Unit 08 moles and stoichiometry
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Unit 08 – Moles and Stoichiometry. Molar Conversions. A. What is the Mole?. VERY. A large amount!!!!. A counting number (like a dozen) Avogadro’s number ( 6.02  10 23 particles ) (SI unit) 1 mol = molar mass. A. What is the Mole?. HOW LARGE IS IT???.

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Unit 08 – Moles and Stoichiometry

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Unit 08 – Moles and Stoichiometry

  • Molar Conversions


A. What is the Mole?

VERY

A large amount!!!!

  • A counting number (like a dozen)

  • Avogadro’s number (6.02  1023 particles) (SI unit)

  • 1 mol = molar mass


A. What is the Mole?

HOW LARGE IS IT???

  • 1 mole of pennies would cover the Earth 1/4 mile deep!

  • 1 mole of hockey pucks would equal the mass of the moon!

  • 1 mole of basketballs would fill a bag the size of the earth!


B. Molar Mass

  • Molar Mass- the mass of a mole of any element or compound (in grams)

    • Round to 2 decimal places

  • Also called:

    • Gram Formula mass – sum of the atomic masses of all the atoms in a formula of a compound

    • Formula weight


  • B. Molar Mass Examples

    • water

    • sodium chloride

    • H2O

    • (1.01g x 2) + 16.00g = 18.02 g

    • NaCl

    • 22.99g + 35.45g = 58.44 g


    B. Molar Mass Examples

    • sodium hydrogen carbonate

    • sucrose

    • NaHCO3

    • 22.99g + 1.01g + 12.01g + (16.00g x 3) = 84.01 g

    • C12H22O11

    • (12.01g x12) + (1.01g x 22) + (16.00g x11)= 342.34 g


    C. Number of Particles in a Mole

    1 mole = 6.02 × 10 23representative particles

    (also called Avogadro’s Number)

    Atom- rep. particle for most elements

    Ions – if atom is charged

    Molecule- rep. particle for covalent compounds and diatomic molecules “BrINCl HOF”

    Formula unit- rep. particle for ionic compounds

    What is a representative particle?

    How the substance normally exists:


    D. Volume of a Mole of Gas

    • The Volume of a gas varies with a change in temperature or pressure.

    • Measured at standard temperature and pressure (STP)

      • 0°C at 1 atmosphere (atm)

  • 1 mole of any gas occupies a volume of 22.4L


  • The MoleRoad Map

    Molecule

    Atoms (ions)

    Formula unit


    E. Molar Conversion Examples

    • How many moles of carbon are in 26 g of carbon?

    26 g C

    1 mol C

    12.01 g C

    = 2.2 mol C


    E. Molar Conversion Examples

    • How many molecules are in 2.50 moles of C12H22O11?

    6.02  1023

    molecules

    1 mol

    2.50 mol

    = 1.51  1024

    molecules

    C12H22O11


    E. Molar Conversion Examples

    • Find the number of molecules of 12.00 L of O2 gas at STP.

    6.02 x 1023 molecules

    1 mol

    1 mol

    22.4 L

    12.00 L

    = 3.225 x 1023 molecules


    II. Stoichiometric Calculations


    A. Proportional Relationships

    Ratio of eggs to cookies

    • I have 5 eggs. How many cookies can I make?

    2 1/4 c. flour

    1 tsp. baking soda

    1 tsp. salt

    1 c. butter

    3/4 c. sugar

    3/4 c. brown sugar

    1 tsp vanilla extract

    2 eggs

    2 c. chocolate chips

    Makes 5 dozen cookies.

    5 eggs

    5 doz.

    2 eggs

    = 12.5 dozen cookies


    A. Proportional Relationships

    • Stoichiometry

      • mass relationships between substances in a chemical reaction

      • based on the mole ratio

    • Mole Ratio

      • indicated by coefficients in a balanced equation

    2 Mg + O2 2 MgO


    B. Stoichiometry Steps

    1. Write a balanced equation.

    2. Identify known & unknown.

    3. Convert known to moles (IF NECESSARY) Line up conversion factors.

    4. Use Mole ratio – from equation

    5. Convert moles to unknown unit

    (IF NECESSARY)

    6. Calculate and write units.


    C. Stoichiometry Problems

    • Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen.

      N2 + 3H2→ 2NH3


    III. Stoichiometry in the Real World


    A. Limiting Reactants

    • Available Ingredients

      • 4 slices of bread

      • 1 jar of peanut butter

      • 1/2 jar of jelly

    • Limiting Reactant

      • bread

    • Excess Reactants

      • peanut butter and jelly


    A. Limiting Reactants

    • Limiting Reactant

      • used up in a reaction

      • determines the amount of product

    • Excess Reactant

      • added to ensure that the other reactant is completely used up

      • cheaper & easier to recycle


    A. Limiting Reactants

    1. Write a balanced equation.

    2. For each reactant, calculate the amount of product formed.

    3. Smaller answer indicates:

    • limiting reactant

    • amount of product


    A. Limiting Reactants

    Using the following equation identify the limiting reagent.

    • How many moles of ammonia (NH3) can be produced from the reaction of 28.2 L of nitrogen and 25.3 L of hydrogen?

    N2 + 3H2 2NH3

    28.2 L

    ? mol

    25.3 L


    A. Limiting Reactants

    N2 + 3H2 2NH3

    28.2 L

    ? mol

    25.3 L

    28.2

    L N2

    1 mol

    N2

    22.4

    L N2

    2 mol

    NH3

    1 mol

    N2

    = 2.5 mol NH3


    A. Limiting Reactants

    N2 + 3H2 2NH3

    28.2 L

    ? mol

    25.3 L

    25.3

    L H2

    1 mol

    H2

    22.4 L H2

    2 mol

    NH3

    3 mol

    H2

    = 0.753 mol NH3


    A. Limiting Reactants

    N2: 2.5 mol NH3 H2: 0.753 mol NH3

    Limiting reactant: H2

    Excess reactant: N2

    Product Formed: 0.753 mol NH3


    Limiting Reactants

    Mg + 2HCl → MgCl2 + H2

    How many grams of magnesium chloride are produced from the reaction of 2.08 mol of Mg and 2.08 mol of HCl?


    B. Percent Yield

    measured in lab

    calculated on paper


    B. Percent Yield

    • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.

    K2CO3 + 2HCl  2KCl + H2O + CO2

    45.8 g

    ? g

    actual: 46.3 g


    B. Percent Yield

    K2CO3 + 2HCl  2KCl + H2O + CO2

    Theoretical Yield:

    45.8 g

    ? g

    actual: 46.3 g

    45.8 g

    K2CO3

    1 mol

    K2CO3

    138.21 g

    K2CO3

    2 mol

    KCl

    1 mol

    K2CO3

    74.55

    g KCl

    1 mol

    KCl

    = 49.4

    g KCl


    B. Percent Yield

    46.3 g

    49.4 g

    K2CO3 + 2HCl  2KCl + H2O + CO2

    Theoretical Yield = 49.4 g KCl

    45.8 g

    49.4 g

    actual: 46.3 g

     100 =

    93.7%

    % Yield =


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