Stoichiometry and Solution Concentrations: Mass Relationships in Chemical Reactions

Stoichiometry and Mass Relationships in Chemical Reactions

Introduction to Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction, based on the balanced chemical equation.

• Stoichiometry uses the mole concept and molar masses to relate quantities of reactants and products.

• All calculations must be performed in moles, regardless of the starting units (grams, volumes, etc.).

• Since moles cannot be directly measured, conversions between grams and moles are essential.

Balanced Chemical Equations and Mole Ratios

A balanced chemical equation provides the exact ratio in which reactants combine and products form. These ratios are used to convert between moles of different substances in the reaction.

• Example: For the decomposition of potassium chlorate:

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

• 2 moles of KClO3 produce 2 moles of KCl and 3 moles of O2.

Analogous to a recipe, the coefficients in a balanced equation indicate the proportions of each substance involved.

Stoichiometric Calculations: Step-by-Step

To determine the amount of product formed or reactant required:

1. Balance the chemical equation.

2. Convert all given quantities to moles.

3. Use mole ratios from the balanced equation to relate moles of one substance to another.

4. Convert moles back to grams or other units if required.

Example 1: Combustion of Octane

• Balanced equation: $2\ \mathrm{C}_8\mathrm{H}_{18}(l) + 25\ \mathrm{O}_2(g) \rightarrow 16\ \mathrm{CO}_2(g) + 18\ \mathrm{H}_2\mathrm{O}(g)$

• If 11.0 moles of C8H18 are burned, how many moles of CO2 are produced?

Calculation:

$11.0\ \text{mol}\ \mathrm{C}_8\mathrm{H}_{18} \times \frac{16\ \text{mol}\ \mathrm{CO}_2}{2\ \text{mol}\ \mathrm{C}_8\mathrm{H}_{18}} = 88.0\ \text{mol}\ \mathrm{CO}_2$

• To convert moles to grams: $\text{moles} \times \frac{\text{g}}{\text{mol}} = \text{g}$

• For C8H18: $11.0\ \text{mol} \times 114.232\ \text{g/mol} = 1.26 \times 10^3\ \text{g}$

• For CO2: $88.0\ \text{mol} \times 44.011\ \text{g/mol} = 3.87 \times 10^3\ \text{g}$

Example 2: Photosynthesis

• Balanced equation: $6\ \mathrm{CO}_2(g) + 6\ \mathrm{H}_2\mathrm{O}(g) \rightarrow \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(aq) + 6\ \mathrm{O}_2(g)$

• If a plant consumes 37.8 g CO2, what mass of glucose can be synthesized?

1. Convert 37.8 g CO2 to moles: $37.8\ \text{g} \times \frac{1\ \text{mol}}{44.011\ \text{g}} = 0.8589\ \text{mol}$

2. Relate to glucose: $0.8589\ \text{mol}\ \mathrm{CO}_2 \times \frac{1\ \text{mol}\ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6}{6\ \text{mol}\ \mathrm{CO}_2} = 0.14315\ \text{mol}\ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$

3. Convert to grams: $0.14315\ \text{mol} \times 180.2\ \text{g/mol} = 25.8\ \text{g}$

Limiting Reactant and Theoretical Yield

When reactants are not present in exact stoichiometric proportions, the limiting reactant is the one that is completely consumed first, thus determining the maximum amount of product (the theoretical yield).

• To identify the limiting reactant, calculate the amount of product each reactant can produce; the smallest value indicates the limiting reactant.

Example: Limiting Reactant with Molecules

• Reaction: $\mathrm{CH}_4 + 2\ \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\ \mathrm{H}_2\mathrm{O}$

• Given: 5 CH4 molecules and 8 O2 molecules.

• From CH4: $5\ \text{CH}_4 \times \frac{1\ \text{CO}_2}{1\ \text{CH}_4} = 5\ \text{CO}_2$

• From O2: $8\ \text{O}_2 \times \frac{1\ \text{CO}_2}{2\ \text{O}_2} = 4\ \text{CO}_2$

• O2 is limiting; theoretical yield is 4 CO2 molecules.

Example: Limiting Reactant with Masses

• Reaction: $2\ \mathrm{NO}(g) + 5\ \mathrm{H}_2(g) \rightarrow 2\ \mathrm{NH}_3(g) + 2\ \mathrm{H}_2\mathrm{O}(g)$

• Given: 86.3 g NO and 25.6 g H2

• Convert to moles: $86.3\ \text{g NO} \times \frac{1\ \text{mol}}{30.01\ \text{g}} = 2.8757\ \text{mol NO}$

• $25.6\ \text{g H}_2 \times \frac{1\ \text{mol}}{2.02\ \text{g}} = 12.6733\ \text{mol H}_2$

• Calculate NH3 from each:

  • From NO: $2.8757\ \text{mol NO} \times \frac{2\ \text{mol NH}_3}{2\ \text{mol NO}} = 2.8757\ \text{mol NH}_3$

  • From H2: $12.6733\ \text{mol H}_2 \times \frac{2\ \text{mol NH}_3}{5\ \text{mol H}_2} = 5.0693\ \text{mol NH}_3$

• NO is limiting; theoretical yield is 2.8757 mol NH3.

