UNIT_5

Unit 5: Moles and Stoichiometry

5.01: Mole Concepts and Molar Mass

Learning Objectives:
- Express and manipulate chemical quantities using scientific conventions and mathematical procedures, including dimensional analysis, scientific notation, and significant figures.
- Analyze, evaluate, and critique scientific explanations.
- Classify matter as pure substances or mixtures through investigation of their properties.
- Define and use the concept of the mole.
- Calculate the number of atoms or molecules in a sample of material using Avogadro’s number.
The Mole:
- The mole (mol) is the SI unit for measuring the amount of a substance.
- It is a counting unit, similar to a "dozen," but for much larger quantities.
- 1 mole = $6.022 × 10^{23}$ representative particles (atoms, molecules, etc.). This is Avogadro's number.
- This number is exactly the number of atoms in 12 g of carbon-12.
Representative Particles:
- Element: Atoms or molecules (e.g., Ne, F2).
- Covalent Compound: Molecules (e.g., H2O).
- Ionic Compound: Formula units (e.g., NaCl).
- Ion: Charged atoms (e.g., $Al^{+3}$ , Cl^{-}$).
Ions vs. Atoms:
- Atoms have an equal number of protons and electrons, resulting in no electrical charge.
- Ions are atoms that have gained or lost electrons, resulting in a charge.
- Positive charge: Lost electrons (e.g., Al^{+3} $has lost 3 electrons).</li> <li>Negative charge: Gained electrons (e.g.,$ Cl^{-} $has gained 1 electron).</li> <li>The mass difference between an atom and its ion is negligible because electrons have insignificant mass (approximately$ 9.1 × 10^{-31} $kg) compared to protons (approximately$ 1.7 × 10^{-27} $kg).</li></ul></li> <li>Mole Analogy:<ul> <li>Just as a dozen always means 12, a mole always means$ 6.022 × 10^{23} $.</li> <li>Examples:<ul> <li>$ 6.022 × 10^{23} $watermelon seeds would make a melon larger than the moon.</li> <li>$ 6.022 × 10^{23} pennies would make stacks reaching the moon.

Counting Units & Conversion Factors:

Counting units like “ream” (500 sheets) and “mole” facilitate easier counting by grouping items into larger, discrete parts.
Example:
- 2 molecules of C6H{12}O_6 $: 2 x 6 = 12 C atoms, 2 x 12 = 24 H atoms, 2 x 6 = 12 O atoms, totaling 48 atoms.</li> <li>Similarly, 2 moles of$ C6H{12}O_6 $contains 12 moles of C atoms, 24 moles of H atoms, and 12 moles of O atoms, totaling 48 moles of atoms.</li></ul></li></ul></li> <li>Ions in Compounds:<ul> <li>Treat ions like atoms when counting.</li> <li>5 formula units of$ Fe2O3 $contain 10$ Fe^{3+} $ions and 15$ O^{2-} $ions. It also contains 10 Fe atoms and 15 O atoms.</li></ul></li> <li>Molar Mass:<ul> <li>The mass of one mole of a substance.</li> <li>It is an intensive physical property.</li> <li>The molar mass of an element is found on the periodic table (g/mol) and equals the average atomic mass (amu).</li> <li>Different elements have different molar masses due to the different masses of their atoms (different numbers of nucleons).</li> <li>A mole of iron and a mole of carbon have the same number of atoms ($ 6.022 × 10^{23} $), but different masses.</li></ul></li> <li>Calculating Molar Mass of Compounds:<ul> <li>Add up the molar masses of the elements in the compound, multiplying each element's molar mass by its subscript.