Stoichiometry, Moles, and Chemical Calculations Study Guide

Concentration and the Mole

• Definition of Concentration:     - Concentration is defined as the amount of solute dissolved in a measured volume of solution.     - It can be expressed and measured in units of $g/dm^3$ or $mol/dm^3$.

• Converting Between Units (Extended Only):     - To convert between $g/dm^3$ and $mol/dm^3$, the mass must be converted to moles (or vice versa) using the relationship between mass, moles, and relative formula mass ($M_r$).     - The formula triangle involved includes $M_r$, amount in moles ($mol$), and mass ($g$).

• The Mole Concept:     - The unit for the amount of substance is the mole, abbreviated as $mol$.     - One mole contains exactly $6.02 \times 10^{23}$ particles (e.g., atoms, ions, or molecules).     - This specific number, $6.02 \times 10^{23}$, is termed the Avogadro constant (Extended only).

• Molar Mass vs. Relative Molecular Mass ($M_r$):     - Numerically, the molar mass is identical to the $M_r$ (relative molecular mass) or $A_r$ (relative atomic mass).     - The distinction lies in the units: molar mass has the unit $g/mol$, whereas $M_r$ is a unitless quantity.     - Example: The $M_r$ of calcium is $40$, but the molar mass of calcium is $40\,g/mol$.

• Fundamental Calculation Formula:     - The relationship between mass, moles, and molar mass is defined as: $\text{amount of substance (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$.

Basic Quantitative Relationships and Calculations

• (a) Calculating the Amount of Substance ($mol$):     - Formula: $\text{Amount of substance (mol)} = \text{Mass} \div M_r$     - Example: Calculate the amount of substance in $426\,g$ of sodium sulfate ($Na_2SO_4$).         - 1. Calculate the $M_r$ of $Na_2SO_4$: $(23 \times 2) + 32 + (16 \times 4) = 142$         - 2. Divide Mass by $M_r$: $426\,g \div 142 = 3\,moles$ of $Na_2SO_4$

• (b) Calculating Mass ($g$):     - Formula: $\text{Mass of a substance (g)} = M_r \times mol$     - Example: Calculate the mass of $0.5\,moles$ of sodium ($Na$).         - 1. Find the $M_r$ of $Na$ on the periodic table: $23$         - 2. Calculate: $23 \times 0.5 = 11.5\,g$

• (c) Calculating Molar Mass ($g/mol$):     - Formula: $\text{Molar mass} = \text{mass} \div mol$     - Example: Find the molar mass of $AlCl_3$ where $moles = 4$ and $mass = 534\,g$.         - 1. Calculate: $534\,g \div 4\,mol = 133.5\,g/mol$         - 2. To verify using periodic table atomic masses ($A_r$): $27 + (3 \times 35.5) = 133.5$

• (d) Calculating Relative Atomic Mass ($A_r$) or Relative Molecular/Formula Mass ($M_r$):     - Use the same process as finding molar mass ($\text{mass} \div mol$), but express the final result without units.     - Example: Find the $M_r$ of $CaCOH$ given $2\,moles$ and a mass of $138\,g$.         - 1. Divide: $138\,g \div 2\,moles = 69\,g/mol$         - 2. Note: $M_r$ is unitless, so the $M_r$ of $CaCOH$ is $69$.         - 3. Verification via periodic table: $40 + 12 + 16 + 1 = 69$

Avogadro's Constant and Gas Volume Calculations

• (e) Calculating Number of Particles:     - The number of particles (atoms, molecules, or ions) is calculated using the Avogadro constant ($6.02 \times 10^{23}$) and the amount of substance (moles).     - Formula: $\text{Number of particles} = \text{Avogadro constant} \times \text{amount of substance}$     - Example 1: Molecules in $1.5\,moles$ of $CO_2$.         - $\text{Number of } CO_2 \text{ molecules} = 6.02 \times 10^{23} \times 1.5 = 9.03 \times 10^{23}$     - Example 2: Atoms in $1.5\,moles$ of $CO_2$ (Extended extra step).         - There are $3$ atoms in each $CO_2$ molecule ($2$ oxygen atoms and $1$ carbon atom).         - $\text{Number of atoms} = (6.02 \times 10^{23} \times 1.5) \times 3 = 2.71 \times 10^{24}\text{ atoms}$