• Convert to grams: $2.8757\ \text{mol} \times 17.03\ \text{g/mol} = 49.0\ \text{g NH}_3$

• Calculate excess H2: $2.8757\ \text{mol NO} \times \frac{5\ \text{mol H}_2}{2\ \text{mol NO}} = 7.1892\ \text{mol H}_2$ used

• Excess: $12.6733 - 7.1892 = 5.4841\ \text{mol H}_2$; $5.4841\ \text{mol} \times 2.02\ \text{g/mol} = 11.07\ \text{g H}_2$

Theoretical Yield, Actual Yield, and Percent Yield

The theoretical yield is the maximum amount of product predicted by stoichiometry. The actual yield is the amount actually obtained from the reaction, which is usually less due to various factors.

• Reasons for lower actual yield:

  • Reversibility of reactions

  • Side reactions

  • Difficulty in recovering products

  • Product loss due to solubility or further reactions

• Percent yield is calculated as:

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

Example: Iron Extraction

• Reaction: $\mathrm{Fe}_2\mathrm{O}_3(s) + 3\ \mathrm{CO}(g) \rightarrow 2\ \mathrm{Fe}(s) + 3\ \mathrm{CO}_2(g)$

• Given: 167 g Fe2O3, 85.8 g CO, actual yield 72.3 g Fe

• Convert to moles:

  • Fe2O3: $167\ \text{g} \times \frac{1\ \text{mol}}{159.7\ \text{g}} = 1.0457\ \text{mol}$

  • CO: $85.8\ \text{g} \times \frac{1\ \text{mol}}{28.011\ \text{g}} = 3.0631\ \text{mol}$

• Fe from Fe2O3: $1.0457\ \text{mol} \times \frac{2\ \text{mol Fe}}{1\ \text{mol Fe}_2\mathrm{O}_3} = 2.0914\ \text{mol Fe}$

• Fe from CO: $3.0631\ \text{mol} \times \frac{2\ \text{mol Fe}}{3\ \text{mol CO}} = 2.0421\ \text{mol Fe}$

• CO is limiting; theoretical yield: $2.0421\ \text{mol} \times 55.85\ \text{g/mol} = 114\ \text{g Fe}$

• Percent yield: $\frac{72.3}{114} \times 100 = 63.5\%$

Solution Concentrations

Molarity (M)

Molarity is the most common unit of concentration, defined as moles of solute per liter of solution.

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

Dilution of Solutions

When diluting a solution, the number of moles of solute remains constant. The relationship is:

$M_1V_1 = M_2V_2$

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

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

• Take 52.10 mL of 0.178 M solution and dilute to 70.26 mL to obtain 0.132 M.

Other Units for Solution Concentration

• Mass percentage: $\frac{\text{mass of component}}{\text{mass of solution}} \times 100$

• Volume percentage: $\frac{\text{volume of component}}{\text{volume of solution}} \times 100$

• Mass-volume percentage: $\frac{\text{mass of solute}}{\text{volume of solution}} \times 100$

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

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

Example: Calculating ppm and ppb

• Given: 200.0 g solution contains 390 mg sodium ion.

• Convert 390 mg to g: $390\ \text{mg} \times \frac{1.00\ \text{g}}{1000\ \text{mg}} = 0.390\ \text{g}$

• ppm: $\frac{0.390}{200.0} \times 10^6 = 1.95 \times 10^3$ ppm

• ppb: $\frac{0.390}{200.0} \times 10^9 = 1.95 \times 10^6$ ppb

Summary Table: Common Solution Concentration Units

Unit	Definition	Formula
Molarity (M)	Moles of solute per liter of solution	$M = \frac{\text{moles}}{\text{liters}}$
Mass %	Mass of component per mass of solution, times 100	$\frac{\text{mass of component}}{\text{mass of solution}} \times 100$
Volume %	Volume of component per volume of solution, times 100	$\frac{\text{volume of component}}{\text{volume of solution}} \times 100$
ppm	Parts per million (mass basis)	$\frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$
ppb	Parts per billion (mass basis)	$\frac{\text{mass of solute}}{\text{mass of solution}} \times 10^9$

Additional info: The notes above include all key steps and concepts for stoichiometry, limiting reactant, theoretical and percent yield, and solution concentration units, as relevant to a general chemistry course.