</li></ul></li> </ul> <h4 id="502moleconversions">5.02: Mole Conversions</h4> <ul> <li>Learning Objectives:<ul> <li>Plan and implement investigative procedures</li> <li>Collect data and make measurements with accuracy and precision</li> <li>Express and manipulate chemical quantities using scientific conventions and mathematical procedures, including dimensional analysis, scientific notation, and significant figures</li> <li>Define and use the concept of a mole</li> <li>Calculate the number of atoms or molecules in a sample of material using Avogadro’s number</li></ul></li> <li>Determining Avogadro’s Number:<ul> <li>Robert Millikan measured the charge of an electron in Coulombs.</li> <li>Dividing the charge of a mole of electrons by the charge of a single electron yields Avogadro’s number ($ 6.022 × 10^{23} $).</li> <li>This connects the mole concept to the number of particles.</li></ul></li> <li>Mole Conversions:<ul> <li>The mole is the counting unit in chemistry.</li> <li>1 mole =$ 6.022 × 10^{23} $representative particles (atoms, ions, formula units, or molecules).</li> <li>$ N_A $is the symbol for Avogadro’s number.</li> <li>It’s the number of$ ^{12}C $atoms in exactly 12 g of the C-12 isotope.</li> <li>It is also the number of Cu atoms in 63.55 g of copper, the number of water molecules in 18.02 g of water, and the number of formula units in 58.44 g of sodium chloride, NaCl.</li> <li>Scientists convert between the number of particles, the number of moles, and mass.</li></ul></li> <li>Mole Map:<ul> <li>The mole map helps to visualize the steps for mole conversions.</li> <li>1 mole =$ 6.022 × 10^{23} $atoms, ions, molecules, or formula units.</li></ul></li> <li>Example 1: Moles to Molecules<ul> <li>Problem: How many molecules are in 4.75 moles of sulfur dioxide?</li> <li>Dimensional Analysis:<ul> <li>$ 4.75 \text{ mol } SO2 × \frac{6.022 ×10^{23} \text{ molecules } SO2}{1 \text{ mol } SO2} = 2.86 × 10^{24} \text{ molecules of } SO2 $</li></ul></li> <li>Proportion Method:<ul> <li>$ \frac{4.75 \text{ moles of } SO2}{x} = \frac{1 \text{ mole of } SO2}{6.022 × 10^{23} \text{ molecules of } SO_2} $</li> <li>$ x = 2.86 × 10^{24} \text{ molecules } SO_2 $</li></ul></li></ul></li> <li>Example 2: Moles to Atoms<ul> <li>Problem: How many atoms are in 2.6 moles of copper?</li> <li>Dimensional Analysis:<ul> <li>$ 2.6 \text{ mol Cu } × \frac{6.022 ×10^{23} \text{ atoms } Cu}{1 \text{ mol } Cu} = 1.6 × 10^{24} \text{ atoms of Cu} $</li></ul></li> <li>Proportion Method:<ul> <li>$ \frac{2.6 \text{ mol Cu}}{x \text{ atoms Cu}} = \frac{1 \text{ mol of Cu}}{6.022 × 10^{23} \text{ atoms of Cu}} $</li> <li>$ x = 1.6 × 10^{24} \text{ atoms } Cu $</li></ul></li></ul></li> <li>Example 3: Formula Units to Atoms<ul> <li>Problem: How many S atoms are in$ 1.45 × 10^{23} $formula units of iron (III) sulfate?</li> <li>The formula for iron (III) sulfate is$ Fe2(SO4)_3 $.</li> <li>Dimensional Analysis:<ul> <li>$ 1.45 × 10^{23} \text{ formula units of } Fe2(SO4)3 × \frac{3 \text{ S atoms}}{1 \text{ formula unit } Fe2(SO4)3} = 4.35 × 10^{23} \text{ S atoms} $</li></ul></li> <li>Proportion Method:<ul> <li>$ \frac{1.45 × 10^{23} \text{ formula units of } Fe2(SO4)3}{x \text{ atoms of S}} = \frac{1 \text{ formula unit of } Fe2(SO4)3}{3 \text{ atoms of S}} $</li> <li>$ x = 4.35 × 10^{23} \text{ atoms } S $</li></ul></li></ul></li> <li>Example 4: Mass to Moles<ul> <li>Problem: How many moles are in 50.0 g of copper (II) chloride?</li> <li>The chemical formula for copper (II) chloride is$ CuCl_2 $.</li> <li>Molar mass of$ CuCl_2 $= 134.446 g/mol.</li> <li>Dimensional Analysis:<ul> <li>$ 50.0 \text{ g } CuCl2 × \frac{1 \text{ mol } CuCl2}{134.446 \text{ g } CuCl2} = 0.372 \text{ mol } CuCl2 $</li></ul></li> <li>Proportion Method:<ul> <li>$ \frac{50.0 \text{ g } CuCl2}{x} = \frac{134.446 \text{ g } CuCl2}{1 \text{ mol } CuCl_2} $</li> <li>$ x = 0.372 \text{ mol } CuCl_2 $</li></ul></li></ul></li> <li>Example 5: Multi-Step Conversion (Mass to Atoms)<ul> <li>Problem: Calculate the number of hydrogen atoms in 100.0 g of glucose,$ C6H{12}O_6 $.