Chemistry: Calculations with Chemical Formulas and Equations

Chapter 3: Calculations with Chemical Formulas and Equations

Overview of Chapter 3

• Topics Covered:
    - Molecular Weight and Formula Weight
    - The Mole Concept
    - Determining Formulas
      - Mass percentage
      - Empirical Formula
      - Molecular Formula
    - Stoichiometric Calculation
      - Molar Interpretation of a Chemical Equation
      - Amounts of Substances in a Chemical Reaction
      - Limiting Reactant; Theoretical and Percentage Yields

Molecular Weight and Formula Weight

• Atomic Mass (or Atomic Weight): Mass of a single atom of an element.

• Molecular Mass (or Molecular Weight): Mass of a single molecule of a substance.
    - Definition: Molecular Mass = sum of Atomic Masses in a molecule.
    - Example: The Molecular Mass of water (H₂O) is 18.0 amu.

• Ionic Compounds: For ionic substances, we use Formula Mass (or Formula Weight).
    - Definition: Formula Mass = sum of the Atomic Masses in its formula.
    - Example: The Formula Mass of CaF₂ is 78.1 amu.

Example Calculations

• Example 1: Calculate formula mass of:
    - a. Chloroform (CHCl₃)
      - Calculation:
        - 1 × AM of C (12.0 amu) + 1 × AM of H (1.0 amu) + 3 × AM of Cl (3 × 35.45 amu)
        - Total FM = 12.0 + 1.0 + 106.35 = 119.4 amu → Answer: 119 amu.
    - b. Iron(III) sulfate (Fe₂(SO₄)₃)
      - Calculation:
        - 2 × AM of Fe (2 × 55.8 amu) + 3 × AM of S (3 × 32.1 amu) + 12 × AM of O (12 × 16.00 amu)
        - Total FM = 111.6 + 96.3 + 192.0 = 399.9 amu → Answer: 4.00 × 10² amu.

The Mole Concept

• Definition of a Mole: A mole is a unit that represents a specific quantity of particles (atoms, molecules, etc.).
    - Exact Value: 1 mole = 6.022 × 10²³ particles (Avogadro's Number).
    - 1 mole of C atoms weighs 12.01 g and contains 6.022 × 10²³ atoms.

Molar Mass

• Definition: The molar mass is the mass in grams of one mole of a compound (units: g/mol).

• Calculating Molar Mass:
    - Example (H₂O): 2(1.008 g) + 16.00 g = 18.02 g/mol.
    - Example (CO₂): 12.01 g + 2 × 16.00 g = 44.01 g/mol.

Significant Concepts Related to Moles

• Number of Moles Formula:
    - $n = rac{m}{M}$
    - where:
      - n = number of moles
      - m = mass in grams
      - M = molar mass in g/mol.

Example Calculations With Moles

• Example 2: Zinc iodide (ZnI₂) formation with given moles.
    - Molar mass of ZnI₂ = 319 g/mol.
    - Calculation:
      - $0.0654 ext{ mol ZnI₂} imes 319 ext{ g ZnI₂/mol} = 20.9 ext{ g ZnI₂}.$

• Example 3: Mass of chlorine atom and hydrogen chloride molecule.
    - a. Chlorine (Cl):
      - Molar mass = 35.5 g/mol
      - Mass of one atom: $rac{35.5 ext{ g}}{6.022 imes 10^{23}} = 5.90 imes 10^{-23} ext{ g}.$
    - b. Hydrogen Chloride (HCl): 36.5 g/mol
      - Mass of one molecule: $rac{36.5 ext{ g}}{6.022 imes 10^{23}} = 6.06 imes 10^{-23} ext{ g}.$

Determining Chemical Formulas

Percent Composition

• Definition: Percentage of each element in a compound by mass.

• How to Calculate: Use either the compound's formula or experimental mass analysis.

• Note: Percentages may not total 100% due to rounding errors.

Example 7: Calculate Percent Composition of C₂H₅OH

• Component Masses:
    - C: 2 moles × 12.01 g/mol = 24.02 g,
    - H: 6 moles × 1.008 g/mol = 6.048 g,
    - O: 1 mole × 16.00 g/mol = 16.00 g.

• Molar Mass of C₂H₅OH: 24.02 + 6.048 + 16.00 = 46.07 g/mol.

• Calculations:
    - rac{24.02}{46.07} imes 100 ext{%} ext{ C} = 52.14 ext{% C},
    - rac{6.048}{46.07} imes 100 ext{%} ext{ H} = 13.13 ext{% H},
    - rac{16.00}{46.07} imes 100 ext{%} ext{ O} = 34.73 ext{% O}.

Empirical and Molecular Formulas

• Empirical Formula: Simplest whole-number ratio of atoms in a molecule.

• Molecular Formula: A multiple of the empirical formula.

• Example 8: Determine Empirical Formula of Benzopyrene (C20H12).
    - GCF of 20 and 12 = 4.
    - Result: C₅H₃ is the empirical formula.

Stoichiometric Calculations

• Stoichiometry: Involves calculation of masses of reactants and products in a chemical reaction.

