one atom of the element to 1/12th the mass of an atom of carbon-12. It is a dimensionless quantity [7](#page=7) [8](#page=8). * Relative Atomic Mass $= \frac{\text{Mass of one atom of an element}}{\frac{1}{12} \times \text{Mass of one atom of }^{12}\text{C}}$ [7](#page=7). * Examples of relative atomic masses: Na = 23, Ca = 40, Mg = 24, O = 16, H = 1 [8](#page=8). * **Molar Mass (MM):** The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g mol$^{-1}$) [10](#page=10) [9](#page=9). * For an element, the molar mass in grams is numerically equal to its atomic mass. For example, the molar mass of Oxygen (O), with an atomic mass of 16, is 16 g mol$^{-1}$ [9](#page=9). * For a molecule, the molar mass is the sum of the atomic masses of all atoms in the molecule. For example, the molar mass of water (H$_2$O) is $2 \times (\text{atomic mass of H}) + (\text{atomic mass of O}) = 2 + 16 = 18$ g mol$^{-1}$ [10](#page=10) [1](#page=1). ### 1.2 Calculations involving moles #### 1.2.1 Calculating the number of moles The number of moles ($n$) can be calculated using the given mass and molar mass: $$ n = \frac{\text{Given mass}}{\text{Molar mass}} $$ [11](#page=11) [12](#page=12) [13](#page=13). #### 1.2.2 Calculating the number of particles (atoms, molecules) * **Number of molecules:** $$ \text{Number of molecules} = n \times N_A $$ [11](#page=11) [13](#page=13) [25](#page=25). * **Number of atoms:** $$ \text{Number of atoms} = n \times N_A \times \text{atomicity} $$ [11](#page=11) [23](#page=23). Alternatively, $$ \text{Number of atoms} = \text{Number of molecules} \times \text{atomicity} $$ [11](#page=11). > **Tip:** Atomicity refers to the number of atoms present in one molecule of a substance (e.g., atomicity of O$_2$ is 2, O$_3$ is 3, and S$_8$ is 8) [11](#page=11). #### 1.2.3 Relationships between mass, moles, and particles The relationships can be summarized as follows: * Mass $\stackrel{\text{divide by MM}}{\longrightarrow}$ Moles $\stackrel{\text{multiply by } N_A}{\longrightarrow}$ Number of Particles [13](#page=13). * Number of Particles $\stackrel{\text{divide by } N_A}{\longrightarrow}$ Moles $\stackrel{\text{multiply by MM}}{\longrightarrow}$ Mass [13](#page=13). > **Example:** To find the number of atoms in 36g of H$_2$O: > 1. Molar mass of H$_2$O = 18 g/mol [10](#page=10). > 2. Number of moles of H$_2$O = $\frac{36 \text{ g}}{18 \text{ g/mol}} = 2$ moles [11](#page=11). > 3. Number of H$_2$O molecules = $2 \times N_A$ [11](#page=11). > 4. Atomicity of H$_2$O is 3 (2 Hydrogen + 1 Oxygen). > 5. Number of atoms = $2 \times N_A \times 3 = 6 N_A$ [11](#page=11). #### 1.2.4 Moles of gases at STP * At Standard Temperature and Pressure (STP), one mole of any ideal gas occupies a volume of 22.4 liters [21](#page=21) [22](#page=22) [25](#page=25). $$ n = \frac{\text{Volume of gas at STP (L)}}{22.4 \text{ L/mol}} $$ [25](#page=25). ### 1.3 Empirical and Molecular Formulas These formulas describe the composition of a chemical compound. #### 1.3.1 Empirical Formula (EF) The empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound [33](#page=33). #### 1.3.2 Molecular Formula (MF) The molecular formula represents the actual number of atoms of each element in one molecule of the compound [33](#page=33). #### 1.3.3 Relationship between MF and EF The molecular formula is always a whole-number multiple of the empirical formula: $$ \text{MF} = (\text{EF}) \times x $$ [35](#page=35). where $x$ is a positive integer, calculated as: $$ x = \frac{\text{Molecular mass of compound}}{\text{Empirical formula mass}} $$ [35](#page=35). #### 1.3.4 Determination of Empirical Formula The empirical formula can be determined from the percentage composition of elements in a compound by following these steps [34](#page=34): 1. Convert the percentage of each element to grams (assuming a total mass of 100g) [37](#page=37) [38](#page=38) [39](#page=39). 2. Convert the mass of each element to moles by dividing by its atomic mass [34](#page=34) [37](#page=37) [38](#page=38). 3. Find the simplest whole-number ratio of these moles by dividing each mole value by the smallest mole value obtained [34](#page=34) [38](#page=38). 4. If the ratios are not whole numbers, multiply them by a small integer (e.g., 2, 3, 4) to obtain whole numbers [36](#page=36). 5. The whole numbers obtained represent the subscripts in the empirical formula [33](#page=33) [36](#page=36). > **Example:** A compound contains 74% Carbon, 8.7% Hydrogen, and 17.3% Nitrogen by mass [37](#page=37). > 1. Assume 100g: 74g C, 8.7g H, 17.3g N. > 2. Moles: > * C: $\frac{74 \text{ g}}{12 \text{ g/mol}} \approx 6.17$ moles [38](#page=38). > * H: $\frac{8.7 \text{ g}}{1 \text{ g/mol}} = 8.7$ moles [38](#page=38). > * N: $\frac{17.3 \text{ g}}{14 \text{ g/mol}} \approx 1.24$ moles [38](#page=38). > 3. Simplest ratio: > * C: $\frac{6.17}{1.24} \approx 5$ [38](#page=38). > * H: $\frac{8.7}{1.24} \approx 7$ [38](#page=38). > * N: $\frac{1.24}{1.24} = 1$ [38](#page=38). > 4. Empirical Formula: C$_5$H$_7$N [37](#page=37). > 5. If the molecular mass is 162 g/mol [37](#page=37): > * Empirical formula mass = $5 + 7 + 1 = 60 + 7 + 14 = 81$ g/mol [12](#page=12) [14](#page=14) [1](#page=1) [37](#page=37). > * $x = \frac{162}{81} = 2$ [37](#page=37). > * Molecular Formula = (C$_5$H$_7$N)$_2$ = C$_{10}$H$_{14}$N$_2$ [37](#page=37). ### 1.4 Stoichiometry and Chemical Equations Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions [73](#page=73). #### 1.4.1 Balanced chemical equations A balanced chemical equation represents a chemical reaction where the number of atoms of each element is the same on both the reactant and product sides. This is in accordance with the law of conservation of mass. The coefficients in a balanced equation represent the molar ratios of reactants and products [2](#page=2) [3](#page=3) [73](#page=73). > **Example:** The reaction $2\text{A} + 3\text{B} \rightarrow 4\text{C} + 5\text{D}$ [73](#page=73). > This equation indicates that 2 moles of A react with 3 moles of B to produce 4 moles of C and 5 moles of D [73](#page=73). #### 1.4.2 Limiting Reagent The limiting reagent (or limiting reactant) is the reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed [75](#page=75) [76](#page=76) [79](#page=79). * **Identification:** To find the limiting reagent, calculate the moles of each reactant. Then, divide the moles of each reactant by its stoichiometric coefficient in the balanced equation. The reactant with the smallest resulting value is the limiting reagent [76](#page=76) [79](#page=79). > **Example:** In the reaction $2\text{A} + 3\text{B} \rightarrow 4\text{C}$ [75](#page=75). > If we have 10 moles of A and 30 moles of B: > * For A: $\frac{10 \text{ moles}}{2} = 5$ > * For B: $\frac{30 \text{ moles}}{3} = 10$ > Since 5 is smaller than 10, A is the limiting reagent [75](#page=75). #### 1.4.3 Calculations based on limiting reagent Once the limiting reagent is identified, all calculations for the amount of product formed are based on the amount of the limiting reagent [76](#page=76) [79](#page=79). * **Mass of product formed:** 1. Determine the moles of the limiting reagent. 2. Use the mole ratio from the balanced equation to find the moles of the product. 3. Convert moles of product to mass using its molar mass. > **Example:** In the reaction N$_2$(g) + 3H$_2$(g) $\rightarrow$ 2NH$_3$(g) [81](#page=81). > If 20g of N$_2$ (molar mass = 28 g/mol) and 5g of H$_2$ (molar mass = 2 g/mol) are reacted: > 1. Moles of N$_2$ = $\frac{20 \text{ g}}{28 \text{ g/mol}} \approx 0.71$ moles [81](#page=81). > 2. Moles of H$_2$ = $\frac{5 \text{ g}}{2 \text{ g/mol}} = 2.5$ moles [81](#page=81). > 3. Limiting Reagent: > * For N$_2$: $\frac{0.71}{1} = 0.71$ > * For H$_2$: $\frac{2.5}{3} \approx 0.83$ > N$_2$ is the limiting reagent [81](#page=81). > 4. Moles of NH$_3$ formed: From the equation, 1 mole of N$_2$ produces 2 moles of NH$_3$. > Moles of NH$_3$ = $0.71 \text{ moles N}_2 \times \frac{2 \text{ moles NH}_3}{1 \text{ mole N}_2} = 1.42$ moles [81](#page=81) [82](#page=82). #### 1.4.4 Percentage Yield The percentage yield is a measure of the efficiency of a chemical reaction, comparing the actual amount of product obtained to the theoretical maximum amount that could be formed [87](#page=87) [88](#page=88). $$ \text{% Yield} = \frac{\text{Observed yield}}{\text{Theoretical yield}} \times 100\% $$ [87](#page=87) [89](#page=89) [92](#page=92). * The theoretical yield is calculated based on the stoichiometry of the reaction and the limiting reagent [87](#page=87). * The observed yield is the actual amount of product measured experimentally [87](#page=87). > **Example:** If a reaction is expected to produce 20g of a product (theoretical yield) but only 10g is obtained (observed yield), the percentage yield is: > % Yield = $\frac{10 \text{ g}}{20 \text{ g}} \times 100\% = 50\%$ [87](#page=87). #### 1.4.5 Sequential and Parallel Reactions * **Sequential Reactions:** Reactions that occur one after another, where the product of one reaction becomes the reactant for the next. Calculations involve carrying over quantities from one step to the next . * **Parallel Reactions (Mixture Problems):** A single reactant can undergo multiple reactions simultaneously, or a mixture of reactants can lead to multiple products. The distribution of products often depends on reaction conditions or the relative amounts of reactants [100](#page=100) [98](#page=98) [99](#page=99). #### 1.4.6 Principle of Atom Conservation (POAC) The Principle of Atom Conservation states that in any chemical reaction, the total number of atoms of each element remains constant before and after the reaction, provided no atoms are lost or gained from the system. This principle is directly related to the law of conservation of mass . * **Application:** It can be used to calculate the amount of elements in reactants or products, especially in combustion analysis or when dealing with complex reactions where direct stoichiometric calculations might be tedious . $$ \text{Moles of atoms of element (reactant)} = \text{Moles of atoms of element (product)} $$ . $$ n_{\text{reactant}} \times \text{atomicity}_{\text{reactant}} = n_{\text{product}} \times \text{atomicity}_{\text{product}} $$ . > **Example:** In the combustion of an organic compound (OC) to