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

Unit 08 – Moles and Stoichiometry

  • Molar Conversions


A what is the mole

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 mole1

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

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

    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 examples1

    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

    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

    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 mole road map

    The MoleRoad Map

    Molecule

    Atoms (ions)

    Formula unit


    E molar conversion examples

    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 examples1

    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 examples2

    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

    II. Stoichiometric Calculations


    A proportional relationships

    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 relationships1

    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

    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.


    Unit 08 moles and stoichiometry

    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

    III. Stoichiometry in the Real World


    A limiting reactants

    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 reactants1

    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 reactants2

    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 reactants3

    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 reactants4

    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 reactants5

    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 reactants6

    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

    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

    B. Percent Yield

    measured in lab

    calculated on paper


    B percent yield1

    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 yield2

    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 yield3

    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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