</li> <li>Molar mass of$ C6H{12}O_6 $= 180.156 g/mol.</li> <li>Dimensional Analysis:<ul> <li>$ 100.0 \text{ g } C6H{12}O6 × \frac{1 \text{ mol } C6H{12}O6}{180.156 \text{ g } C6H{12}O6} × \frac{6.022 × 10^{23} \text{ molecules } C6H{12}O6}{1 \text{ mol } C6H{12}O6} × \frac{12 \text{ H atoms}}{1 \text{ molecule } C6H{12}O6} = 4.011 × 10^{24} \text{ H atoms} $</li></ul></li></ul></li> <li>Important Notes:<ul> <li>Maintain metric conversions in dimensional analysis.</li> <li>A 1:1 ratio is a mole conversion.</li> <li>For mixtures, calculate moles of atoms for each component separately and then add them.</li></ul></li> </ul> <h4 id="503percentcompositionempiricalandmolecularformulas">5.03: Percent Composition, Empirical, and Molecular Formulas</h4> <ul> <li>Learning Objectives: <ul> <li>Plan and implement investigative procedures.</li> <li>Collect data and make measurements with accuracy and precision.</li> <li>Express and manipulate chemical quantities using scientific conventions and mathematical procedures, including dimensional analysis, scientific notation, and significant figures.</li> <li>Organize, analyze, evaluate, make inferences, and predict trends from data.</li> <li>Communicate valid conclusions supported by the data.</li> <li>Communicate and apply scientific information extracted from various sources.</li> <li>Calculate the percent composition of compounds.</li> <li>Differentiate between empirical and molecular formulas.</li></ul></li> <li>Percent Composition: <ul> <li>Joseph Proust’s law of definite proportions states that the percentages of elements by mass in a compound are always the same.</li> <li>The percent composition is always the same, regardless of how a sample is prepared.</li></ul></li> <li>Percent Composition Calculations: <ul> <li>Formula:</li> <li>$ \text{% composition of element} = \frac{\text{mass of element in one mole of compound}}{\text{molar mass of compound}} × 100\% $</li></ul></li> <li>Example: Calculate the percent composition of potassium in potassium oxide ($ K_2O $). <ul> <li>$ \text{% comp of K in } K_2O = \frac{(2 \text{ mol } × 39.098 \frac{g}{mol})}{((2 \text{ mol } × 39.098 \frac{g}{mol}) + (1 \text{ mol } × 15.999 \frac{g}{mol}))} × 100\% = 83.015\% $</li></ul></li> <li>Empirical and Molecular Formulas: <ul> <li>Molecular Formula:<ul> <li>The actual number of each element in a formula (e.g.,$ C6H{12}O_6 $for glucose).</li></ul></li> <li>Empirical Formula:<ul> <li>The simplest positive integer ratio of atoms in a compound (e.g.,$ CH_2O $for glucose).</li></ul></li> <li>Different compounds can have the same empirical formula but are not the same compound.</li></ul></li> <li>Examples: <ul> <li>CO (Molecular Formula) and CO (Empirical Formula) - Carbon Monoxide</li> <li>$ C2H4O2 $(Molecular Formula) and$ CH2O $(Empirical Formula) - Vinegar</li> <li>$ CH2O $(Molecular Formula) and$ CH2O $(Empirical Formula) - Formaldehyde</li> <li>$ C6H{12}O6 $(Molecular Formula) and$ CH2O $(Empirical Formula) - Glucose</li></ul></li> <li>Calculating Empirical Formula from Lab Data: <ul> <li>Step 1: Calculate the mass of each element.<ul> <li>If given percentages, assume 100 g of the substance and convert the percentages to grams.</li></ul></li> <li>Step 2: Convert to moles by dividing the mass in grams by the molar mass of the element.