produce CO$_2$ and H$_2$O : > Using POAC for Carbon: > Moles of C atoms in OC = Moles of C atoms in CO$_2$ > $$ \frac{\text{mass of OC}}{\text{Molar mass of OC}} \times \text{atomicity of C in OC} = \frac{\text{mass of CO}_2}{\text{Molar mass of CO}_2} \times \text{atomicity of C in CO}_2 $$ > If the atomicity of C in OC is 1, and atomicity of C in CO$_2$ is 1: > $$ \frac{w_{\text{OC}}}{M_{\text{OC}}} = \frac{w_{\text{CO}_2}}{44} \times 1 $$ > This allows calculation of the mass of carbon in the organic compound. ### 1.5 Laws of Chemical Combination These laws describe the quantitative relationships in chemical reactions: * **Law of Conservation of Mass (Lavoisier):** Mass is neither created nor destroyed in a chemical reaction. Total mass of reactants equals total mass of products . * **Law of Constant Composition (Proust):** A chemical compound always contains the same elements in fixed proportions by mass, regardless of its source . * **Law of Multiple Proportions (Dalton):** When two elements combine to form two or more compounds, the masses of one element that combine with a fixed mass of the other element are in simple whole-number ratios . * **Law of Reciprocal Proportions (Richter):** If two different elements combine separately with the same mass of a third element, the ratio in which they combine with the third element is either the same or a simple multiple of the ratio in which they combine with each other . * **Gay-Lussac's Law of Gaseous Volumes:** When gases react or are produced in a chemical reaction, their volumes are in simple whole-number ratios, provided they are at the same temperature and pressure . --- # Ideal gas law and its applications This section delves into the ideal gas law, covering its foundational equation, necessary unit conversions, calculations involving gas density, and the definition of standard temperature and pressure (STP) conditions, along with practical problem-solving [14](#page=14). ### 2.1 The ideal gas equation The ideal gas law is a fundamental equation that describes the behavior of ideal gases. It relates the macroscopic properties of a gas: pressure (P), volume (V), the amount of gas in moles (n), and temperature (T) [14](#page=14). The equation is expressed as: $$PV = nRT$$ where: * $P$ represents the pressure of the gas [14](#page=14). * $V$ represents the volume of the gas [14](#page=14). * $n$ represents the number of moles of the gas [14](#page=14). * $R$ is the universal gas constant [14](#page=14). * $T$ represents the absolute temperature of the gas in Kelvin [14](#page=14). #### 2.1.1 Unit conversions for constants and units Accurate calculations using the ideal gas law require consistent units for all variables. Various units for pressure, volume, and temperature are commonly used, and their conversions are crucial. **Pressure Units:** * 1 atm = 1013.25 N/m$^2$ [15](#page=15). * 1 atm = 101325 Pa [15](#page=15). * 1 atm = 1.01325 bar [15](#page=15). * 1 atm = 760 torr [15](#page=15). * 1 atm = 76 cm Hg [15](#page=15). **Volume Units:** * 1 litre (l) = 1 dm$^3$ [15](#page=15). * 1 litre (l) = 1000 ml [15](#page=15). * 1 litre (l) = 1000 cm$^3$ (cc) [15](#page=15). * 1 litre (l) = 10$^{-3}$ m$^3$ [15](#page=15). * 1 m$^3$ = 1000 l [15](#page=15). * 12 = (10$^{-1}$ m)$^3$ [15](#page=15). * 1000 ml = 1000 cc = 1 dm$^3$ = 1 l [15](#page=15). **Temperature Units:** * The ideal gas law requires temperature to be in Kelvin (K). To convert from Celsius (°C) to Kelvin, add 273.15: $T(K) = T(°C) + 273.15$ [17](#page=17). * 0 °C = 273.15 K [17](#page=17). * 25 °C = 298.15 K [17](#page=17). #### 2.1.2 Values of the universal gas constant (R) The value of the universal gas constant, $R$, depends on the units used for pressure and volume. Common values for $R$ include: * $R$ = 0.0821 L atm K$^{-1}$ mol$^{-1}$ [16](#page=16). * $R$ = 8.314 J K$^{-1}$ mol$^{-1}$ [16](#page=16). * $R$ = 2 cal K$^{-1}$ mol$^{-1}$ [16](#page=16). > **Tip:** Always ensure the units of $R$ are consistent with the units used for pressure, volume, and temperature in your calculations. ### 2.2 Density of an ideal gas The ideal gas law can be rearranged to calculate the density ($\rho$ or $d$) of an ideal gas. Density is defined as mass per unit volume ($d = \frac{m}{V}$) [17](#page=17). Starting with $PV = nRT$ we can substitute $n = \frac{m}{M}$, where $m$ is the mass of the gas and $M$ is its molar mass [14](#page=14) [17](#page=17): $$PV = \left(\frac{m}{M}\right)RT$$ Rearranging for density ($d = \frac{m}{V}$): $$P = \frac{m}{V} \frac{RT}{M}$$ $$P = d \frac{RT}{M}$$ This leads to the formula for density: $$d = \frac{PM}{RT}$$ where: * $d$ is the density of the gas [17](#page=17). * $P$ is the pressure of the gas [17](#page=17). * $M$ is the molar mass of the gas [17](#page=17). * $R$ is the universal gas constant [17](#page=17). * $T$ is the absolute temperature in Kelvin [17](#page=17). > **Tip:** The density of a gas is directly proportional to its pressure and molar mass, and inversely proportional to its temperature. > **Example:** Calculate the density of CO$_2$ gas at 27°C and 1 atm. > Given: $P$ = 1 atm, $T$ = 27°C = 300 K, $M$ (CO$_2$) = 44 g/mol, $R$ = 0.0821 L atm K$^{-1}$ mol$^{-1}$. > $d = \frac{PM}{RT} = \frac{(1 \text{ atm})(44 \text{ g/mol})}{(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})(300 \text{ K})}$ > $d \approx 1.79 \text{ g/L}$ [18](#page=18). ### 2.3 Standard temperature and pressure (STP) conditions Standard Temperature and Pressure (STP) are reference conditions used for comparing gas properties. There are different conventions for STP. #### 2.3.1 Old convention for STP * Standard Temperature: 0 °C (273.15 K) [19](#page=19). * Standard Pressure: 1 atm [19](#page=19). Under the old STP conditions, one mole of any ideal gas occupies a volume of approximately 22.4 liters. This value can be derived from the ideal gas law: $$V = \frac{nRT}{P} = \frac{(1 \text{ mol})(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})(273.15 \text{ K})}{1 \text{ atm}} \approx 22.4 \text{ L}$$ [18](#page=18) [20](#page=20). #### 2.3.2 New convention for STP (IUPAC) The International Union of Pure and Applied Chemistry (IUPAC) has adopted a newer standard. * Standard Temperature: 0 °C (273.15 K) [19](#page=19). * Standard Pressure: 1 bar [19](#page=19). Note that 1 bar is slightly less than 1 atm (1 atm = 1.01325 bar) [15](#page=15). Under the new STP conditions, one mole of any ideal gas occupies a volume of approximately 22.7 liters. This can be calculated as: $$V = \frac{nRT}{P} = \frac{(1 \text{ mol})(8.314 \text{ J K}^{-1} \text{ mol}^{-1})(273.15 \text{ K})}{100000 \text{ Pa}} \approx 0.0227 \text{ m}^3 \approx 22.7 \text{ L}$$ [20](#page=20). #### 2.3.3 Volume calculations at STP The number of moles of an ideal gas ($n$) can be directly calculated from its volume ($V$) at STP: * Using the old convention (1 atm, 22.4 L/mol): $n = \frac{V(l)}{22.4}$ [20](#page=20). * Using the new IUPAC convention (1 bar, 22.7 L/mol): $n = \frac{V(l)}{22.7}$ [20](#page=20). > **Example:** If 3 moles of CO$_2$ gas are present at STP (old convention), what is their volume? > $V = n \times 22.4 \text{ L/mol} = 3 \text{ mol} \times 22.4 \text{ L/mol} = 67.2 \text{ L}$ [20](#page=20). > **Example:** How many moles of an ideal gas occupy 5 liters at STP (old convention)? > $n = \frac{5 \text{ L}}{22.4 \text{ L/mol}} \approx 0.223 \text{ mol}$ [20](#page=20). ### 2.4 Related problems and applications The ideal gas law is applied in numerous scenarios to determine unknown gas properties. > **Example:** Calculate the volume occupied by 36 grams of CO$_2$ gas at 0°C and 1 atm. > First, convert the mass of CO$_2$ to moles: > Molar mass of CO$_2$ ($M$) = 12 (C) + 2 * 16 (O) = 44 g/mol. > Number of moles ($n$) = $\frac{\text{mass}}{\text{molar mass}} = \frac{36 \text{ g}}{44 \text{ g/mol}} \approx 0.818 \text{ mol}$ [19](#page=19). > Now, use the ideal gas law: $PV = nRT$ > $P$ = 1 atm, $T$ = 0°C = 273.15 K, $R$ = 0.0821 L atm K$^{-1}$ mol$^{-1}$. > $(1 \text{ atm}) \times V = (0.818 \text{ mol}) \times (0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}) \times (273.15 \text{ K})$ > $V \approx 18.37 \text{ L}$ [19](#page=19). > **Tip:** Always pay close attention to whether the problem specifies the old or new STP convention if the volume of 1 mole is required. The slight difference can be significant in precise calculations. --- # Concentration terms of solutions This section details various methods for quantifying the concentration of solutions, including mole fraction, mass percentages, volume percentages, parts per million/billion, molarity, and molality, along with their temperature dependence [41-70. ### 3.1 Mole fraction ($\chi$) Mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in a solution. It is a dimensionless quantity [41](#page=41). For a solution with components A, B, and C, the mole fractions are expressed as: $$ \chi_A = \frac{n_A}{n_A + n_B + n_C} $$ $$ \chi_B = \frac{n_B}{n_A + n_B + n_C} $$ $$ \chi_C = \frac{n_C}{n_A + n_B + n_C} $$ The sum of mole fractions of all components in a solution is always equal to 1: $$ \sum \chi_i = 1 $$ [41](#page=41). **Example:** If a solution has 1 mole of urea and 9 moles of another component, the mole fraction of urea is: $$ \chi_{urea} = \frac{1}{1 + 9} = \frac{1}{10} = 0.1 $$ [41](#page=41). ### 3.2 Percentage by mass (% m/m) Percentage by mass, also known as % by mass or % weight/weight, represents the mass of the solute in grams divided by the total mass of the solution in grams, multiplied by 100 [43](#page=43). The formula is: $$ \% \frac{m}{m} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 100 $$ [43](#page=43). **Example:** A solution with 49 g of $H_2SO_4$ in 100 g of solution has a % m/m of: $$ \% \frac{m}{m} = \frac{49 \text{ g}}{100 \text{ g}} \times 100 = 49\% $$ [44](#page=44). In this case, the mass of the solvent (water) would be $100 \text{ g} - 49 \text{ g} = 51 \text{ g}$ [44](#page=44). ### 3.3 Percentage mass by volume (% m/v) Percentage mass by volume expresses the mass of the solute in grams per 100 milliliters of solution [45](#page=45). The formula is: $$ \% \frac{m}{v} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 $$ [45](#page=45). **Example:** A 25% m/v urea solution means there are 25 grams