• Molar Gas Volume (Extended Only):     - Equal amounts in moles of all gases occupy the same volume under identical conditions of temperature and pressure.     - At Room Temperature and Pressure (RTP: $20^{\circ}C$ and $1\text{ atmosphere pressure}$), the volume of $1\,mol$ of any gas is $24\,dm^3$.     - Formula Triangles for Gas Volume:         - $\text{Volume (dm}^3) = \text{moles} \times 24$         - $\text{Volume (cm}^3) = \text{moles} \times 24000$     - Example (Volume in $dm^3$): Calculate the volume of $0.75\,mol$ of oxygen at RTP.         - $\text{Volume} = 0.75 \times 24 = 18\,dm^3$     - Example (Volume in $cm^3$): Calculate the volume of $0.5\,mol$ of hydrogen at RTP.         - $\text{Volume} = 0.5 \times 24000 = 12000\,cm^3$

Advanced Stoichiometry and Limiting Reactants

• Calculating Stoichiometric Reacting Masses (Extended Only):     - Example: Calculate the mass of oxygen needed to react with $24\,g$ of magnesium to form magnesium oxide ($2Mg\text{ (s)} + O_2\text{ (g)} \rightarrow 2MgO\text{ (s)}$).         - 1. Find $M_r$ of oxygen ($O_2$): $16 \times 2 = 32$         - 2. Find $M_r$ of magnesium ($Mg$): $24$ (the coefficient $2$ is ignored for $M_r$).         - 3. Find moles of $Mg$: $24\,g \div 24 = 1\,mol$         - 4. Determine molar ratio from balancing numbers: $2\,mol\,Mg : 1\,mol\,O_2$. Therefore, $1\,mol\,Mg$ reacts with $0.5\,mol\,O_2$.         - 5. Calculate mass of oxygen: $32\,g/mol \times 0.5\,mol = 16\,g$

• Calculating Limiting Reactants (Extended Only):     - A reaction concludes when one reactant is entirely consumed; this is the "limiting reactant" and determines the product yield.     - The leftover reactant is said to be "in excess."     - Example: $0.96\,g$ of magnesium reacts with $2.19\,g$ of hydrochloric acid ($Mg + 2HCl \rightarrow MgCl_2 + H_2$). Find the mass of $MgCl_2$ formed.         - 1. Moles of reactants:             - $\text{Moles of } Mg = 0.96 \div 24 = 0.04\,mol$             - $\text{Moles of } HCl = 2.19 \div 36.5 = 0.06\,mol$         - 2. Determine limiting reactant using molar ratio ($Mg : HCl = 1:2$):             - $0.04\,mol$ of $Mg$ would require $0.08\,mol$ of $HCl$.             - Since only $0.06\,mol$ of $HCl$ is available, $HCl$ is the limiting reactant.         - 3. Determine moles of product using ratio of limiting reactant to product ($HCl : MgCl_2 = 2:1$):             - $\text{Moles of } MgCl_2 = 0.06 \div 2 = 0.03\,mol$         - 4. Calculate mass of product:             - $M_r\text{ of } MgCl_2 = 24 + (35.5 \times 2) = 95$             - $\text{Mass of } MgCl_2 = 0.03 \times 95 = 2.85\,g$

Solutions, Volume Conversions, and Titrations

• Converting Between $cm^3$ and $dm^3$:     - Because these are cubic units, the conversion factor is $1000$.     - Visualization: A cube measuring $1\,dm \times 1\,dm \times 1\,dm$ has a volume of $1\,dm^3$.     - Since $1\,dm = 10\,cm$, the same cube is $10\,cm \times 10\,cm \times 10\,cm = 1000\,cm^3$.     - Conversion rules:         - $cm^3 \to dm^3$: Divide by $1000$         - $dm^3 \to cm^3$: Multiply by $1000$

• Concentration and Solution Volume Calculations (Extended Only):     - Formula Triangle: $n = c \times v$ where:         - $n = \text{number of moles (mol)}$         - $c = \text{concentration (mol/dm}^3)$         - $v = \text{volume of solution (dm}^3)$     - Concentration units can also be denoted as $M$ (molar).

• Conversion between $g/dm^3$ and $mol/dm^3$:     - $mol/dm^3 \to g/dm^3$: Multiply by the $M_r$     - $g/dm^3 \to mol/dm^3$: Divide by the $M_r$

• Solution Example: $200\,cm^3$ of hydrochloric acid contains $0.25\,mol$ of hydrogen chloride. Calculate concentration in $mol/dm^3$ and $g/dm^3$.     - 1. Convert volume: $200\,cm^3 \div 1000 = 0.2\,dm^3$     - 2. Calculate $mol/dm^3$: $0.25 \div 0.2 = 1.25\,mol/dm^3$     - 3. Calculate $g/dm^3$: $M_r\text{ of } HCl = 36.5$. Conversion: $1.25 \times 36.5 = 45.625\,g/dm^3$