</li> <li>Step 3: Divide all mole values by the smallest mole value to obtain subscripts.</li> <li>Step 4: If any result is a decimal mixed number, multiply all values by a factor to make them whole numbers.</li> <li>Example: A 10.0 g sample containing only iron and oxygen contains 6.994 g iron. What is the empirical formula?<ul> <li>Mass of O = 10.0 g - 6.994 g = 3.006 g</li> <li>Moles of Fe =$ \frac{6.994 \text{ g Fe}}{55.845 \text{ g/mol Fe}} = 0.12523 \text{ mol Fe} $</li> <li>Moles of O =$ \frac{3.006 \text{ g O}}{15.999 \text{ g/mol O}} = 0.18788 \text{ mol O} $</li> <li>Divide by smallest (0.12523): Fe = 1, O = 1.5</li> <li>Multiply by 2 to get whole numbers: Fe2O3</li></ul></li></ul></li> <li>Calculating Molecular Formula: <ul> <li>You need the empirical formula and the molar mass of the molecular compound.</li> <li>Step 1: Calculate the molar mass of the empirical formula (EF).</li> <li>Step 2: Divide the known molar mass of the molecular formula (MF) by the molar mass of the empirical formula to get a multiplier, n.</li> <li>Step 3: Multiply the subscripts of the empirical formula by n to get the subscripts of the molecular formula.</li> <li>Example: The molar mass of a compound is 181.50 g/mol, and the empirical formula is$ C_2HCl $. What is the molecular formula?<ul> <li>Molar Mass of EF = (2 x 12.011) + (1 x 1.008) + (1 x 35.45) = 60.48 g/mol</li> <li>$ n = \frac{181.50 \text{ g/mol}}{60.48 \text{ g/mol}} = 3 $</li> <li>$ C{2×3}H{1×3}Cl{1×3} = C6H3Cl3

5.04: Hydrate Calculations

Learning Objectives:
- Communicate and apply scientific information extracted from various sources.
- Express and manipulate chemical quantities using scientific conventions and mathematical procedures, including dimensional analysis, scientific notation, and significant figures.
- Define and use the concept of the mole.
- Calculate the number of atoms or molecules in a sample of material using Avogadro's number.
- Calculate the percent composition of compounds.
- Differentiate between empirical and molecular formulas.
Properties of Hydrates:
- Hydrates are ionic compounds with water molecules trapped within their crystal lattice structure.
- Anhydrous ionic compounds do not contain water.
- Water can be removed from a hydrate by heating.
- The mass of water is determined by subtracting the mass of the anhydrous crystal from the mass of the hydrate.
Determining the Formula & Molar/Formula Weight of a Hydrate
- The formula for the hydrate indicates the number of molecules of water in the structure
- In one formula unit of CuSO4 \cdot 5H2O $, 1$ CuSO4 $and 5$ H2O $are present; in one mole, 1 mole of$ CuSO4 $and 5 moles of$ H2O $are present</li></ul></li> <li>Molar Mass of Hydrates:<ul> <li>Add the molar masses of the elements in the first compound, multiplying the molar mass by its subscripts.</li> <li>Add the molar mass of the elements in the water molecules, also accounting for subscripts</li> <li>Multiply the coefficient in front of the water by the molar mass of water.</li> <li>Add the mass of water to the mass of the original compound to get the total molar mass.</li></ul></li> <li>Example: What is the molar mass of copper(II) sulfate pentahydrate ($ CuSO4 \cdot 5H2O $)?<ul> <li>Molar mass of$ CuSO_4 $= (1 x 63.546) + (1 x 32.06) + (4 x 15.999) = 159.60 g/mol</li> <li>Molar mass of$ 5H_2O $= 5 x [(2 x 1.008) + (1 x 15.999)] = 5 x 18.015 = 90.075 g/mol</li> <li>Molar mass of$ CuSO4 \cdot 5H2O = 159.60 + 90.075 = 249.68 g/mol
Determining a Hydrate Formula Using Experimental Data:
- Measure the mass of the hydrate sample both before & after heating.
