Stoichiometry

Stoichiometry Overview

Chapter 5a Overview

• Key concepts in stoichiometry include:

  • Limiting Reagent

  • Percent Yield

Introduction to Stoichiometry

• Stoichiometry is the method of calculating the proportions of reactants and products in chemical reactions.

• Example:

  • Recipe for cookies:

  • 2 ¼ cups Gold Medal all-purpose flour

  • 1 cup butter, softened

  • 1 ½ cup sugar

  • 1 egg

  • 2 cups semisweet chocolate chips

  • 1 teaspoon baking soda

  • ½ teaspoon salt

  • 1 teaspoon vanilla

  • Makes 24 cookies

  • Questions related to stoichiometry:

  • What if I only had 1 cup of chocolate chips?

  • What if I wanted to make 48 cookies?

  • If I had 4 ½ cups flour, 3 cups butter, 2 cups sugar, 1 egg, and 8 cups chocolate chips?

Fundamental Stoichiometric Concepts

Chemical Reactions

• Example of a chemical reaction:

  • $ext{Pb(NO}<em>3 ext{)}</em>2(aq) + 2 ext{NaCl}(aq) <br>ightarrow ext{PbCl}<em>2(s) + 2 ext{NaNO}</em>3(aq)$

• Important to note:

  • We can weigh reactants in grams, but stoichiometric calculations involve molecules/moles.

  • Relation on a molecular level:

  • 1 molecule Pb(NO3)2 + 2 molecules NaCl → 1 molecule PbCl2 + 2 molecules NaNO3

• Conversion to moles:

  • $1 ext{ mol Pb(NO}<em>3 ext{)}</em>2 + 2 ext{ mol NaCl} <br>ightarrow 1 ext{ mol PbCl}<em>2 + 2 ext{ mol NaNO}</em>3$

• Using Avogadro's number:

  • $6.022 imes 10^{23}$ molecules.

Mole Calculations

Definition of Stoichiometry

• Stoichiometry allows calculation of the ratio of reactants and products in a reaction.

Example Reaction: Water Formation

• $2 ext{H}<em>2 + ext{O}</em>2 ightarrow 2 ext{H}_2 ext{O}$

  • For every 2 molecules of hydrogen, 1 molecule of oxygen forms 2 molecules of water.

• Calculation example:

  • From 3 mol O₂, how many mol H₂O can be formed?

  • Calculation:

  • $ext{mol O}<em>2 imes rac{2 ext{mol H}</em>2 ext{O}}{ ext{mol O}<em>2} = 3 ext{mol O}</em>2 imes 2 = 6 ext{ mol H}_2 ext{O}$

Additional Mole Calculation Example

• Given:

  • 1.42 mol NH₃ reacting with sufficient phosphoric acid.

• Reaction: $3 ext{NH}<em>3 + ext{H}</em>3 ext{PO}<em>4 ightarrow ( ext{NH}</em>4)<em>3 ext{PO}</em>4$

• Need to find mol of ammonium phosphate produced.

Practice Problem:

• Given reaction: $ext{KIO}<em>3 + 5 ext{KI} + 6 ext{HCl} ightarrow 3 ext{I}</em>2 + 3 ext{H}_2 ext{O} + 6 ext{KCl}$

• Questions:

  • How many mol of P needed to react with 6 mol Cl₂?

  • How many mol of I₂ from 4 mol HCl reacting with sufficient KIO3 and KI?

Mass Calculations

Mass of Ammonium Phosphate

• Reaction: $3 ext{NH}<em>3 + ext{H}</em>3 ext{PO}<em>4 ightarrow ( ext{NH}</em>4)<em>3 ext{PO}</em>4$

• Given mass of ammonia: 12.4 g.

  • Find mass of ammonium phosphate produced.

Practice Problems

1. For PCl₅ + 4 H₂O → 5 HCl + H₃PO₄, calculate the mass of water consumed by 20.83 g of phosphorous pentachloride.

2. For 8 HNO₃ + 3 Cu → 2 NO + 3 Cu(NO₃)₂ + 4 H₂O, calculate the mass of NO produced from 14.0 g of Cu.

Limiting Reactant

Concept of Limiting Reactant

• The limiting reactant determines the extent of a chemical reaction, regulating how much of the reactants can be used and how much product can form.

• Example with cookie recipe:

  • If only sufficient eggs are available, the recipe cannot produce more cookies than dictated by the egg quantity despite having other ingredients.

Reaction Example:

• Reaction: $ext{N}<em>2(g) + 3 ext{H}</em>2(g) ightarrow 2 ext{NH}_3(g)$

  • Importance of identifying limiting reactant to inform stoichiometric calculations.

• Example Calculation:

  • Given 1.5 mol of CH₃OH and 0.5 mol of O₂:

  • Reaction: $2 ext{CH}<em>3 ext{OH}(g) + 3 ext{O}</em>2(g) <br>ightarrow 2 ext{CO}<em>2(g) + 4 ext{H}</em>2O(g)$

  • Questions:

  1. Is the chemical equation balanced?

  2. Identify limiting reactant based on given moles.

Practice Problem with NaHCO₃ and H₂SO₄

• For the reaction: $ext{H}<em>2 ext{SO}</em>4 + 2 ext{NaHCO}<em>3 ightarrow ext{Na}</em>2 ext{SO}<em>4 + 2 ext{H}</em>2 ext{O} + 2 ext{CO}_2$

  • Given: 14.4 g NaHCO₃ and 4.9 g H₂SO₄, determine mass of Na₂SO₄ produced.

Theoretical vs Actual Yield

Definitions:

• Theoretical Yield: This is the maximum amount of product that can be produced based on stoichiometric calculations.

• Actual Yield: This is the amount of product actually obtained from a chemical reaction in the lab.

• Percent Yield: The efficiency of a reaction, expressed mathematically as:
  ext{Percent Yield} = rac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100 ext{%}

Example of Yield Calculation:

• If the theoretical yield is 7.44 g and the actual yield is 7.02 g, then:

  • Calculation: ext{Percent Yield} = rac{7.02 ext{ g}}{7.44 ext{ g}} imes 100 ext{%} = 94.4 ext{%}

• Practical implications of yield issues:

  • Possible reasons for yields less than 100% include:

  1. Incomplete reactions

  2. Side reactions

  3. Incomplete recovery of product

Revisiting Yield Calculation with Sulfuric Acid Reaction:

• For the reaction: $H<em>2SO</em>4 + 2 ext{NaHCO}<em>3 ightarrow ext{Na}</em>2 ext{SO}<em>4 + 2 ext{H}</em>2 ext{O} + 2 ext{CO}_2$

  • If the actual yield is 6.42 g Na₂SO₄, calculate the percentage yield from theoretical yield calculations.