• Titration Calculations (Extended Only):     - Data from titrations are used to find moles, concentration, or volume using the $n = c \times v$ formula triangle.     - Example: $25\,cm^3$ of dilute $HCl$ is neutralized by $20\,cm^3$ of $0.5\,mol/dm^3\,NaOH$. Find the concentration of $HCl$.         - 1. Convert volumes: $25\,cm^3 = 0.025\,dm^3$; $20\,cm^3 = 0.020\,dm^3$         - 2. Moles of $NaOH$: $0.5 \times 0.02 = 0.01\,mol$         - 3. Reaction Ratio ($HCl + NaOH \rightarrow NaCl + H_2O$): $1:1$ ratio. Moles of $HCl = 0.01\,mol$.         - 4. Concentration of $HCl$: $0.01 \div 0.025 = 0.4\,mol/dm^3$

Determining Empirical and Molecular Formulae

• Definitions (Extended Only):     - Empirical Formula: The simplest whole number ratio of atoms of each element in a compound.     - Molecular Formula: The actual number of atoms of each element in a compound.

• Finding Empirical Formula Example 1 (Percentage mass):     - Data: $Al = 20.2\%$, $Cl = 79.8\%$     - 1. $A_r$: $Al = 27$, $Cl = 35.5$     - 2. Molar ratio: $Al = 20.2 \div 27 = 0.748$; $Cl = 79.8 \div 35.5 = 2.248$     - 3. Simplify ratio: $Al = 0.748 \div 0.748 = 1$; $Cl = 2.248 \div 0.748 = 3$     - Result: $AlCl_3$

• Finding Empirical Formula Example 2 (Mass values):     - Data: $Fe = 7.83\,g$, $O = 3.37\,g$     - 1. Moles: $Fe = 7.83 \div 56 = 0.14$; $O = 3.37 \div 16 = 0.21$     - 2. Divide by smallest ($0.14$): $Fe = 1$; $O = 1.5$     - 3. Multiply to get whole numbers ($x 2$): $Fe = 2$, $O = 3$     - Result: $Fe_2O_3$

• Finding Molecular Formula (Extended Only):     - Example: Empirical formula is $CH_2O$, $M_r$ is $180$.     - 1. Find $M_r$ of empirical formula: $12 + (2 \times 1) + 16 = 30$     - 2. Scale factor: $180 \div 30 = 6$     - 3. Multiply atoms: $C_{1\times 6}H_{2\times 6}O_{1\times 6} = C_6H_{12}O_6$

Quantitative Analysis: Yield, Composition, and Purity

• Percentage Yield (Extended Only):     - It is often impossible to reach a $100\%$ theoretical yield for three main reasons:         - 1. Reaction is reversible and may not go to completion.         - 2. Product is lost during separation from the mixture.         - 3. Reactants engage in unexpected side reactions.     - Formula: $\text{Percentage yield} = \frac{\text{Actual amount of product produced}}{\text{Theoretical amount of product possible}} \times 100$     - Example: $32\,g$ of $NaOH$ reacts with $HCl$ to produce $16.2\,g$ of $NaCl$.         - 1. Find $M_r$: $NaOH = 40$, $NaCl = 58.5$         - 2. Calculate Theoretical yield: $\text{Moles } NaOH = 32 \div 40 = 0.8\,mol$. Molar ratio is $1:1$, so $0.8\,mol\,NaCl$ is possible. $\text{Theoretical mass} = 0.8 \times 58.5 = 46.8\,g$         - 3. Calculate Percentage Yield: $\frac{16.2\,g}{46.8\,g} \times 100 = 34.6\%$

• Percentage Composition by Mass (Extended Only):     - Formula: $\text{Percentage mass} = \frac{\text{Total } A_r \text{ of the element}}{M_r \text{ of the compound}} \times 100$     - Example: Calculate percentage of $Mg$ in $MgCO_3$.         - 1. $M_r\text{ of compound} = 24 + 12 + (16 \times 3) = 84$         - 2. Total $A_r\text{ of } Mg = 24 \times 1 = 24$         - 3. Calculate: $\frac{24}{84} \times 100 = 28.6\%$

• Percentage Purity (Extended Only):     - Formula: $\text{Percentage purity} = \frac{\text{Mass of the pure substance}}{\text{Total mass of sample}} \times 100$     - Example: A solution of sodium chloride contains $0.64\,g$ of $NaCl$ in $100\,g$ of water.         - Calculate: $\frac{0.64\,g}{100\,g} \times 100 = 0.64\%$