of urea in 100 mL of solution [46](#page=46). ### 3.4 Percentage by volume (% v/v) Percentage by volume indicates the volume of the solute in milliliters per 100 milliliters of solution [47](#page=47). The formula is: $$ \% \frac{v}{v} = \frac{\text{volume of solute (mL)}}{\text{volume of solution (mL)}} \times 100 $$ [47](#page=47). ### 3.5 Parts per million (ppm) Parts per million is used for very dilute solutions and represents the parts of solute per million parts of solution, typically by mass [49](#page=49). The formula is: $$ \text{ppm} = \frac{\text{parts of solute}}{\text{parts of solution}} \times 10^6 $$ Or, when expressed in mass: $$ \text{ppm} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^6 $$ [49](#page=49). ### 3.6 Parts per billion (ppb) Parts per billion is used for extremely dilute solutions and represents the parts of solute per billion parts of solution, typically by mass [50](#page=50). The formula is: $$ \text{ppb} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^9 $$ [50](#page=50). ### 3.7 Molarity (M) Molarity is defined as the number of moles of solute per liter of solution. It is expressed in units of moles per liter (mol/L) or molar (M) [51](#page=51). The formula is: $$ M = \frac{\text{moles of solute}}{\text{volume of solution (L)}} $$ [51](#page=51). **Example:** A 3M glucose solution means there are 3 moles of glucose dissolved in 1 liter of solution [52](#page=52). ### 3.8 Molality (m) Molality is defined as the number of moles of solute per kilogram of solvent. It is expressed in units of moles per kilogram (mol/kg) or molal (m) [53](#page=53). The formula is: $$ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} $$ [53](#page=53). **Example:** A 3 molal $HNO_3$ solution means there are 3 moles of $HNO_3$ dissolved in 1 kilogram (1000 g) of solvent [54](#page=54). ### 3.9 Temperature dependence of concentration terms The dependence of concentration terms on temperature varies based on whether the term involves volume in its definition [55](#page=55). * **Temperature Independent:** Concentration terms that depend on mass or moles are generally temperature independent. These include: * Percentage by mass (% m/m) [55](#page=55). * Mole fraction ($\chi$) [55](#page=55). * Parts per million (ppm) [55](#page=55). * Parts per billion (ppb) [55](#page=55). * Molality (m) [55](#page=55). * **Temperature Dependent:** Concentration terms that involve volume are temperature dependent because volume changes with temperature. These include [55](#page=55): * Percentage mass by volume (% m/v) [55](#page=55). * Percentage by volume (% v/v) [55](#page=55). * Molarity (M) [55](#page=55). **Tip:** When dealing with changes in temperature, remember that molarity will change, while molality will remain constant for the same amount of solute and solvent [55](#page=55). ### 3.10 Dilution of solutions Dilution involves adding more solvent to a solution, which decreases the concentration but does not change the number of moles of solute. The principle of conservation of moles is applied, leading to the equation $M_1V_1 = M_2V_2$, where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume, respectively [56](#page=56). $$ M_1V_1 = M_2V_2 $$ [56](#page=56). ### 3.11 Mixing of solutions When solutions of the same solute are mixed, the total number of moles of solute is the sum of the moles of solute in each individual solution. The final molarity can be calculated by dividing the total moles of solute by the total volume of the mixed solution [58](#page=58). The total number of moles of solute is $n_{total} = M_1V_1 + M_2V_2$ [58](#page=58). The total volume of the solution is $V_{total} = V_1 + V_2$ [58](#page=58). The final molarity ($M_{final}$) is: $$ M_{final} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2} $$ [58](#page=58). ### 3.12 Practice Problems and Examples * **Question:** The quantity which changes with temperature is: (a) Molarity (b) Mass percentage (c) Molality (d) Mole fraction [60](#page=60). * **Answer:** (a) Molarity [60](#page=60). * **Question:** A solution is prepared by adding 1 mole of ethyl alcohol in 9 moles of water. The mass percent of solute in the solution is (Given: Molar mass of ethyl alcohol = 46 g/mol, water = 18 g/mol) [61](#page=61) [62](#page=62). * Mass of solute (ethyl alcohol) = 1 mole $\times$ 46 g/mol = 46 g [61](#page=61) [62](#page=62). * Mass of solvent (water) = 9 moles $\times$ 18 g/mol = 162 g [61](#page=61) [62](#page=62). * Mass of solution = Mass of solute + Mass of solvent = 46 g + 162 g = 208 g [61](#page=61) [62](#page=62). * % m/m = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100 = \frac{46 \text{ g}}{208 \text{ g}} \times 100 \approx 22.11\%$, which rounds to 22% [61](#page=61) [62](#page=62). * **Question:** The mole fraction of a solute in a 100 molal aqueous solution is X. Calculate X. (Given: Atomic masses H: 1.0u, O: 16.0u) [63](#page=63) [64](#page=64). * 100 molal means 100 moles of solute in 1 kg (1000 g) of solvent (water) [63](#page=63). * Moles of water ($n_A$) = $\frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.55$ moles [63](#page=63). * Moles of