L2 - Mass Le C

Molecular Mass and Monsieur Le Chatelier Who is Henri Louis Le Chatelier?

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Today’s Quote It is possible to store the mind with a million facts and still be entirely uneducated. - Alec Bourne

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Formula Definitions • Empirical Formula – ratio of each

atom in each molecule of a given compound: Glucose = CH2O

• Molecular Formula – actual number of each of the atoms present in a compound: Glucose = C6H12O6

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Molecular Mass • Molecular Mass = sum of atomic

masses of all the atoms present in one molecule of a compound. • Example: H2O 2 x the atomic mass of H (2 x 1) = 2 1 x the atomic mass of 0 (1 x 16) = 16 Molecular Mass =

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Molecular Mass of Glucose • Formula = C6H12O6 6 Carbons (Atomic Mass of 12) 6 x 12= 72 12 Hydrogens (Atomic Mass of 1) 12x1=12 6 Oxygens (Atomic Mass of 16) 6 x 16= 96 5

Collection Terms 1 trio

=

3 singers

1 six-pack Cola

=

6 cans Cola drink

1 dozen donuts =

12 donuts

1 gross of pencils

=

144 pencils

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The Mole Calculations Using Molar Mass

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A Mole of Particles Contains 6.02 x 1023 particles 1 mole C

= 6.02 x 1023 C atoms

1 mole H2O = 6.02 x 1023 H2O molecules 1 mole NaCl and

= 6.02 x 1023 Na+ ions 8

More Examples of Moles Moles of elements 1 mole Mg

= 6.02 x 1023 Mg atoms

1 mole Au

= 6.02 x 1023 Au atoms

Moles of compounds 1 mole NH3

= 6.02 x 1023 NH3 molecules

1 mole C9H8O4 = 6.02 x 1023 aspirin molecules

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Avogadro's Number 6.02 x 1023 particles 1 mole or 1 mole 6.02 x 1023 particles 10

• An education isn't how much you

have committed to memory, or even how much you know. It's being able to differentiate between what you do know and what you don't. • Anatole France (1844 - 1924)

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What is the….? 1. Number of atoms in 0.500 mole of Al 1) 500 Al atoms 2) 6.02 x 1023 Al atoms 3) 3.01 x 1023 Al atoms

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2.Number of moles of S in 1.8 x 1024 S atoms 1) 1.0 mole S atoms 2) 3.0 mole S atoms 3) 1.1 x 1048 mole S atoms

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Solution 1. Number of atoms in 0.500 mol of Al 3) 3.01 x 1023 Al atoms 2. Number of moles of S if a sample of S contains 1.8 x 1024 S atoms 2) 3.0 mole S atoms

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Molar Mass • Number of grams in 1 mole • Equal to the numerical value of the atomic mass 1 mole of C atoms

=

12.0

1 mole of Mg atoms =

24.3 g

1 mole of Cu atoms

63.5 g

g =

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Molar Mass of Compounds Mass in grams of 1 mole equal numerically to the sum of the atomic masses 1 mole of CaCl2 = 111.1 g/mole 1 mole Ca x 40.1 g/mole + 2 moles Cl x 35.5 g/mole 1 mole of N2O4 = 92.0 g/mole 2 moles N x 14.0 g/mole

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Calculations A. 1 mole of K2O

= 94.2 g

2 K x 39.1 g/mole + 1 O x 16.0 g/mole B. 1 mole of antacid Al(OH)3 = 78.0 g 1 Al x 27.0 g/mole + 3 O x 16.0 g/mole

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Calculation Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. It has a molar mass of 1) 40.0 g/mole 2) 262 g/mole 3) 309 g/mole 18

Solution Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. It has a molar mass of 3) 309 g/mole 17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) 19

Molar Mass Factors for CH4

Methane CH4 known as natural gas is used in gas stoves and gas heaters. Express the molar mass of methane in the form of conversion factors. Molar mass of CH4 = 16.0 g CH4

and

16.0 g 1 mole

CH4 1 mole CH4

16.0 20

Calculations with Molar Mass molar mass Grams

Moles

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Moles and Grams Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al

? g Al

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1. Molar mass of Al 1 mole Al = 27.0 g Al 2. Conversion factors for Al 27.0g Al or 1 mol Al 1 mol Al 27.0 g Al 3. Setup 3.00 moles Al x

Answer

27.0 g Al 1 mole Al = 81.0 g Al 23

Reaction Energy Pathway • Shows the change in energy during a chemical reaction

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Exothermic Reaction • Reaction that

releases energy • Products have lower PE than reactants

energy released

2H2(l) + O2(l) → 2H2O(g) + energy

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Endothermic Reaction • Reaction that

absorbs energy • Reactants have lower PE than products

energy absorbed

2Al2O3 + energy → 4Al + 3O2 26

Collision Theory • Reaction rate depends on the

collisions between reacting particles

• Successful collisions occur if the particles...

