topic 13

The Mole

1. Introduction

• All molecules in this video were generated using HyperChem by HYPERCUBE, INC

2. Overview of Content Covered

• We have previously focused on atoms and their behaviors in Topics 1-12.

• This section will return to chemical reactions, concentrating on quantifying the amounts of reactants and products.

• Introduction to Stoichiometry, a mathematically intensive unit involving calculations and algebraic thinking.

• For additional help, refer to the MATH READER.

3. Fundamental Concepts

A. Chemical Reactions

• Chemical reactions are represented by chemical equations in the format:
  REACTANT(S) -> PRODUCT(S)

• The law of mass conservation states that the number of each type of atom must be the same on both sides of a balanced equation.

• Coefficients in the equation change to achieve balance.

• A balanced chemical equation acts as a recipe for a chemical reaction.

B. Molecule Ratios in Reactions

• Example question: How many molecules of H$2$ are needed to react with 18 molecules of O$2$? The reaction is represented as: 2H2(g) + 1O2(g) -> 2H_2O(l)

  • The combining ratio of reactants to products is established through the balanced equation.

  • The ratio of H$2$: O$2$: H$_2$O is 2:1:2.

• Calculation example: For 9.2 x 10$^{11}$ molecules of O$2$, you can expect to produce 18.4 x 10$^{11}$ molecules of H$2$O.

4. Understanding the Mole

A. Concept of the Mole

• The mole is a unit representing a very large number, making it impractical to think in terms of individual atoms or molecules.

• Background: In a laboratory, reactions occur on a scale involving at least 10$^{16}$ molecules.

• The mole allows scientists to work with grouped units of particles.

B. Practical Application of the Mole

• 1 mole is defined as: $1 ext{ mole} = 6.022 imes 10^{23} ext{ particles}$

  • This is known as Avogadro's number.

• Every equation includes a fixed mole ratio that can be utilized for stoichiometric calculations.

• The mole (mol) is the fundamental grouping unit in chemistry.

• In the reaction: $2H<em>2 + 1O</em>2 ightarrow 2H_2O$

  • 2 moles of H$2$ react with 1 mole of O$2$ to produce 2 moles of H$_2$O.

5. Practice Problems

A. Practice Problem 1

• Scenario: Magnesium reacts with oxygen to produce magnesium oxide. If you start with 3.2 moles of Mg, how many moles of O$_2$ do you need to react completely?

• Steps to Solution:

  1. Write the balanced chemical equation:
     $2Mg(s) + O_2(g) <br>ightarrow 2MgO(s)$

  2. Calculate moles of O$2$: $ext{Moles of O}</em>2 = 3.2 imes (1/2) = 1.6 ext{ moles of O}_2$

B. Practice Problem 2

• Given: 44.6 moles of C$3$H$8$. Find how many moles of CO$2$ are produced when it reacts with O$2$.

  • Reaction:
    $C<em>3H</em>8 + 5O<em>2 ightarrow 3CO</em>2 + 4H_2O$

  • Calculation:
    $ext{Moles of CO}<em>2 = 44.6 imes (3/1) = 133.8 ext{ moles of CO}</em>2$

C. Practice Problem 3

• If 1.87 moles of diphosphorous pentoxide reacts with water, how many moles of H$_2$O are consumed?

  • Reaction:
    $P<em>2O</em>5(s) + 3H<em>2O(l) ightarrow 2H</em>3PO_4(aq)$

  • Calculation:
    $ext{Moles of H}<em>2O = 1.87 imes (3) = 5.61 ext{ moles of H}</em>2O$

6. Avogadro’s Number and Molar Mass

A. Definition

• The mole represents not just a count of particles but also correlates to a specific mass of a substance which is called molar mass.

• 1 mole of a substance will have a definite number of grams based upon its atomic mass from the periodic table.

• Example: 1 mole of Carbon = 12.01 grams; 1 mole of Oxygen = 16.00 grams.

B. Use of Molar Mass in Calculations

• Molar mass is essential for converting between moles, grams, and particles.

• The formula:
  Number of grams = Number of moles x Molar mass (g/mol).

C. Example Calculation of Molar Mass

• Water (H$_2$O) calculation:

  • Molar Mass: 2(H) + 1(O) = 2(1.008 g/mol) + 1(16.00 g/mol) = 18.016 g/mol.

• For a compound like C${16}H{16}F_3NO:

  • Molar Mass: = 16(12.01 g/mol) + 16(1.008 g/mol) + 3(19.00 g/mol) + 1(14.01 g/mol) + 1(16.00 g/mol) = 295.3 g/mol.

7. Recap

A. The Significance of the Mole

1. Represents a fixed number of particles (6.022 x 10$^{23}$) and a specific mass of substance.

2. The shape of stoichiometric equations helps to calculate amounts in chemical reactions.

B. Further Exploration

• Practice additional problems surrounding mole conversions and use of molar mass in various scenarios.

8. Conclusion

• Understanding the mole and its implications is crucial in both chemical calculations and in comprehending the fundamental nature of matter in reactions and compounds.