chapter 3 lec 2

Mass Relationships in Chemical Reactions

Empirical Formulas

The empirical formula of a compound represents the simplest whole-number ratio of the elements present. Determining the empirical formula from percent composition involves several systematic steps.

• Step 1: Assume 100 g of the compound to directly convert percentages to grams.

• Step 2: Convert grams to moles for each element using their atomic masses.

• Step 3: Divide each mole value by the smallest number of moles to obtain the simplest ratio.

• Step 4: Write the empirical formula using these ratios as subscripts.

Example: For a compound with 40.1% C, 6.6% H, and 53.3% O:

• Assume 100 g sample: 40.1 g C, 6.6 g H, 53.3 g O

• Convert to moles:

  • C: $40.1 \ \mathrm{g} \times \frac{1 \ \mathrm{mol}}{12.011 \ \mathrm{g}} = 3.339 \ \mathrm{mol}$

  • H: $6.6 \ \mathrm{g} \times \frac{1 \ \mathrm{mol}}{1.008 \ \mathrm{g}} = 6.548 \ \mathrm{mol}$

  • O: $53.3 \ \mathrm{g} \times \frac{1 \ \mathrm{mol}}{16.00 \ \mathrm{g}} = 3.331 \ \mathrm{mol}$

• Divide by smallest (3.331):

  • C: $\frac{3.339}{3.331} = 1.002$

  • H: $\frac{6.548}{3.331} = 1.966$

  • O: $\frac{3.331}{3.331} = 1.000$

• Empirical formula: CH2O

Combustion Analysis

Combustion analysis is a laboratory technique used to determine the empirical formula of compounds, especially those containing carbon and hydrogen (and sometimes oxygen). The compound is burned in excess oxygen, and the resulting CO2 and H2O are collected and weighed.

• All C in the sample is converted to CO2.

• All H is converted to H2O.

• Oxygen in the original compound is determined by difference (total mass minus mass of C and H).

Key reactions:

• All C: $\text{C (in compound)} + \mathrm{O}_2 (g) \rightarrow \mathrm{CO}_2 (g)$

• All H: $\text{H (in compound)} + \mathrm{O}_2 (g) \rightarrow \mathrm{H}_2\mathrm{O} (g)$

Example: A hydrocarbon sample produces 1.60 g CO2 and 0.819 g H2O upon combustion. The original sample mass is 0.8233 g.

1. Convert CO2 and H2O to moles:

   • CO2: $1.60 \ \mathrm{g} \times \frac{1 \ \mathrm{mol}}{44.01 \ \mathrm{g}} = 0.03636 \ \mathrm{mol}$

   • H2O: $0.819 \ \mathrm{g} \times \frac{1 \ \mathrm{mol}}{18.01 \ \mathrm{g}} = 0.04547 \ \mathrm{mol}$

2. Find moles of C and H:

   • C: $0.03636 \ \mathrm{mol}$ (1 mol C per 1 mol CO2)

   • H: $0.04547 \ \mathrm{mol} \times 2 = 0.09094 \ \mathrm{mol}$ (2 mol H per 1 mol H2O)

3. Convert moles to grams:

   • C: $0.03636 \ \mathrm{mol} \times 12.011 \ \mathrm{g/mol} = 0.4366 \ \mathrm{g}$

   • H: $0.09094 \ \mathrm{mol} \times 1.008 \ \mathrm{g/mol} = 0.0917 \ \mathrm{g}$

4. Oxygen by difference: $0.8233 \ \mathrm{g} - (0.4366 \ \mathrm{g} + 0.0917 \ \mathrm{g}) = 0.2949 \ \mathrm{g}$

5. Convert O to moles: $0.2949 \ \mathrm{g} \times \frac{1 \ \mathrm{mol}}{16.00 \ \mathrm{g}} = 0.01843 \ \mathrm{mol}$

6. Write empirical formula: C0.03636H0.09094O0.01843

7. Divide by smallest (0.01843): C1.97H4.93O1 ≈ C2H5O

Additional info: The numbers in the original example were inconsistent; the above values are recalculated for clarity.

Determination of Molecular Formulas

The molecular formula shows the actual number of atoms of each element in a molecule. It is a whole-number multiple of the empirical formula. To determine the molecular formula, the molar mass of the compound must be known.

• Calculate the molar mass of the empirical formula ($M_{emp}$).

• Divide the compound's molar mass ($M_{mol}$) by $M_{emp}$ to find the integer $n$:

$n = \frac{M_{mol}}{M_{emp}}$

• The molecular formula is then $(\text{empirical formula})_n$.

Example 1: Empirical formula C2H3O, molar mass = 86.09 g/mol.

• Molar mass of empirical formula: $(2 \times 12.011) + (3 \times 1.008) + 16.00 = 43.046$ g/mol

• $n = \frac{86.09}{43.046} = 2$

• Molecular formula: C4H6O2

Example 2: Empirical formula CH, molar mass = 78.11 g/mol.

• Molar mass of CH: $12.011 + 1.008 = 13.019$ g/mol

• $n = \frac{78.11}{13.019} = 6$

• Molecular formula: C6H6

Solution Concentrations

Molarity

Molarity (M) is the most common unit of concentration in chemistry, defined as the number of moles of solute per liter of solution.

$M = \frac{\text{moles of solute}}{\text{liters of solution}}$

Dilution of Solutions

When a solution is diluted, the amount of solute remains constant, but the volume increases. The relationship is:

$M_1 V_1 = M_2 V_2$

where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume.

