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 should 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 mole ratios of reactants and products, which are used to calculate the quantities involved in a reaction.

• Example: $2\ \mathrm{KClO_3}(s) \rightarrow 2\ \mathrm{KCl}(s) + 3\ \mathrm{O_2}(g)$

• This equation tells us that 2 moles of potassium chlorate produce 2 moles of potassium chloride and 3 moles of oxygen gas.

• Analogous to a recipe: just as a pizza recipe specifies the ratio of ingredients, a chemical equation specifies the ratio of reactants and products.

Stoichiometric Calculations: Step-by-Step

To determine the amount of product formed or reactant required, follow these steps:

1. Balance the chemical equation.

2. Convert all given quantities to moles.

3. Use mole ratios from the balanced equation to relate reactants and products.

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

Example: Combustion of octane

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

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

• Calculation: $11.0\ \mathrm{mol}\ \mathrm{C_8H_{18}} \times \frac{16\ \mathrm{mol}\ \mathrm{CO_2}}{2\ \mathrm{mol}\ \mathrm{C_8H_{18}}} = 88.0\ \mathrm{mol}\ \mathrm{CO_2}$

• To convert to grams: $88.0\ \mathrm{mol}\ \mathrm{CO_2} \times 44.011\ \mathrm{g/mol} = 3.87 \times 10^3\ \mathrm{g}\ \mathrm{CO_2}$ (3 significant figures)

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 (theoretical yield) that can be formed.

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

• The excess reactant is the one that remains after the reaction is complete.

Example: Synthesis of ammonia from NO and H2

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

• Given: 86.3 g NO and 25.6 g H2

• Convert to moles: $86.3\ \mathrm{g}\ \mathrm{NO} \times \frac{1\ \mathrm{mol}}{30.01\ \mathrm{g}} = 2.88\ \mathrm{mol}\ \mathrm{NO}$

• $25.6\ \mathrm{g}\ \mathrm{H_2} \times \frac{1\ \mathrm{mol}}{2.02\ \mathrm{g}} = 12.67\ \mathrm{mol}\ \mathrm{H_2}$

• Calculate moles of NH3 from each: NO yields 2.88 mol NH3, H2 yields 5.07 mol NH3

• NO is limiting; theoretical yield = 2.88 mol NH3

• Convert to grams: $2.88\ \mathrm{mol}\ \mathrm{NH_3} \times 17.03\ \mathrm{g/mol} = 49.0\ \mathrm{g}\ \mathrm{NH_3}$

Percent Yield

The percent yield compares the actual amount of product obtained to the theoretical yield, indicating the efficiency of a reaction.

• Formula: $\%\ \mathrm{yield} = \left( \frac{\mathrm{actual\ yield}}{\mathrm{theoretical\ yield}} \right) \times 100$

• Actual yield is usually less than theoretical yield due to incomplete reactions, side reactions, or loss of product.

Example: Iron production from iron oxide

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

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

• Find limiting reagent and theoretical yield (see above for calculations): theoretical yield = 114 g Fe

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

Solution Concentrations

Molarity (M)

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

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

Dilution of Solutions

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

• Formula: $M_1V_1 = M_2V_2$

• Example: To dilute 52.10 mL of 0.178 M HCl to 0.132 M, calculate final volume:

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

Other Units for Solution Concentration

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

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

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

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

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

Example: Calculating ppm and ppb

• Given: 390 mg Na+ in 200.0 g solution

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

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

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

Summary Table: Common Solution Concentration Units

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

Additional info: In practice, solution concentrations can also be expressed in molality (moles of solute per kilogram of solvent) and normality (equivalents per liter), but these are less common in introductory general chemistry.