CH 4-4: Chemical Reactions and Stoichiometry: Yields, Concentration, Dilution, and Solution Stoichiometry

Chapter 4: Chemical Reactions and Stoichiometry

Yields in Chemical Reactions

In chemical reactions, the concept of yield is used to quantify the efficiency of a reaction. There are several types of yields that chemists consider:

• Theoretical Yield: The maximum amount of product that can be produced from a given amount of reactants, assuming perfect conditions and complete conversion. It is calculated based on stoichiometry.

• Actual Yield: The amount of product actually obtained from a chemical reaction, which is often less than the theoretical yield due to losses, side reactions, or incomplete reactions.

• Percentage Yield: A measure of the efficiency of a reaction, calculated as the ratio of actual yield to theoretical yield, multiplied by 100.

Formula for Percentage Yield:

$\text{\% yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100$

Example: If the theoretical yield of P4 is 1.20 g and the actual yield obtained is 0.90 g, then:

$\text{\% yield} = \left( \frac{0.90\ \text{g}}{1.20\ \text{g}} \right) \times 100 = 75\%$

Concentration of Solutions

Concentration describes how much solute is dissolved in a given quantity of solvent or solution. The most common unit of concentration in chemistry is molarity (M), defined as moles of solute per liter of solution.

• Molarity (M): $M = \frac{n}{V}$, where n is the number of moles of solute and V is the volume of solution in liters.

• Units: mol L-1 or mol/dm3 (1 L = 1 dm3)

Example: To prepare 1.00 L of 0.150 mol L-1 AgNO3 solution:

$n_{\text{AgNO}_3} = 0.150\ \text{mol L}^{-1} \times 1.00\ \text{L} = 0.150\ \text{mol}$ $m_{\text{AgNO}_3} = 0.150\ \text{mol} \times 169.88\ \text{g mol}^{-1} = 25.5\ \text{g}$

Ion Concentrations: When ionic compounds dissolve, they dissociate into their constituent ions. For example:

• 1 mol L-1 KCl(aq) contains 1 mol L-1 K+(aq) and 1 mol L-1 Cl-(aq)

• 1 mol L-1 Ca(NO3)2(aq) contains 1 mol L-1 Ca2+(aq) and 2 mol L-1 NO3-(aq)

Dilution of Solutions

Dilution is the process of reducing the concentration of a solution by adding more solvent. Importantly, dilution does not change the number of moles of solute present; it only increases the total volume.

• Key Principle: The number of moles of solute before and after dilution remains constant.

• Formula: $M_i V_i = M_f V_f$ Where Mi and Vi are the initial molarity and volume, and Mf and Vf are the final molarity and volume after dilution.

Example: To prepare 0.500 L of 0.100 mol L-1 K2CrO4 from a 1.75 mol L-1 stock solution:

$M_i V_i = M_f V_f$ $1.75\ \text{mol L}^{-1} \times V_i = 0.100\ \text{mol L}^{-1} \times 0.500\ \text{L}$ $V_i = 0.0286\ \text{L} = 28.6\ \text{mL}$

Steps for Dilution:

1. Measure the required volume of concentrated solution.

2. Transfer to a volumetric flask.

3. Add solvent (usually water) up to the calibration mark.

4. Mix thoroughly to ensure uniform concentration.

Solution Stoichiometry

Stoichiometry in solutions involves using molarity and volume to determine the amounts of reactants and products in chemical reactions occurring in solution. The same principles of stoichiometry (moles, limiting reactants, yields) apply, but concentrations and volumes are used instead of masses.

• Key Steps:

  1. Write the balanced chemical equation.

  2. Calculate moles of reactants using $n = M \times V$.

  3. Use stoichiometric ratios to find moles of products or other reactants.

  4. Convert moles to mass or concentration as required.

Example: The reaction of butanethiol (C4H9SH) with NaOCl to deodorize skunk odor:

$2\ \mathrm{C_4H_9SH} (l) + \mathrm{NaOCl} (aq) \rightarrow \mathrm{C_4H_9S-SC_4H_9} (l) + \mathrm{NaCl} (aq) + \mathrm{H_2O} (l)$

Given 5.00 mL of 0.0985 mol L-1 NaOCl, calculate the mass of butanethiol that can be deodorized:

• Moles NaOCl: $n = 0.00500\ \text{L} \times 0.0985\ \text{mol L}^{-1} = 4.925 \times 10^{-4}\ \text{mol}$

• Stoichiometry: $2\ \text{mol NaOCl} : 1\ \text{mol butanethiol}$

• Moles butanethiol: $4.925 \times 10^{-4}\ \text{mol} \times \frac{1}{2} = 2.463 \times 10^{-4}\ \text{mol}$

• Mass butanethiol: $2.463 \times 10^{-4}\ \text{mol} \times 90.18\ \text{g mol}^{-1} = 0.0222\ \text{g}$

Ion Concentration Example: Mixing solutions of Ca(NO3)2 and NaNO3:

• Calculate moles of each ion using $n = M \times V$ for each solution.

• Add moles from each source for ions present in both solutions.

• Divide total moles by total volume to get final concentration.

Beaker Concentration Comparison: When comparing solutions visually, concentration is determined by the number of solute particles per unit volume. For example, a beaker with 10 particles in 1 unit volume has a higher concentration than one with 5 particles in 1 unit volume.

Summary Table: Key Solution Concepts

Concept	Definition	Key Formula	Example
Theoretical Yield	Maximum possible amount of product from given reactants	Stoichiometry from balanced equation	1.20 g P4
Actual Yield	Amount of product actually obtained	Measured experimentally	0.90 g P4
Percentage Yield	Efficiency of reaction	$\text{\% yield} = \left( \frac{\text{Actual}}{\text{Theoretical}} \right) \times 100$	75%
Molarity (M)	Moles of solute per liter of solution	$M = \frac{n}{V}$	0.150 mol L-1 AgNO3
Dilution	Decreasing concentration by adding solvent	$M_i V_i = M_f V_f$	28.6 mL of stock to make 0.500 L diluted solution