– collide with each other – have the correct orientation – have enough kinetic energy to break bonds

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Collision Theory • Particle Orientation

Required Orientation

Successful Collision

Unsuccessful Collisions 28

Collision Theory • Activation Energy (Ea) – minimum energy required for a reaction to occur

Activation Energy

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Collision Theory • Activation Energy – depends on reactants – low Ea = fast rxn rate Ea

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Factors Affecting Rxn Rate • Surface Area – – –

high SA = fast rxn rate more opportunities for collisions Increase surface area by… • using smaller particles • dissolving in water

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Factors Affecting Rxn Rate • Concentration – –

high conc = fast rxn rate more opportunities for collisions

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Factors Affecting Rxn Rate • Temperature – high temp = fast rxn rate – high Kinetic Energy of particles • fast-moving particles • more likely to reach activation energy

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Factors Affecting Rxn Rate • Temperature

5 mph “fender bender”

50 mph “high-speed crash”

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Factors Affecting Rxn Rate • Catalyst – substance that increases rxn rate without being consumed in the rxn – lowers the activation energy

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Factors Affecting Rxn Rate • Enzyme Catalysis

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Ways to change rate of reaction: 1. Change concentration 2. Change temperature 3. Add a catalyst 4. Increase the surface area

These actions also change the rate of reaction of an equilibrium reaction. A+B C+D 37

The four most commons changes to make for equilibrium reactions are: 1. Concentration changes for reactants 2. Concentration changes for products 3. Temperature changes for reaction 4. Volume changes for gaseous reactions

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Concentration changes for Reactants and Products

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Le Chatelier’s Principle

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Think of equilibrium as a big seesaw. At equilibrium, the seesaw is balanced. A + B

C + D

To the eye, no changes are occurring to the amount of reactants on the left or to the amount of product on the right. 41

When you increase the concentration of a reactant A, you are adding weight to the left side of the seesaw. A + B

C + D

How can you re-balance the seesaw? How can you achieve equilibrium again?

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By shifting some of the weight toward the right! A + B A + B

C + D C + D

Equilibrium has been re-established when concentrations stop changing.

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Before A was added, the system was at equilibrium. A + B

C + D

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At the moment that A was added, the [A] went up. In our example, increased [A] is symbolized as more weight on the left side of the seesaw.

A + B

C + D

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In the process of re-establishing equilibrium, the concentration of C and D went up. In our example, increased concentration is symbolized as more weight on the right side of the seesaw.

A + B A + B

C + D C + D

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Let’s look at the overall A + Bprocess C +one D more time. System was at equilibrium. [A] increased. System not at equilibrium.

A + B

System regains equilibrium.

A + B

C + D

C + D

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What would the seesaw look like if we increased the [D]? A + B

C + D

In which direction do we need to shift “weight” in order to regain equilibrium? Shift to the LEFT!

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A + B

C + D

Caused by increasing [A]. Shift RIGHT to regain equilibrium.

Result: [C] and [D] increase. A + B

C + D

Caused by increasing [D].

Shift LEFT to regain equilibrium. Result: [A] and [B] increase. 49

What if we removed D as it was formed? This would be the same as decreasing [D]. What would the seesaw look like? A + B

C + D

When you decrease the [D], you are removing “weight” from that side of the seesaw. 50

How would you re-establish equilibrium? A + B

C + D

Shift “weight” to the right. More products will form. This is a common way to make an equilibrium reaction go to “completion.” 51

Change in Temperature

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What happens when you change the temperature of a reaction? It will depend on whether the reaction is exothermic or endothermic. Exothermic reaction:

A + B <--> C + D + heat

Endothermic reaction:

A + B + heat <--> C + D

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Treat “heat” like a reactant or product. If you increase the heat, you are adding “weight” to the seesaw. If you decrease the heat, you are removing “weight from the seesaw. If the reaction is exothermic, the change in “weight” occurs on the product side. If the reaction is endothermic, the change in “weight” occurs on the reactant side. 54

Exothermic Reaction with increased temperature. A + B

C + D + Heat Shift Left to Produce More Reactants

Exothermic Reaction with decreased temperature. A + B

C + D + heat Shift Right to Produce More Products 55

Endothermic Reaction with increased temperature. A + B + heat

C + D Shift Right! More Products

Endothermic Reaction with decreased temperature. A + B + heat

C + D

Shift Left! More Reactants 56

Changing Volume

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When you have gaseous reactants or products and you change volume, you are changing Concentration. .

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Let’s look at a reaction with gaseous components: 2A(g) + B(g) <--> 2C(g) There are THREE moles of gas on the reactant side and TWO moles of gas on the product side. 3 moles

2 moles

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If we cut the volume in half, the concentration will double. This means that the concentration of ALL gases went up. The side of the reaction with the most moles of gas, will be most disturbed by the increased concentration.

3 moles

2 moles

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If we shift the reaction toward the side with fewer moles of gas, the effect of cutting the volume in half will be minimized. For this reaction, cutting the volume in half results in MORE product.

3 3 moles moles

2 moles 2 moles

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Is it always true that cutting the volume in half will cause more products to form? NO! You have to examine each reaction with gaseous components to see, first, which side has more moles of gas. Cutting the volume in half, increases concentration. Reaction will shift toward side with FEWER moles of gas. Doubling the volume, decreases concentration. Reaction will shift toward side with MOST moles of gas. 62

3A(s) + B(g) <--> 2C(g) 1 mole of gas

2 moles of gas

If volume is increased, which direction will reaction shift? If volume increases, concentration decreases. The side with most moles has the greatest reduction in concentration. Product side loses “weight.” 2 moles 1 mole Reaction will shift to the right. 63

Let’s see if you can put it all together,

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2 A(g) + B(g) + heat <--> C(g) 1. In which direction will reaction shift if A is doubled? Increase [A], shift right. More products formed. 2. In which direction will reaction shift if C is removed? Decrease [C], shift right. More products formed. 3. In which direction will reaction shift if temp goes up? Endothermic reaction. Heat is a reactant. Shift right. More products formed. 4. In which direction will reaction shift if volume goes up? Increase V, decrease A, B and C. Side with most gas moles loses “weight.” Shift left. More reactants formed. 65