solute ($n_B$) = 100 moles [63](#page=63). * Mole fraction of solute ($\chi_B$) = $\frac{n_B}{n_A + n_B} = \frac{100}{55.55 + 100} = \frac{100}{155.55} \approx 0.642$ [64](#page=64). * X = 64.2 $\times 10^{-2}$, which rounds to 64 [64](#page=64). * **Question:** The density of a 3 M solution of NaCl is 1.0 g/mL. Calculate the molality of the solution. (Molar mass of Na = 23 g/mol, Cl = 35.5 g/mol) [65](#page=65) [66](#page=66). * 3 M NaCl solution means 3 moles of NaCl in 1 liter (1000 mL) of solution [65](#page=65). * Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol [65](#page=65). * Mass of NaCl = 3 moles $\times$ 58.5 g/mol = 175.5 g [65](#page=65). * Mass of solution = Density $\times$ Volume = 1.0 g/mL $\times$ 1000 mL = 1000 g [65](#page=65). * Mass of solvent (water) = Mass of solution - Mass of solute = 1000 g - 175.5 g = 824.5 g = 0.8245 kg [65](#page=65). * Molality ($m$) = $\frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{3 \text{ moles}}{0.8245 \text{ kg}} \approx 3.64$ molal [66](#page=66). * Molality = 3.64 $\times 10^{-2}$ m (rounded to the nearest integer) [66](#page=66). * **Question:** 100 mL of a $Na_3PO_4$ solution contains 3.45 g of sodium. Calculate the molarity of the solution. (Atomic masses: Na: 23.0u, O: 16.0u, P: 31.0u) [67](#page=67) [68](#page=68). * The molar mass of $Na_3PO_4$ = 3(23.0) + 31.0 + 4(16.0) = 69.0 + 31.0 + 64.0 = 164 g/mol [67](#page=67). * Mass of Na in 100 mL solution = 3.45 g [67](#page=67). * Moles of Na = $\frac{3.45 \text{ g}}{23.0 \text{ g/mol}} = 0.15$ moles [67](#page=67). * In $Na_3PO_4$, there are 3 moles of Na per mole of $Na_3PO_4$. Therefore, moles of $Na_3PO_4$ = $\frac{\text{moles of Na}}{3} = \frac{0.15 \text{ moles}}{3} = 0.05$ moles [67](#page=67) [68](#page=68). * Volume of solution = 100 mL = 0.1 L [67](#page=67). * Molarity = $\frac{0.05 \text{ moles}}{0.1 \text{ L}} = 0.5$ M [68](#page=68). * Molarity = 50.0 $\times 10^{-2}$ M (rounded to the nearest integer) [68](#page=68). * **Question:** The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. The density of the solution is 1.2 g/cm³. Calculate the molarity of the urea solution. (Molar masses: urea = 60 g/mol, water = 18 g/mol) [69](#page=69) [70](#page=70). * Mole fraction of urea ($\chi_{urea}$) = 0.05 [69](#page=69) [70](#page=70). * Mass of water = 900 g [69](#page=69). * Moles of water ($n_{water}$) = $\frac{900 \text{ g}}{18 \text{ g/mol}} = 50$ moles [70](#page=70). * Let $n_{urea}$ be the moles of urea. $\chi_{urea} = \frac{n_{urea}}{n_{water} + n_{urea}}$ [70](#page=70). * $0.05 = \frac{n_{urea}}{50 + n_{urea}}$ [70](#page=70). * $0.05 \times (50 + n_{urea}) = n_{urea}$ [70](#page=70). * $2.5 + 0.05 n_{urea} = n_{urea}$ [70](#page=70). * $2.5 = 0.95 n_{urea}$ [70](#page=70). * $n_{urea} = \frac{2.5}{0.95} \approx 2.63$ moles [70](#page=70). * Mass of urea = $n_{urea} \times \text{Molar mass of urea} = 2.63 \text{ moles} \times 60 \text{ g/mol} \approx 157.8$ g [70](#page=70). * Mass of solution = Mass of urea + Mass of water = 157.8 g + 900 g = 1057.8 g [70](#page=70). * Density of solution = 1.2 g/mL [69](#page=69). * Volume of solution = $\frac{\text{Mass of solution}}{\text{Density of solution}} = \frac{1057.8 \text{ g}}{1.2 \text{ g/mL}} \approx 881.5$ mL = 0.8815 L [70](#page=70). * Molarity = $\frac{\text{moles of urea}}{\text{volume of solution (L)}} = \frac{2.63 \text{ moles}}{0.8815 \text{ L}} \approx 2.98$ M [70](#page=70). --- # Average atomic mass The average atomic mass is a weighted average of the masses of an element's isotopes, considering their natural abundance. ### 4.1 Isotopes and atomic mass Atoms of the same element can exist as isotopes, which have the same number of protons and electrons but different numbers of neutrons. This difference in neutron number leads to different atomic masses for each isotope [27](#page=27). ### 4.2 When is average atomic mass applicable? The concept of average atomic mass is primarily defined for elements that exist as two or more isotopes with comparable occurrences in nature. If an element has isotopes with non-comparable natural abundances (e.g., one isotope is extremely rare), the average atomic mass is not typically applicable or meaningful in the same way [29](#page=29). > **Tip:** Hydrogen is an example where isotopes exist with vastly different abundances (protium ~99%, deuterium and tritium are much less abundant), making a simple average atomic mass calculation based on all isotopes less common than for elements like chlorine or boron. ### 4.3 Calculating average atomic mass The average atomic mass is calculated as a weighted average. This means that the mass of each isotope is multiplied by its fractional abundance (percentage abundance divided by 100), and then these products are summed up [30](#page=30). The general formula for average atomic mass ($M_{avg}$) is: $$M_{avg} = \sum_{i=1}^{n} (x_i \cdot M_i)$$ Where: - $M_{avg}$ is the average atomic mass of the element. - $n$ is the number of isotopes of the element. - $x_i$ is the fractional abundance of the $i$-th isotope. - $M_i$ is the atomic mass of the $i$-th isotope. An alternative way to express