• Chemical Equation Representation: Reactants on the left and products on the right.
    - Example: $CH_4 + 2O_2 <br>ightarrow CO_2 + 2H_2O$.

Stoichiometric Coefficients

• They represent the ratio of moles of reactants to products.

• Example 11: Calculate moles of Carbon Monoxide (CO) from Oxygen (O₂).
    - Equation: $2CO + O_2 <br>ightarrow 2CO_2$
      - For 3.2 moles of O₂:
        - $3.2 ext{ moles O₂} imes rac{2 ext{ moles CO}}{1 ext{ mole O₂}} = 6.4 ext{ moles CO}.$

Limiting and Excess Reactants

• Limiting Reactant: Completely consumed in the reaction.

• Excess Reactant: Not fully consumed.

• Theoretical Yield: Maximum product obtained when limiting reactant is exhausted.

Example 13: Determining Ammonia Production

• Balanced Reaction: $N_2(g) + 3H_2(g) <br>ightarrow 2NH_3(g)$.

• Given: 25 kg N₂ and 5 kg H₂. To determine mass of ammonia:
    - Calculate moles of each reactant:
      - $n_{N_2} = rac{25000 ext{ g}}{28} = 893 ext{ moles N₂}$.
      - $n_{H_2} = rac{5000 ext{ g}}{2.016} = 2481 ext{ moles H₂}$.
    - Identify limiting reactant via ratios.

• Conclusion: Calculate mass of NH₃ produced by reacting H₂ as the limiting reagent.

Percent Yield

• Definition: Actual yield to theoretical yield expressed as a percentage.

• Formula: ext{Percent Yield} = rac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100 ext{%} .

• Example 14: If 72.0 g CO₂ is produced with a theoretical yield of 88.0 g:
    - ext{Percent Yield} = rac{72.0 ext{ g}}{88.0 ext{ g}} imes 100 ext{%} = 81.82 ext{%}.

Problems Section

Key Sample Problems

1. Problem 46: Determine number of Fe atoms in 500.0 g of Fe.
      - Solution: $500.0 ext{ g} Fe imes rac{1 ext{ mol}}{55.85 ext{ g}} = 8.953 ext{ mol Fe}.$, leading to $8.953 ext{ mol Fe} imes 6.022 imes 10^{23} ext{ atoms} = 5.391 imes 10^{24} ext{ atoms Fe}.$

2. Problem 51: Calculate the molar mass of $(NH_4)_2Cr_2O_7$:
      - Solution: $2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 ext{ g/mol}.$

3. Problem 83: Empirical formula from mass percentages.
      - Compounds mass = 48.64% C, 8.16% H, 43.20% O yield the empirical formula C₃H₆O₂ after analysis.

Overview of Chapter 3 - Topics Covered:

• Molecular Weight and Formula Weight

• The Mole Concept

• Determining Formulas

• Mass Percentage

• Empirical Formula

• Molecular Formula

• Stoichiometric Calculation

• Molar Interpretation of a Chemical Equation

• Amounts of Substances in a Chemical Reaction

• Limiting Reactant; Theoretical and Percentage Yields

Molecular Weight and Formula Weight - Atomic Mass (or Atomic Weight): Mass of a single atom of an element.

• Molecular Mass (or Molecular Weight): Mass of a single molecule of a substance.

• Definition: Molecular Mass = sum of Atomic Masses in a molecule.

• Example: The Molecular Mass of water (H₂O) is 18.0 amu.

• Ionic Compounds: For ionic substances, we use Formula Mass (or Formula Weight).

• Definition: Formula Mass = sum of the Atomic Masses in its formula.

• Example: The Formula Mass of CaF₂ is 78.1 amu.

Example Calculations - Example 1: Calculate the formula mass of:

• a. Chloroform (CHCl₃)

  • Calculation:

  • 1 × AM of C (12.0 amu) + 1 × AM of H (1.0 amu) + 3 × AM of Cl (3 × 35.45 amu)

  • Total FM = 12.0 + 1.0 + 106.35 = 119.4 amu → Answer: 119 amu.

• b. Iron(III) sulfate (Fe₂(SO₄)₃)

  • Calculation:

  • 2 × AM of Fe (2 × 55.8 amu) + 3 × AM of S (3 × 32.1 amu) + 12 × AM of O (12 × 16.00 amu)

  • Total FM = 111.6 + 96.3 + 192.0 = 399.9 amu → Answer: 4.00 × 10² amu.

The Mole Concept - Definition of a Mole: A mole is a unit that represents a specific quantity of particles (atoms, molecules, etc.).

• Exact Value: 1 mole = 6.022 × 10²³ particles (Avogadro's Number).

• 1 mole of C atoms weighs 12.01 g and contains 6.022 × 10²³ atoms.

Molar Mass - Definition: The molar mass is the mass in grams of one mole of a compound (units: g/mol).

• Calculating Molar Mass:

  • Example (H₂O): 2(1.008 g) + 16.00 g = 18.02 g/mol.

  • Example (CO₂): 12.01 g + 2 × 16.00 g = 44.01 g/mol.

