The Mole and Stoichiometry

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The Mole and Stoichiometry

• Chemistry gets Real.
• Tough that is

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The mole

• A term for a certain number of something.
• Brainstorm other counting words!
• Dozen
• Pair
• Gross
• A mole of something is 6.02 x 1023 of
  something.
• 602, 000, 000, 000, 000, 000, 000, 000

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2
144
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Molecular Weight

• M.W. the weight (in grams) of a mole of
  substance
• On your periodic tables
• Round to the nearest tenth
• Hydrogen is 1.00797 ? 1.0 g/mol
• Mole Weight is an INTENSIVE propertydoesnt
  depend on amount

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Try these MWs

• Ca
• 40.1
• H2
• 2 (1.0) 2.0
• BaF2
• 137.3 2(19.0) 175.3
• MW of 2BaF2 is still 175.3

1 mole of Ca weighs 40.1 grams 1 mole of H2
weighs 2.0 grams
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• 1 penny 2.68
• 6 pennies g
• 6 pennies

2.68 X 6 16.08
2.68
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Avogadros Number
Avogadro (1776-1856)                            
                                                  
                                                  
                  

• NA 6.02 x 1023 of anything

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• 1 mole K grams
• MW of potassium
• 1 dozen K atoms
• 1 mole K atoms

39.10
39.1
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6.02 x 1023
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• MW of CO2
• 3 moles of CO2
• 3 moles of CO2 ____ g/mol
• MW of nitrogen gas
• N2 (g)

44.0
44.0
2 ( 14.0) 28.0
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Which elements exist as diatomic molecules?

• H2 N2 O2 F2 Cl2 Br2 I2

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How Big Is The Mole?

• One mole of marbles would cover the entire Earth
  to a depth of fifty miles
• One mole of hockey pucks would equal the mass of
  the moon.
• One mole of rice grains is more than the number
  of grains of all crops grown since the beginning
  of time.
• If one mole of pennies was divided up equally
  between all the people on Earth, you would have
  enough money to spend a million dollars every
  hour, 24 hours a day, for your entire life. When
  you died, you would have spent less than half of
  your riches.

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Lab 11
MW
X NA
Xatoms
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Percent Composition

• A. Determined from Formulas (Accepted Value)
• Is NaCl 50.0 Na by weight?
• No, Na is 23.0 g/mole and Cl is 35.5 g/mole
• To Prove, Na

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Percent Compositionfrom Formula

• oxygen in CaCO3?
• Grams of Mg in 4.00 grams of MgO?

First Calculate Mg in MgO
Second Calculate g Mg in 4.00 g MgO
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B. Experimental Percent Composition

• FROM DATA (Experimental Value)
• 4.00g of Ag2O is decomposed to yield 3.65g Ag.
• The Experimental ?
• The Accepted ?
• Your Experimental Error?

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Experimental Ag

• Equation
• Ag2O ?
• Experimental Ag

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Ag O2
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4.00 g
3.65 g
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True (Accepted) Ag
Percent Error?
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Empirical Formula

• DefinitionThe simplest formula indicating the
• mole ratio of elements in a compound
• Examples
• H2O2? HO
• C6H6?CH
• N2O4??
• CO2 ? ?

NO2
CO2
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Empirical Formula STEPS

• Change grams to moles
• Divide by the least moles for a RATIO
• Apply ratio to the formula

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Solving Empirical Formuladetermined from gram
composition

• A compound contains 0.90 g Ca and 1.60 g Cl

CaCl2
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Try This

• 0.556 g Carbon and 0.0933 g Hydrogen

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Why is the Empirical Formula a ratio of small
WHOLE numbers?

• Cant have half of an atom
• Atoms combine as whole units
• Shows the simplest way that atoms can pair

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What formula would this ratio give?

• K 0.26 moles
• N 0.25 moles
• O 0.78 moles
• Yields an Empirical Formula of
• KNO3

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Try This 70.5 Fe and 29.5 O

• ___ moles Fe, ___ moles O
• Assume 100g of substance

1.26
1.84
1.26 moles
1.26 moles
Fe2O3
FeO1.5
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Try This 40.0 C 6.7 H 53.3 O

• ___ moles C, ___ moles H, ___ moles O
• Assume 100g of substance

3.3
6.7
3.3
After dividing by the least moles yields
CH2O
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• Why doesnt the ratio of the give the empirical
  formula?
• Must account for differing masses of elements.
• Why does the ratio of the moles give the
  empirical formula?
• The ratio of the of atoms normalizes for mass
  differences.

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Molecular Formulas

• Definiton Formula of an actual compound
• as it exists in molecules.
• Benzene exists as C6H6 not CH.
• Hydrogen Peroxide exists as H2O2 not HO.

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17 g/mol
H2O2
100 g/mol
C2H2O2Cl4
75.5 g/mol
C3Cl3N6
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• Why is the M.W. needed to determine the molecular
  formula?
• Need M.W. of actual compound to find how many
  each type of atom is in a molecule.

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Stoichiometry The Big Leagues

• A. DefineProblem Solving involving mass-mass
  relationships in chemical changes
• Ex. How many grams of rust are formed when 12.00
  g of Fe reacts with oxygen.
• B. Must use balanced equations for the correct
  mole ratios

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• C. Coefficients yield the mole ratio!!!
• 2 H2 O2 ? 2 H2O
• 2 1 2

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D. Example

• 4 Fe 3 O2 ? 2 Fe2O3
• 4 3 2
• If 4 moles of iron rust, moles of Fe2O3 will
  form
• If 8 moles of iron rust,
• moles of Fe2O3 will form

2
4
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Solving Mass-Mass Problems
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28.00 g of iron yields ? g of rust?
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0.50 moles
0.25 moles

• 4 Fe 3 O2 ? 2 Fe2O3

28.00 g Fe
40.00 g Fe2O3
? g Fe2O3
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36.00 g of water resulted from ? g of methane?
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1.00 moles
2.00 moles
Balance
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2

• CH4 O2 ? CO2 H2O

? g CH4
36.00 g H2O
16.00g of methane
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Variation12 moles of oxygen combusting will
yield how many grams of CO2?
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12 moles
6 moles
2
2

• CH4 O2 ? CO2 H2O

? g CO2
264.00 g CO2