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).
- Negative charge: Gained electrons (e.g., Cl^{-} has gained 1 electron).
- 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).
Mole Analogy:
- Just as a dozen always means 12, a mole always means 6.022 × 10^{23}.
- Examples:
  - 6.022 × 10^{23} watermelon seeds would make a melon larger than the moon.
  - 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.
  - 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.
Ions in Compounds:
- Treat ions like atoms when counting.
- 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.
Molar Mass:
- The mass of one mole of a substance.
- It is an intensive physical property.
- The molar mass of an element is found on the periodic table (g/mol) and equals the average atomic mass (amu).
- Different elements have different molar masses due to the different masses of their atoms (different numbers of nucleons).
- A mole of iron and a mole of carbon have the same number of atoms (6.022 × 10^{23}), but different masses.
Calculating Molar Mass of Compounds:
- Add up the molar masses of the elements in the compound, multiplying each element's molar mass by its subscript.

5.02: Mole Conversions

Learning Objectives:
- Plan and implement investigative procedures
- Collect data and make measurements with accuracy and precision
- 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 a mole
- Calculate the number of atoms or molecules in a sample of material using Avogadro’s number
Determining Avogadro’s Number:
- Robert Millikan measured the charge of an electron in Coulombs.
- Dividing the charge of a mole of electrons by the charge of a single electron yields Avogadro’s number (6.022 × 10^{23}).
- This connects the mole concept to the number of particles.
Mole Conversions:
- The mole is the counting unit in chemistry.
- 1 mole = 6.022 × 10^{23} representative particles (atoms, ions, formula units, or molecules).
- N_A is the symbol for Avogadro’s number.
- It’s the number of ^{12}C atoms in exactly 12 g of the C-12 isotope.
- 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.
- Scientists convert between the number of particles, the number of moles, and mass.
Mole Map:
- The mole map helps to visualize the steps for mole conversions.
- 1 mole = 6.022 × 10^{23} atoms, ions, molecules, or formula units.
Example 1: Moles to Molecules
- Problem: How many molecules are in 4.75 moles of sulfur dioxide?
- Dimensional Analysis:
  - 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
- Proportion Method:
  - \frac{4.75 \text{ moles of } SO2}{x} = \frac{1 \text{ mole of } SO2}{6.022 × 10^{23} \text{ molecules of } SO_2}
  - x = 2.86 × 10^{24} \text{ molecules } SO_2
Example 2: Moles to Atoms
- Problem: How many atoms are in 2.6 moles of copper?
- Dimensional Analysis:
  - 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}
- Proportion Method:
  - \frac{2.6 \text{ mol Cu}}{x \text{ atoms Cu}} = \frac{1 \text{ mol of Cu}}{6.022 × 10^{23} \text{ atoms of Cu}}
  - x = 1.6 × 10^{24} \text{ atoms } Cu
Example 3: Formula Units to Atoms
- Problem: How many S atoms are in 1.45 × 10^{23} formula units of iron (III) sulfate?
- The formula for iron (III) sulfate is Fe2(SO4)_3.
- Dimensional Analysis:
  - 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}
- Proportion Method:
  - \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}}
  - x = 4.35 × 10^{23} \text{ atoms } S
Example 4: Mass to Moles
- Problem: How many moles are in 50.0 g of copper (II) chloride?
- The chemical formula for copper (II) chloride is CuCl_2.
- Molar mass of CuCl_2 = 134.446 g/mol.
- Dimensional Analysis:
  - 50.0 \text{ g } CuCl2 × \frac{1 \text{ mol } CuCl2}{134.446 \text{ g } CuCl2} = 0.372 \text{ mol } CuCl2
- Proportion Method:
  - \frac{50.0 \text{ g } CuCl2}{x} = \frac{134.446 \text{ g } CuCl2}{1 \text{ mol } CuCl_2}
  - x = 0.372 \text{ mol } CuCl_2
Example 5: Multi-Step Conversion (Mass to Atoms)
- Problem: Calculate the number of hydrogen atoms in 100.0 g of glucose, C6H{12}O_6.
- Molar mass of C6H{12}O_6 = 180.156 g/mol.
- Dimensional Analysis:
  - 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}
Important Notes:
- Maintain metric conversions in dimensional analysis.
- A 1:1 ratio is a mole conversion.
- For mixtures, calculate moles of atoms for each component separately and then add them.