Significant Concepts Related to Moles - Number of Moles Formula:

• $n = \frac{m}{M}$

• where:

  • n = number of moles

  • m = mass in grams

  • M = molar mass in g/mol.

Example Calculations With Moles - Example 2: Zinc iodide (ZnI₂) formation with given moles.

• Molar mass of ZnI₂ = 319 g/mol.

• Calculation:

• $0.0654 \text{ mol ZnI₂} \times 319 \text{ g ZnI₂/mol} = 20.9 \text{ g ZnI₂}.$

• Example 3: Mass of chlorine atom and hydrogen chloride molecule.

• a. Chlorine (Cl):

  • Molar mass = 35.5 g/mol

  • Mass of one atom: $\frac{35.5 \text{ g}}{6.022 \times 10^{23}} = 5.90 \times 10^{-23} \text{ g}.$

• b. Hydrogen Chloride (HCl): 36.5 g/mol

  • Mass of one molecule: $\frac{36.5 \text{ g}}{6.022 \times 10^{23}} = 6.06 \times 10^{-23} \text{ g}.$

Determining Chemical Formulas #### Percent Composition - Definition: Percentage of each element in a compound by mass.

• How to Calculate: Use either the compound's formula or experimental mass analysis.

• Note: Percentages may not total 100% due to rounding errors.

Example 7: Calculate Percent Composition of C₂H₅OH - Component Masses:

• C: 2 moles × 12.01 g/mol = 24.02 g,

• H: 6 moles × 1.008 g/mol = 6.048 g,

• O: 1 mole × 16.00 g/mol = 16.00 g.

• Molar Mass of C₂H₅OH: 24.02 + 6.048 + 16.00 = 46.07 g/mol.

• Calculations:

  • \frac{24.02}{46.07} \times 100 \text{% C} = 52.14 \text{% C},

  • \frac{6.048}{46.07} \times 100 \text{% H} = 13.13 \text{% H},

  • \frac{16.00}{46.07} \times 100 \text{% O} = 34.73 \text{% O}.

Empirical and Molecular Formulas - Empirical Formula: Simplest whole-number ratio of atoms in a molecule.

• Molecular Formula: A multiple of the empirical formula.

• Example 8: Determine Empirical Formula of Benzopyrene (C20H12).

• GCF of 20 and 12 = 4.

• Result: C₅H₃ is the empirical formula.

Stoichiometric Calculations - Stoichiometry: Involves calculation of masses of reactants and products in a chemical reaction.

• Chemical Equation Representation: Reactants on the left and products on the right.

• Example: $CH₄ + 2O₂ \rightarrow CO₂ + 2H₂O$.

Stoichiometric Coefficients - They represent the ratio of moles of reactants to products.

• Example 11: Calculate moles of Carbon Monoxide (CO) from Oxygen (O₂).

• Equation: $2CO + O₂ \rightarrow 2CO₂$

• For 3.2 moles of O₂:

• $3.2 \text{ moles O₂} \times \frac{2 \text{ moles CO}}{1 \text{ mole O₂}} = 6.4 \text{ moles CO}.$

Limiting and Excess Reactants - Limiting Reactant: Completely consumed in the reaction.

• Excess Reactant: Not fully consumed.

• Theoretical Yield: Maximum product obtained when limiting reactant is exhausted.

Example 13: Determining Ammonia Production - Balanced Reaction: $N₂(g) + 3H₂(g) \rightarrow 2NH₃(g)$.

• Given: 25 kg N₂ and 5 kg H₂. To determine the mass of ammonia:

• Calculate moles of each reactant:

  • n_{N₂} = \frac{25000 \text{ g}}{28} = 893 \text{ moles N₂} $.

  • n_{H₂} = \frac{5000 \text{ g}}{2.016} = 2481 \text{ moles H₂} $.

• Identify limiting reactant via ratios.

• Conclusion: Calculate mass of NH₃ produced by reacting H₂ as the limiting reagent.

Percent Yield - Definition: Actual yield to theoretical yield expressed as a percentage.

• Formula: \text{Percent Yield} = \frac{ \text{Actual Yield}}{ \text{Theoretical Yield}} \times 100 \text{%} $.

• Example 14: If 72.0 g CO₂ is produced with a theoretical yield of 88.0 g:

• \text{Percent Yield} = \frac{72.0 \text{ g}}{88.0 \text{ g}} \times 100 \text{%} = 81.82 \text{%}.

Problems Section #### Key Sample Problems 1. Problem 46: Determine the number of Fe atoms in 500.0 g of Fe.

• Solution: 500.0 \text{ g} Fe \times \frac{1 \text{ mol}}{55.85 \text{ g}} = 8.953 \text{ mol Fe}. $, leading to$ 8.953 \text{ mol Fe} \times 6.022 \times 10^{23} \text{ atoms} = 5.391 \times 10^{24} \text{ atoms Fe}.

1. Problem 51: Calculate the molar mass of $(NH₄)2Cr2O_7$:

• Solution: 2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 \text{ g/mol}. $$

1. Problem 83: Empirical formula from mass percentages.

• Compounds mass = 48.64% C, 8.16% H, 43.20% O yield the empirical formula C₃H₆O₂ after analysis.