- Calculate how many grams of water are present in the sample by subtracting the mass after heating from the mass before heating.
- Divide the number of moles of water by the number of moles of the anhydrous compound to calculate the moles of hydrations.
- The chemical formula is written as the chemical formula of the anhydrous compound, then a raised dot, then the number of water hydrations.
Example: A 15.67 g sample of magnesium carbonate hydrate was heated with mass reduced to 7.58 g. What is the formula and name of the hydrate?
- Mass of H_2O $= 15.67 g - 7.58 g = 8.09 g</li> <li>Moles of$ H_2O $=$ \frac{8.09 \text{ g}}{18.015 \text{ g/mol}} = 0.44907 \text{ mol} $</li> <li>Moles of$ MgCO_3 $=$ \frac{7.58 \text{ g}}{84.313 \text{ g/mol}} = 0.089903 \text{ mol} $</li> <li>Ratio of$ H2O $to$ MgCO3 = \frac{0.44907}{0.089903} \approx 5 $</li> <li>Chemical formula:$ MgCO3 \cdot 5H2O $</li></ul></li> <li>Mass Percent of Water:<ul> <li>$ \text{mass % of water} = (\frac{\text{mass of water}}{\text{mass of original sample}}) × 100 $</li> <li>$ \text{mass % of water} = (\frac{8.09 \text{ g } H2O}{15.67 \text{ g } MgCO3 \cdot 5 H_2O}) × 100 = 51.6\% $</li></ul></li> </ul> <h4 id="505stoichiometrythebasics">5.05: Stoichiometry: The Basics</h4> <ul> <li>Learning Objectives: <ul> <li>Communicate and apply scientific information extracted from various sources.</li> <li>Express and manipulate chemical quantities using scientific conventions and mathematical procedures, including dimensional analysis, scientific notation, and significant figures.</li> <li>Calculate the number of atoms or molecules in a sample of material using Avogadro's number.</li> <li>Perform stoichiometric calculations, including determination of mass and gas volume relationships between reactants and products and percent yield.</li></ul></li> <li>Stoichiometry: <ul> <li>Stoichiometry is the study of quantities in chemical reactions.</li> <li>It makes predictions on how much product is made when a specific amount of reactants are used.</li></ul></li> <li>Mole Ratio: <ul> <li>A chemical reaction is like a recipe, and the ratio of reactants and products is important to determine how much product will be made.</li> <li>The ratio is always in terms of moles in a chemical reaction.</li> <li>The numbers for mole ratios come from the coefficients in the balanced chemical equation.</li></ul></li> <li>Example Chemical Equation: 2 Na (s) + 2 H2O(l) → 2 NaOH(aq) + H2 (g) <ul> <li>2 Na atoms react with 2 H2O molecules to yield 2 NaOH formula units and 1 H2 molecule</li> <li>2 dozen Na atoms react with 2 dozen H2O molecules to yield 2 dozen NaOH formula units and 1 H2 molecules</li> <li>2 moles of Na atoms react with 2 moles of H2O molecules to yield 2 moles NaOH formula units and 1 mole H2 molecules</li></ul></li> <li>Other Examples N2 (g) + 3 H2 (g) → 2 NH3 (g) <ul> <li>Mole ratios:</li> <li>Na: H2O = 1:1</li> <li>H2O: H2 = 2:1</li></ul></li> <li>Stoichiometric map: <ul> <li>Determine the given quantity and its unit first. That determines where you start on the map.