chapter 3 lec 3

Stoichiometry: Quantitative Relationships in Chemical Reactions

Definition and Importance

Stoichiometry is the study of quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict how much product will form or how much reactant is needed for a given reaction, based on the balanced chemical equation.

• Mole Concept: Chemical reactions are governed by the number of moles, not by mass or volume. Always convert quantities to moles before performing stoichiometric calculations.

• Balanced Equation: The balanced equation provides the mole ratios between reactants and products.

• Conversion: Since moles cannot be weighed directly, convert between grams and moles using molar mass.

Example: For the reaction $2\ KClO_3(s) \rightarrow 2\ KCl(s) + 3\ O_2(g)$, 2 moles of potassium chlorate yield 2 moles of potassium chloride and 3 moles of oxygen gas when heated.

Stoichiometric Calculations: Recipe Analogy

Stoichiometry can be compared to following a recipe, where ingredients must be combined in specific ratios to produce a desired outcome.

• If you have only 2 cups of cheese and plenty of other ingredients, you can make 1 pizza (using the ratio: 2 cups cheese per pizza).

• To calculate how many pizzas can be made with 6 cups of cheese: $6\ \text{cups cheese} \times \left(\frac{1\ \text{pizza}}{2\ \text{cups cheese}}\right) = 3\ \text{pizzas}$

Worked Example: Combustion of Octane

Suppose 11.0 moles of octane ($C_8H_{18}$) are burned. How many moles of carbon dioxide ($CO_2$) are produced?

• Balanced equation: $2\ C_8H_{18}(l) + 25\ O_2(g) \rightarrow 16\ CO_2(g) + 18\ H_2O(g)$

• Mole ratio: For every 2 moles of octane, 16 moles of $CO_2$ are produced.

• Calculation: $11.0\ \text{mol}\ C_8H_{18} \times \left(\frac{16\ \text{mol}\ CO_2}{2\ \text{mol}\ C_8H_{18}}\right) = 88.0\ \text{mol}\ CO_2$

• Convert moles to grams using molar mass:

  • $11.0\ \text{mol}\ C_8H_{18} \times 114.232\ \text{g/mol} = 1256.55\ \text{g}\ C_8H_{18}$

  • $88.0\ \text{mol}\ CO_2 \times 44.011\ \text{g/mol} = 3872.968\ \text{g}\ CO_2$

Worked Example: Photosynthesis

Plants convert carbon dioxide and water into glucose and oxygen. If a plant consumes 37.8 g $CO_2$ in one week, what mass of glucose can be synthesized?

• Balanced equation: $6\ CO_2(g) + 6\ H_2O(g) \rightarrow C_6H_{12}O_6(aq) + 6\ O_2(g)$

• Mole relationships: 6 moles $CO_2$ produce 1 mole glucose.

• Convert grams $CO_2$ to moles: $37.8\ \text{g}\ CO_2 \times \left(\frac{1\ \text{mol}}{44.011\ \text{g}}\right) = 0.8589\ \text{mol}$

• Calculate moles glucose: $0.8589\ \text{mol}\ CO_2 \times \left(\frac{1\ \text{mol}\ C_6H_{12}O_6}{6\ \text{mol}\ CO_2}\right) = 0.14315\ \text{mol}\ C_6H_{12}O_6$

• Convert moles glucose to grams: $0.14315\ \text{mol}\ C_6H_{12}O_6 \times 180.2\ \text{g/mol} = 25.8\ \text{g}$

Limiting Reagent and Theoretical Yield

Definition and Calculation

The limiting reagent is the reactant that is completely consumed first, thus determining the maximum amount of product that can be formed. The theoretical yield is the maximum amount of product predicted by stoichiometry, while the actual yield is the amount actually obtained.

• To identify the limiting reagent, calculate the amount of product each reactant can produce. The reactant producing the least product is the limiting reagent.

• Excess reagent is the reactant left over after the reaction.

Worked Example: Nitrous Oxide and Hydrogen

Given 86.3 g $NO$ and 25.6 g $H_2$, find the limiting reagent and theoretical yield of ammonia ($NH_3$).

• Balanced equation: $2\ NO(g) + 5\ H_2(g) \rightarrow 2\ NH_3(g) + 2\ H_2O(g)$

• Convert to moles:

  • $86.3\ \text{g}\ NO \times \left(\frac{1\ \text{mol}}{30.01\ \text{g}}\right) = 2.8757\ \text{mol}\ NO$

  • $25.6\ \text{g}\ H_2 \times \left(\frac{1\ \text{mol}}{2.02\ \text{g}}\right) = 12.6733\ \text{mol}\ H_2$

• Calculate moles $NH_3$ from each reactant:

  • $2.8757\ \text{mol}\ NO \times \left(\frac{2\ \text{mol}\ NH_3}{2\ \text{mol}\ NO}\right) = 2.8757\ \text{mol}\ NH_3$

  • $12.6733\ \text{mol}\ H_2 \times \left(\frac{2\ \text{mol}\ NH_3}{5\ \text{mol}\ H_2}\right) = 5.0693\ \text{mol}\ NH_3$

• NO is the limiting reagent; theoretical yield is 2.8757 mol $NH_3$.