this when considering two isotopes is: $$M_{avg} = (x_1 \cdot M_1) + (x_2 \cdot M_2)$$ where $x_1 + x_2 = 1$. If the abundances are given in percentages, the formula can be written as: $$M_{avg} = \frac{(\% \text{abundance}_1 \cdot M_1) + (\% \text{abundance}_2 \cdot M_2)}{100}$$ #### 4.3.1 Example calculation Consider an element that exists as two isotopes with atomic masses $M_1$ and $M_2$, and their natural abundances are 40% and 60% respectively. The average atomic mass would be: $$M_{avg} = \frac{(40 \cdot M_1) + (60 \cdot M_2)}{100}$$ [30](#page=30). Alternatively, if we consider 100 atoms, with 40 atoms of the first isotope and 60 atoms of the second: $$M_{avg} = \frac{(40 \times M_1) + (60 \times M_2)}{100}$$ [30](#page=30). ### 4.4 Working with isotopic ratios When the average atomic mass and the masses of the isotopes are known, one can determine the ratio of the isotopes in nature. #### 4.4.1 Example problem **Question:** The average molar mass of chlorine is 35.5 g mol$^{-1}$. The ratio of $^{35}$Cl to $^{37}$Cl in naturally occurring chlorine is close to: (a) 4:1 (b) 3:1 (c) 2:1 (d) 1:1 [31](#page=31). **Solution:** Let the fractional abundance of $^{35}$Cl be $x$. Then, the fractional abundance of $^{37}$Cl will be $(1-x)$ [32](#page=32). The average atomic mass ($M_{avg}$) is given by: $M_{avg} = (x \cdot M_{^{35}Cl}) + ((1-x) \cdot M_{^{37}Cl})$ [32](#page=32). Given $M_{avg} = 35.5$, $M_{^{35}Cl} = 35$, and $M_{^{37}Cl} = 37$: $35.5 = x + (1-x) $ [32](#page=32) . $35.5 = 35x + 37 - 37x$ [32](#page=32). $35.5 = 37 - 2x$ [32](#page=32). $2x = 37 - 35.5$ [32](#page=32). $2x = 1.5$ [32](#page=32). $x = \frac{1.5}{2} = 0.75$ [32](#page=32). The fractional abundance of $^{35}$Cl is 0.75. The fractional abundance of $^{37}$Cl is $1-x = 1 - 0.75 = 0.25$ [32](#page=32). The ratio of $^{35}$Cl to $^{37}$Cl is therefore: Ratio = $\frac{0.75}{0.25} = \frac{3}{1}$ or 3:1 [32](#page=32). The correct option is (b) 3:1. > **Tip:** Always double-check if the question asks for the ratio of isotopes or their fractional abundances. When working with percentages, ensure they are converted to fractional abundances (divide by 100) before using them in the weighted average formula, or adjust the formula accordingly as shown in the example. ### 4.5 Notation for isotopes Isotopes are often represented by their mass number (total number of protons and neutrons) as a superscript before the element symbol, e.g., $^{12}$C, $^{13}$C, $^{35}$Cl, $^{37}$Cl. The atomic number (number of protons) can also be included as a subscript, though it is often omitted as it is implied by the element symbol, e.g., $_{6}^{12}$C [27](#page=27) [29](#page=29). --- ## Common mistakes to avoid - Review all topics thoroughly before exams - Pay attention to formulas and key definitions - Practice with examples provided in each section - Don't memorize without understanding the underlying concepts

| Term | Definition | |------|------------| | Atom | The basic unit of a chemical element, consisting of a nucleus (protons and neutrons) and electrons orbiting the nucleus. | | Molecule | A group of two or more atoms held together by chemical bonds, forming the smallest particle of a substance that retains all the chemical and physical properties of the substance. | | Element | A pure substance consisting only of atoms that all have the same number of protons in their atomic nuclei. | | Compound | A substance formed when two or more chemical elements are chemically bonded together. | | Dalton's Atomic Theory | A scientific theory of the nature of matter, proposed by John Dalton in the early 19th century, which states that matter is composed of discrete units called atoms. | | Chemical Combination | The process by which atoms or molecules form chemical bonds to create larger structures like molecules or compounds. | | Atomic Mass | The mass of an atom, typically expressed in atomic mass units (amu). It is approximately the sum of protons and neutrons in the nucleus. | | Molecular Mass | The mass of a molecule, calculated by summing the atomic masses of all the atoms in the molecule. | | Mole Concept | A fundamental concept in chemistry that defines a standard quantity of a substance, equal to Avogadro's number of particles (atoms, molecules, ions, etc.). | | Molar Mass | The mass of one mole of a substance, usually expressed in grams per mole (g/mol). | | Percentage Composition | The percentage by mass of each element in a compound. | | Empirical Formula | The simplest whole-number ratio of atoms of each element present in a compound. | | Molecular Formula | The actual number of atoms of each element in one molecule of a compound. | | Chemical Equation | A symbolic representation of a chemical reaction, showing the reactants and products. | | Stoichiometry | The study of the quantitative relationships between reactants and products in chemical reactions. | | Redox Reaction | A type of chemical reaction that involves the transfer of electrons between two species; one species is oxidized (loses electrons) and another is reduced (gains electrons). | | Equilibrium | A state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in