5.03: Percent Composition, Empirical, and Molecular Formulas

Learning Objectives:
- Plan and implement investigative procedures.
- Collect data and make measurements with accuracy and precision.
- Express and manipulate chemical quantities using scientific conventions and mathematical procedures, including dimensional analysis, scientific notation, and significant figures.
- Organize, analyze, evaluate, make inferences, and predict trends from data.
- Communicate valid conclusions supported by the data.
- Communicate and apply scientific information extracted from various sources.
- Calculate the percent composition of compounds.
- Differentiate between empirical and molecular formulas.
Percent Composition:
- Joseph Proust’s law of definite proportions states that the percentages of elements by mass in a compound are always the same.
- The percent composition is always the same, regardless of how a sample is prepared.
Percent Composition Calculations:
- Formula:
- \text{% composition of element} = \frac{\text{mass of element in one mole of compound}}{\text{molar mass of compound}} × 100\%
Example: Calculate the percent composition of potassium in potassium oxide (K_2O).
- \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\%
Empirical and Molecular Formulas:
- Molecular Formula:
  - The actual number of each element in a formula (e.g., C6H{12}O_6 for glucose).
- Empirical Formula:
  - The simplest positive integer ratio of atoms in a compound (e.g., CH_2O for glucose).
- Different compounds can have the same empirical formula but are not the same compound.
Examples:
- CO (Molecular Formula) and CO (Empirical Formula) - Carbon Monoxide
- C2H4O2 $(Molecular Formula) and$ CH2O (Empirical Formula) - Vinegar
- CH2O $(Molecular Formula) and$ CH2O (Empirical Formula) - Formaldehyde
- C6H{12}O6 $(Molecular Formula) and$ CH2O (Empirical Formula) - Glucose
Calculating Empirical Formula from Lab Data:
- Step 1: Calculate the mass of each element.
  - If given percentages, assume 100 g of the substance and convert the percentages to grams.
- Step 2: Convert to moles by dividing the mass in grams by the molar mass of the element.
- Step 3: Divide all mole values by the smallest mole value to obtain subscripts.
- Step 4: If any result is a decimal mixed number, multiply all values by a factor to make them whole numbers.
- Example: A 10.0 g sample containing only iron and oxygen contains 6.994 g iron. What is the empirical formula?
  - Mass of O = 10.0 g - 6.994 g = 3.006 g
  - Moles of Fe = \frac{6.994 \text{ g Fe}}{55.845 \text{ g/mol Fe}} = 0.12523 \text{ mol Fe}
  - Moles of O = \frac{3.006 \text{ g O}}{15.999 \text{ g/mol O}} = 0.18788 \text{ mol O}
  - Divide by smallest (0.12523): Fe = 1, O = 1.5
  - Multiply by 2 to get whole numbers: Fe2O3
Calculating Molecular Formula:
- You need the empirical formula and the molar mass of the molecular compound.
- Step 1: Calculate the molar mass of the empirical formula (EF).
- 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.
- Step 3: Multiply the subscripts of the empirical formula by n to get the subscripts of the molecular formula.
- Example: The molar mass of a compound is 181.50 g/mol, and the empirical formula is C_2HCl. What is the molecular formula?
  - Molar Mass of EF = (2 x 12.011) + (1 x 1.008) + (1 x 35.45) = 60.48 g/mol
  - n = \frac{181.50 \text{ g/mol}}{60.48 \text{ g/mol}} = 3
  - 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
Molar Mass of Hydrates:
- Add the molar masses of the elements in the first compound, multiplying the molar mass by its subscripts.
- Add the molar mass of the elements in the water molecules, also accounting for subscripts
- Multiply the coefficient in front of the water by the molar mass of water.
- Add the mass of water to the mass of the original compound to get the total molar mass.
Example: What is the molar mass of copper(II) sulfate pentahydrate (CuSO4 \cdot 5H2O)?
- Molar mass of CuSO_4 = (1 x 63.546) + (1 x 32.06) + (4 x 15.999) = 159.60 g/mol
- Molar mass of 5H_2O = 5 x [(2 x 1.008) + (1 x 15.999)] = 5 x 18.015 = 90.075 g/mol
- 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
- Moles of H_2O $=$ \frac{8.09 \text{ g}}{18.015 \text{ g/mol}} = 0.44907 \text{ mol}
- Moles of MgCO_3 $=$ \frac{7.58 \text{ g}}{84.313 \text{ g/mol}} = 0.089903 \text{ mol}
- Ratio of H2O $to$ MgCO3 = \frac{0.44907}{0.089903} \approx 5
- Chemical formula: MgCO3 \cdot 5H2O
Mass Percent of Water:
- \text{mass % of water} = (\frac{\text{mass of water}}{\text{mass of original sample}}) × 100
- \text{mass % of water} = (\frac{8.09 \text{ g } H2O}{15.67 \text{ g } MgCO3 \cdot 5 H_2O}) × 100 = 51.6\%