• Convert to grams: $2.8757\ \text{mol}\ NH_3 \times 17.03\ \text{g/mol} = 49.0\ \text{g}$

• Calculate excess $H_2$: $12.6733\ \text{mol} - 7.1892\ \text{mol} = 5.4841\ \text{mol}$; $5.4841\ \text{mol} \times 2.02\ \text{g/mol} = 2.71\ \text{g}$

Worked Example: Iron Production and Percent Yield

Given 167 g $Fe_2O_3$ and 85.8 g $CO$, reaction produces 72.3 g iron. Find limiting reagent, theoretical yield, and percent yield.

• Balanced equation: $Fe_2O_3(s) + 3\ CO(g) \rightarrow 2\ Fe(s) + 3\ CO_2(g)$

• Convert to moles:

  • $167\ \text{g}\ Fe_2O_3 \times \left(\frac{1\ \text{mol}}{159.7\ \text{g}}\right) = 1.0457\ \text{mol}$

  • $85.8\ \text{g}\ CO \times \left(\frac{1\ \text{mol}}{28.011\ \text{g}}\right) = 3.0631\ \text{mol}$

• Calculate moles $Fe$ from each reactant:

  • $1.0457\ \text{mol}\ Fe_2O_3 \times \left(\frac{2\ \text{mol}\ Fe}{1\ \text{mol}\ Fe_2O_3}\right) = 2.0914\ \text{mol}$

  • $3.0631\ \text{mol}\ CO \times \left(\frac{2\ \text{mol}\ Fe}{3\ \text{mol}\ CO}\right) = 2.0421\ \text{mol}$

• CO is the limiting reagent; theoretical yield is 2.0421 mol $Fe$.

• Convert to grams: $2.0421\ \text{mol}\ Fe \times 55.85\ \text{g/mol} = 114\ \text{g}$

• Percent yield: $\text{Percent yield} = \left(\frac{\text{actual yield}}{\text{theoretical yield}}\right) \times 100$

• $\text{Percent yield} = \left(\frac{72.3\ \text{g}}{114\ \text{g}}\right) \times 100 = 63.5\%$

Solution Concentrations

Molarity (M)

Molarity is the most common unit for expressing concentration in chemistry. It is defined as the number of moles of solute per liter of solution.

• Formula: $M = \frac{\text{moles}}{\text{liters of solution}}$

Dilution of Solutions

When a solution is diluted, the number of moles of solute remains constant, but the volume increases, decreasing the concentration.

• Formula: $M_1V_1 = M_2V_2$

• Example: To dilute 52.10 mL of 0.178 M HCl to 0.132 M:

  • $V_2 = \frac{M_1V_1}{M_2} = \frac{0.178 \times 52.10}{0.132} = 70.26\ \text{mL}$

Other Units for Solution Concentrations

Besides molarity, concentrations can be expressed in mass percentage, volume percentage, mass-volume percentage, parts per million (ppm), and parts per billion (ppb).

• Mass percentage: $\text{Mass percentage} = \left(\frac{\text{mass of one component}}{\text{mass of whole solution}}\right) \times 100$

• Volume percentage: $\text{Volume percentage} = \left(\frac{\text{volume of one component}}{\text{volume of whole solution}}\right) \times 100$

• Mass-volume percentage: $\text{Mass-volume percentage} = \left(\frac{\text{mass of solute}}{\text{volume of solution}}\right) \times 100$

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

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

Worked Example: ppm and ppb Calculation

If a 200.0 g sample of solution contains 390 mg sodium ion, what is the concentration in ppm and ppb?

• Convert mg to g: $390\ \text{mg} \times \left(\frac{1\ \text{g}}{1000\ \text{mg}}\right) = 0.390\ \text{g}$

• Calculate ppm: $\left(\frac{0.390\ \text{g}}{200.0\ \text{g}}\right) \times 10^6 = 1.95 \times 10^3\ \text{ppm}$

• Calculate ppb: $\left(\frac{0.390\ \text{g}}{200.0\ \text{g}}\right) \times 10^9 = 1.95 \times 10^6\ \text{ppb}$

Summary Table: Stoichiometric Calculations

Step	Description	Formula/Example
1. Balance Equation	Ensure correct mole ratios	e.g., $2\ NO + 5\ H_2 \rightarrow 2\ NH_3 + 2\ H_2O$
2. Convert to Moles	Use molar mass to convert grams to moles	$\text{moles} = \frac{\text{mass}}{\text{molar mass}}$
3. Use Mole Ratios	Relate reactants and products	$\text{moles product} = \text{moles reactant} \times \frac{\text{mole ratio}}{\text{reactant}}$
4. Convert to Mass	Use molar mass to convert moles to grams	$\text{mass} = \text{moles} \times \text{molar mass}$
5. Calculate Yield	Find theoretical and percent yield	$\text{Percent yield} = \left(\frac{\text{actual yield}}{\text{theoretical yield}}\right) \times 100$

Additional info: The notes also briefly mention reasons for less than 100% yield, including reversibility, side reactions, solubility, and product reactivity, which are important for understanding real-world chemical processes.