reactant or product concentrations. | | Mole Fraction | A concentration term defined as the ratio of the number of moles of a component to the total number of moles of all components in a solution. | | Molarity | A concentration term defined as the number of moles of solute per liter of solution (mol/L). | | Molality | A concentration term defined as the number of moles of solute per kilogram of solvent (mol/kg). | | Normality | A concentration term defined as the number of gram equivalents of solute per liter of solution. | | Ideal Gas | A theoretical gas composed of many non-interacting point particles that are the same but repel at infinitesimal distances and whose only state function is its kinetic energy. | | Pressure (P) | The force exerted per unit area, typically measured in atmospheres (atm), pascals (Pa), or torr. | | Volume (V) | The amount of space occupied by a substance, typically measured in liters (L) or cubic meters ($m^3$). | | Temperature (T) | A measure of the average kinetic energy of the particles in a system, typically measured in Kelvin (K) or Celsius (°C). | | Universal Gas Constant (R) | A physical constant that relates the energy scale to the temperature scale, with different values depending on the units used (e.g., 0.0821 L·atm/mol·K or 8.314 J/mol·K). | | Ideal Gas Equation | A mathematical relationship that describes the behavior of ideal gases: $PV = nRT$. | | Density (d) | The mass of a substance per unit volume, typically expressed in grams per liter (g/L) or kg/$m^3$. | | STP (Standard Temperature and Pressure) | A set of conditions used as a reference point for gas calculations. The "old" convention is 0°C (273.15 K) and 1 atm, while the "new" convention is 0°C (273.15 K) and 1 bar. | | Percentage by Mass (%m/m) | A concentration term defined as the mass of solute divided by the mass of solution, multiplied by 100. | | Mass by Volume Percentage (%m/v) | A concentration term defined as the mass of solute (in grams) divided by the volume of solution (in milliliters), multiplied by 100. | | Volume by Volume Percentage (%v/v) | A concentration term defined as the volume of solute divided by the volume of solution, multiplied by 100. | | Parts Per Million (ppm) | A concentration term used for very dilute solutions, defined as the mass of solute divided by the mass of solution, multiplied by $10^6$. | | Parts Per Billion (ppb) | A concentration term used for extremely dilute solutions, defined as the mass of solute divided by the mass of solution, multiplied by $10^9$. | | Dilution | The process of reducing the concentration of a solute in a solution, usually by adding more solvent. | | Mixing of Solutions | The process of combining solutions, where the final concentration depends on the initial concentrations, volumes, and the total volume after mixing. | | Isotope | Atoms of the same element that have the same number of protons but different numbers of neutrons. | | Average Atomic Mass | The weighted average of the atomic masses of all naturally occurring isotopes of an element, taking into account their relative abundances. | | Empirical Formula Mass | The sum of the atomic masses of the atoms in an empirical formula. | | Limiting Reagent | The reactant in a chemical reaction that is completely consumed first, thereby determining the maximum amount of product that can be formed. | | Percentage Yield | The ratio of the actual yield of a product to the theoretical yield, expressed as a percentage, indicating the efficiency of a chemical reaction. | | Stoichiometric Coefficients | The numbers in front of chemical formulas in a balanced chemical equation that indicate the relative number of moles of reactants and products. | | Principle of Atom Conservation (POAC) | A principle stating that the total number of atoms of each element remains constant in a chemical reaction; atoms are neither created nor destroyed. | | Sequential Reactions | A series of chemical reactions where the product of one reaction becomes the reactant for the next reaction. | | Parallel Reactions | Reactions where a single reactant can undergo multiple different reactions simultaneously. | | Law of Conservation of Mass | A fundamental law stating that in any closed system, mass is conserved over time; mass cannot be created or destroyed. | | Law of Constant Composition | A law stating that a given chemical compound always contains its component elements in fixed ratio (by mass) and thus has identical properties. | | Law of Multiple Proportion | When two elements combine to form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element are simple whole numbers. | | Law of Reciprocal Proportion | If element A combines with element B and element C separately, and the masses of B and C that combine with a fixed mass of A are in the ratio $x:y$, then the masses of B and C that combine with each other will be in the ratio $x:y$ or a simple multiple of it. | | Gay-Lussac's Law of Gaseous Volumes | At constant temperature and pressure, the volume of gaseous reactants and products in a chemical reaction are in simple whole number ratios. |