</li></ul></li> <li>Stoichiometry with Copper and Silver Nitrate Using equation Cu (s) + 2 AgNO3 (aq) → 2 Ag (s) + Cu(NO3)2 (aq) <ul> <li>Example 1: How many moles of copper are needed to react with excess silver nitrate to produce 0.500 moles of silver?<ul> <li>Dimensional analysis:<ul> <li>0. 500 mol of Ag * 1 mol of Cu / 2 mol of Ag = 0.250 mol of Cu</li></ul></li> <li>Proportion Method<ul> <li>0. 500 mol of Ag / 2 mol of Ag = x / 1 mol of Cu<ul> <li>x = 0.250 mol of Cu</li></ul></li></ul></li></ul></li> <li>Example 2: Calculate the mass of sodium metal needed to react with 50.0 g of water using balanced equation 2 Na (s) + 2 HOH(l) → 2 NaOH(aq) + H2 (g)<ul> <li>Dimensional analysis:<ul> <li>50.0 g H2O * 1 mol H2O / 18.015 g H2O * 2 mol Na/ 2 mol H2O * 22.990 g Na / 1 mol Na = 63.8 g Na</li></ul></li> <li>Proportion Method<ul> <li><ol start="50"> <li>0 g H2O / (2 mol H2O)∙(18.015 g/ mol H2O) = x / (2 mol Na)∙(22.990 g /mol 𝑁𝑎)</li></ol><ul> <li>X = 63.8 g of Na</li></ul></li></ul></li></ul></li> <li>Example 3: Calculate the mass of hydrogen that will be produced if you place 60.0 g of sodium in an excess amount of water.<ul> <li>Dimensional analysis:<ul> <li><ol start="60"> <li>0 g Na * 1 mol Na / 22.990 g Na * 1 mol of H2 / 2 mol of Na * 2.016 g H2 / 1 mol H2 = 2.63 g H2</li></ol></li></ul></li> <li>Proportion Method:<ul> <li>60.0 g Na / (2 mol Na) * (22.990 g /mol 𝑁𝑎) = x / (1 mol 𝐻2) * (2.016 g/ mol 𝐻2)<ul> <li>x = 2.63 g 𝐻2</li></ul></li></ul></li></ul></li></ul></li> </ul> <h4 id="506stoichiometrytheoreticalandactualyield">5.06: Stoichiometry: Theoretical and Actual Yield</h4> <ul> <li>Learning Objectives: <ul> <li>Communicate and apply scientific information extracted from various sources</li> <li>Express and manipulate chemical quantities using scientific conventions and mathematical procedures, including dimensional analysis, scientific notation, and significant figures</li> <li>Calculate the number of atoms or molecules in a sample of material using Avogadro’s number</li> <li>Differentiate between empirical and molecular formulas</li> <li>Perform stoichiometric calculations, including determination of mass and gas volume relationships between reactants and products and percent yield</li></ul></li> <li>Theoretical Yield: <ul> <li>The mass of product that should be made if a certain mass of reactant is put in.</li></ul></li> <li>Percent Yield: <ul> <li>The ratio of the actual yield to the theoretical yield and reflects the amount of error that was present.</li> <li>$ \text{Percent Yield} = \frac{\text{actual yield}}{\text{theoretical yield}} * 100$$
- Most common to have percent yields below 100%, but it can be above 100% as well
Example: A chemical company is mass producing ammonia (NH3). The chemists ran the reaction using 64.3 moles of hydrogen gas in the presence of excess nitrogen, with 130 g of ammonia being created. Calculate the percent yield of ammonia using the balanced equation 𝑁2 (𝑔) + 3 𝐻2 (𝑔) → 2 𝑁𝐻3 (𝑔)
- Dimensional analysis:
  - 1. 3 mol H2 * 2 mol NH3 / 3 mol H2 * 17.031 g NH3 / 1 mol NH3 = 730. g NH3
- Proportion method:
  - 1. 3 mol 𝐻2 / 3 mol 𝐻2 = x / (2 mol 𝑁𝐻3) (17.031 g/mol 𝑁𝐻3)
    - x = 730. g 𝑁𝐻3
- Percent Yield:
  - % 𝑌𝑖𝑒𝑙𝑑 = 130. g 𝑁𝐻3 / 730. g 𝑁𝐻3 = 0.1780 * 100 = 17.8%