5.05: Stoichiometry: The Basics

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.
- Calculate the number of atoms or molecules in a sample of material using Avogadro's number.
- Perform stoichiometric calculations, including determination of mass and gas volume relationships between reactants and products and percent yield.
Stoichiometry:
- Stoichiometry is the study of quantities in chemical reactions.
- It makes predictions on how much product is made when a specific amount of reactants are used.
Mole Ratio:
- A chemical reaction is like a recipe, and the ratio of reactants and products is important to determine how much product will be made.
- The ratio is always in terms of moles in a chemical reaction.
- The numbers for mole ratios come from the coefficients in the balanced chemical equation.
Example Chemical Equation: 2 Na (s) + 2 H2O(l) → 2 NaOH(aq) + H2 (g)
- 2 Na atoms react with 2 H2O molecules to yield 2 NaOH formula units and 1 H2 molecule
- 2 dozen Na atoms react with 2 dozen H2O molecules to yield 2 dozen NaOH formula units and 1 H2 molecules
- 2 moles of Na atoms react with 2 moles of H2O molecules to yield 2 moles NaOH formula units and 1 mole H2 molecules
Other Examples N2 (g) + 3 H2 (g) → 2 NH3 (g)
- Mole ratios:
- Na: H2O = 1:1
- H2O: H2 = 2:1
Stoichiometric map:
- Determine the given quantity and its unit first. That determines where you start on the map.
Stoichiometry with Copper and Silver Nitrate Using equation Cu (s) + 2 AgNO3 (aq) → 2 Ag (s) + Cu(NO3)2 (aq)
- Example 1: How many moles of copper are needed to react with excess silver nitrate to produce 0.500 moles of silver?
  - Dimensional analysis:
    - 0. 500 mol of Ag * 1 mol of Cu / 2 mol of Ag = 0.250 mol of Cu
  - Proportion Method
    - 0. 500 mol of Ag / 2 mol of Ag = x / 1 mol of Cu
      - x = 0.250 mol of Cu
- 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)
  - Dimensional analysis:
    - 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
  - Proportion Method
    - 1. 0 g H2O / (2 mol H2O)∙(18.015 g/ mol H2O) = x / (2 mol Na)∙(22.990 g /mol 𝑁𝑎)
      - X = 63.8 g of Na
- 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.
  - Dimensional analysis:
    - 1. 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
  - Proportion Method:
    - 60.0 g Na / (2 mol Na) * (22.990 g /mol 𝑁𝑎) = x / (1 mol 𝐻2) * (2.016 g/ mol 𝐻2)
      - x = 2.63 g 𝐻2

5.06: Stoichiometry: Theoretical and Actual Yield

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
- Calculate the number of atoms or molecules in a sample of material using Avogadro’s number
- Differentiate between empirical and molecular formulas
- Perform stoichiometric calculations, including determination of mass and gas volume relationships between reactants and products and percent yield
Theoretical Yield:
- The mass of product that should be made if a certain mass of reactant is put in.
Percent Yield:
- The ratio of the actual yield to the theoretical yield and reflects the amount of error that was present.
- \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%