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It’s time to learn about. the Mole !. What\'s The Difference Between Roast Beef And Pea Soup?. Anyone Can Roast Beef. Stoichiometry. Stoichiometry : Mole Ratios to Determining Grams of Product At the conclusion of our time together, you should be able to:.

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Stoichiometry : Mole Ratios to Determining Grams of ProductAt the conclusion of our time together, you should be able to:

  • Review the conversion of particles or grams to moles
  • Determine mole ratios from a balanced chemical equation
  • Determine the amount of product produced when given the amount of reactants
review the molar mass of compounds
Review the Molar Mass of Compounds
  • The molar mass (MM) of a compound is determined by adding up all the atomic masses for the molecule (or compound)
    • Ex. Molar mass of CaCl2
    • Avg. Atomic mass of Calcium = 40.08g
    • Avg. Atomic mass of Chlorine = 35.45g
    • Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl2

20

Ca40.08

17Cl 35.45

practice
Practice
  • Calculate the Molar Mass of calcium phosphate
    • Formula =
    • Masses elements:
    • Molar Mass =

310.18 g

Ca3(PO4)2

slide7

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calculations
Calculations

molar mass Avogadro’s number Grams Moles Particles

Everything must go through Moles!!!

flowchart
Flowchart

Atoms or Molecules

Divide by 6.02 X 1023

Multiply by

6.02 X 1023

Multiply by atomic/molar mass from periodic table

Moles

Divide by atomic/molar mass from periodic table

Mass (grams)

chocolate chip cookies
Chocolate Chip Cookies!!

1 cup butter 2 eggs

1/2 cup white sugar 1 teaspoon salt

1 cup packed brown sugar

1 teaspoon vanilla extract

2 1/2 cups all-purpose flour

1 teaspoon baking soda

2 cups semisweet chocolate chips

Makes 3 dozen

How many eggs are needed to make 3 dozen cookies?

How much butter is needed for the amount of chocolate chips used?

How much brown sugar would I need if I had 1½ cups white sugar?

cookies and chemistry huh
Cookies and Chemistry…Huh!?!?
  • Just like chocolate chip cookies have recipes, chemists have recipes as well
  • Instead of calling them recipes, we call them chemical equations
  • Furthermore, instead of using cups and teaspoons, we use moles
  • Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients
chemistry recipes
Chemistry Recipes
  • Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe)
  • Be sure you have a balanced reaction before you start! Balance #1 on your worksheet!!
      • Example: 2 Ag+ Cl2  2 AgCl
      • This reaction tells us that by mixing 2 moles of silver with 1 mole of chlorine we will get 2 moles of silver chloride
      • What if we wanted 4 moles of AgCl?
      • 10 moles?
      • 50 moles?
practice1
Practice
  • Write the balanced reaction for hydrogen gas reacting with oxygen gas.

2 H2 + O2 2 H2O

    • How many moles of reactants are needed?
    • What if we wanted 4 moles of water?
    • What if we had 3 moles of oxygen, how much hydrogen would we need to react, and how much water would we get?
    • What if we had 50 moles of hydrogen, how much oxygen would we need, and how much water would be produced?
mole ratios
Mole Ratios
  • These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical
  • Example: How many moles of chlorine are needed to react with 5 moles of silver (without any silver left over)?

2 Ag + Cl2 2 AgCl

5 Ag + 2.5 Cl2 5 AgCl

question answer

x 1 mol Cl2

2 mol Ag

Question Answer

2 Ag+ Cl2 2 AgCl

# moles of Cl2 =

5 mol Ag

= 2.5 mol Cl2

mole mole conversions

2 mol AgCl

1 mol Cl2

Mole-Mole Conversions
  • How many moles of silver chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of silver metal?

2 Ag + Cl2 2 AgCl

# moles of AgCl =

2.6 mol Cl2 x

= 5.2 mol AgCl

mass mole
Mass-Mole
  • We can also start with mass and convert to moles of product or another reactant
  • We use molar mass and the mole ratio to get to moles of the compound of interest
    • Calculate the number of moles of the combustion of ethane (C2H6) needed to produce 10.0 g of water
    • 2 C2H6 + 7 O2 4 CO2 + 6 H20
mass mole1

x 2 mol C2H6

6 mol H2O

x 1 mol H2O

18.02 g H2O

Mass-Mole
  • 2 C2H6 + 7 O2 4 CO2 + 6 H20

10.0 g H2O

= 0.185 mol C2H6

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Stoichiometry : Mole Ratios to Determining Grams of ProductAt the conclusion of our time together, you should be able to:

  • Review the conversion of particles or grams to moles
  • Determine mole ratios from a balanced chemical equation
  • Determine the amount of product produced when given the amount of reactants
mass mass conversions
Mass-Mass Conversions
  • Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)
  • Now we must go from grams to moles, to mole ratios, and back to grams of the compound we are interested in.
mass mass conversion
Mass-Mass Conversion
  • Ex. Calculate how many grams of ammonia are produced when you react 2.00 g of nitrogen with excess hydrogen.
  • N2 + 3 H2  2 NH3

2.00 g N2 1 mol N2 2 mol NH3 17.04 g NH3

28.02 g N2 1 mol N2 1 mol NH3

= 2.43 g NH3

practice mass mole mass
Practice Mass-Mole-Mass
  • How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?
  • Ca + N2 Ca3N2
  • 3 Ca + N2 Ca3N2
mass mole mass

x 1 mol Ca3N2

x 148.26 g Ca3N2

3 mol Ca

1 mol Ca3N2

x 1 mol Ca

40.08 g Ca

Mass-Mole-Mass
  • 3 Ca + N2 Ca3N2

2.00 g Ca

= 2.47 g Ca3N2

stoichiometry mole ratios to determining grams of product let s see if you can

Stoichiometry : Mole Ratios to Determining Grams of ProductLet’s see if you can:

  • Review the conversion of particles or grams to moles
  • Determine mole ratios from a balanced chemical equation
  • Determine the amount of product produced when given the amount of reactants
mass mole mass1

x 3 mol N2

x 28.02 g N2

2 mol NaN3

1 mol N2

x 1 mol NaN3

65.02 g NaN3

Mass-Mole-Mass
  • NaN3 Na + N2
  • Decomposition
  • 2 NaN3 2 Na + 3 N2
  • 65.02 g 22.99 g 28.02 g

94.5 g NaN3

= 61.1 g N2

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