Other Units for Solution Concentrations

• Mass percentage (% w/w): $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

• Volume percentage (% v/v): $\frac{\text{volume of solute}}{\text{volume of solution}} \times 100$

• Mass-Volume percentage (% w/v): $\frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100$

• Parts per million (ppm): $\frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$

• Parts per billion (ppb): $\frac{\text{mass of solute}}{\text{mass of solution}} \times 10^9$

Chemical Reactions and Chemical Equations

Writing Chemical Equations

A chemical equation represents a chemical reaction, showing the reactants and products, their physical states, and the relative amounts involved. The general form is:

$\text{A} + \text{B} \rightarrow \text{C} + \text{D}$

• Reactants are on the left, products on the right.

• Physical states are indicated: (s) solid, (l) liquid, (g) gas, (aq) aqueous.

Example: Combustion of methane:

$\mathrm{CH}_4 (g) + 2 \mathrm{O}_2 (g) \rightarrow \mathrm{CO}_2 (g) + 2 \mathrm{H}_2\mathrm{O} (g)$

This equation shows that one molecule (or mole) of methane reacts with two molecules (or moles) of oxygen to produce one molecule (or mole) of carbon dioxide and two molecules (or moles) of water.

Balancing Chemical Equations

Balancing ensures the law of conservation of mass is obeyed: the number of atoms of each element is the same on both sides of the equation.

• Start by balancing elements that appear only once on each side.

• Balance hydrogen and oxygen last, as they often appear in multiple compounds.

• Use coefficients to balance atoms; never change subscripts.

Example: Balancing the combustion of methane:

1. Write the unbalanced equation:

   • $\mathrm{CH}_4 (g) + \mathrm{O}_2 (g) \rightarrow \mathrm{CO}_2 (g) + \mathrm{H}_2\mathrm{O} (g)$

2. Balance C: 1 on both sides.

3. Balance H: 4 on left, 2 on right. Place 2 in front of H2O:

   • $\mathrm{CH}_4 (g) + \mathrm{O}_2 (g) \rightarrow \mathrm{CO}_2 (g) + 2\mathrm{H}_2\mathrm{O} (g)$

4. Balance O: 2 on left, 2 (from CO2) + 2 (from 2 H2O) = 4 on right. Place 2 in front of O2:

   • $\mathrm{CH}_4 (g) + 2\mathrm{O}_2 (g) \rightarrow \mathrm{CO}_2 (g) + 2\mathrm{H}_2\mathrm{O} (g)$

Interpretation: 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O.

Stoichiometric Calculations

Balanced equations provide the mole ratios needed for quantitative calculations in chemical reactions.

Example: If 1.234 g CH4 is burned:

• Moles CH4: $1.234 \ \mathrm{g} \times \frac{1 \ \mathrm{mol}}{16.043 \ \mathrm{g}} = 0.076918 \ \mathrm{mol}$

• Moles O2 needed: $2 \times 0.076918 = 0.153836 \ \mathrm{mol}$

• Mass O2: $0.153836 \ \mathrm{mol} \times 32.00 \ \mathrm{g/mol} = 4.9228 \ \mathrm{g}$

• Moles CO2 produced: $0.076918 \ \mathrm{mol}$

• Mass CO2: $0.076918 \ \mathrm{mol} \times 44.011 \ \mathrm{g/mol} = 3.3852 \ \mathrm{g}$

• Moles H2O produced: $0.153836 \ \mathrm{mol}$

• Mass H2O: $0.153836 \ \mathrm{mol} \times 18.016 \ \mathrm{g/mol} = 2.7715 \ \mathrm{g}$

Additional Examples of Balancing Equations

• Decomposition of potassium chlorate:

  • Unbalanced: KClO3(s) → KCl(s) + O2(g)

  • Balanced: $2\mathrm{KClO}_3 (s) \rightarrow 2\mathrm{KCl} (s) + 3\mathrm{O}_2 (g)$

• Combustion of butane:

  • Unbalanced: C4H10(g) + O2(g) → CO2(g) + H2O(g)

  • Balanced: $2\mathrm{C}_4\mathrm{H}_{10} (g) + 13\mathrm{O}_2 (g) \rightarrow 8\mathrm{CO}_2 (g) + 10\mathrm{H}_2\mathrm{O} (g)$

• Reduction of cobalt(III) oxide:

  • Unbalanced: Co2O3(s) + C(s) → Co(s) + CO2(g)

  • Balanced: $2\mathrm{Co}_2\mathrm{O}_3 (s) + 3\mathrm{C} (s) \rightarrow 4\mathrm{Co} (s) + 3\mathrm{CO}_2 (g)$

• Reaction of silicon dioxide and carbon:

  • Unbalanced: SiO2(s) + C(s) → SiC(s) + CO(g)

  • Balanced: $\mathrm{SiO}_2 (s) + 3\mathrm{C} (s) \rightarrow \mathrm{SiC} (s) + 2\mathrm{CO} (g)$

Summary Table: Steps for Empirical and Molecular Formula Determination

Step	Empirical Formula	Molecular Formula
1	Obtain mass or % composition of each element	Obtain empirical formula and molar mass
2	Convert grams to moles	Calculate empirical formula mass
3	Divide by smallest number of moles	Divide molar mass by empirical formula mass
4	Write empirical formula	Multiply subscripts in empirical formula by integer n

Additional info: Some calculations and examples have been clarified and expanded for completeness and accuracy.