Stoichiometry

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Stoichiometry

• Study of mass and chemical amounts.
• 1. Compound stoichiometry
• 2. Reaction stoichiometry

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Connecting the Macro and Atomic Scales The Mole

• Chemical macroscopic counting unit
• 1 mole contains 6.022 x 1023 particles
• 6.022 x 1023 is Avogadros number, named after
• Amedeo Avogadro.
• Avogadros number is not special, like pi, but is
  invented by us for our convenience.

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Why it works.

• The mass of 1 mol of an element is equal to its
  atomic mass in grams
• Molar mass is in g/mol
• Molar mass of Al 26.98 g/mol
• Unit sign mole mol sometimes,
  molar mass M

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How it works

• Two new conversion factors
• Avogadros number
• Molar mass

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How many atoms are in 2.60 mol of Ne?
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How many atoms are in 3.84 mol of B?

1. 3.84
2. 2.31 x 1025
3. 2.31 x 1024
4. 6.38 x 10-24

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How many moles do 4.00 x 1022 atoms of Ne make?
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How many moles do 6.84 x 1021 atoms of B make?

1. 6.84 x 1021 mol
2. 0.0114 mol
3. 4.12 x 1045 mol
4. 6.84 x 10-21 mol

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How many moles do 45.0 g of Ne make?
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How many moles do 45.0 g of Al make?

1. 1.67 mol
2. 1214 mol
3. 7.47 x 10-23 mol
4. 2.71 x 1025 mol

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What is the mass of 2.60 mol of C?
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What is the mass of 2.60 mol of O?

1. 8.90 g
2. 0.163 g
3. 2.60 g
4. 41.6 g

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How many moles are represented by 1.00 g of C?
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How many atoms are in 1.00 g of O?
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What is the mass of one Au atom?
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How many C atoms are in a 1.00 caret diamond?
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Compounds and Moles

• 1 mole of a compound contains 6.022 x 1023
  molecules or units
• Molar mass sum of atomic molar masses
• H2O
• CO2
• Fe(NO3)3

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Molecules, Atoms and Moles

• Consider UF6
• 0.50 mol UF6 contains

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Language for nonmolecular compounds

• 1 mol CO2 contains 6.022 x 1023 molecules
• 1 mol NaCl contains 6.022 x 1023 formula units

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Percent Composition

• What fraction of a compound is made up of one of
  its elements?
• Consider FeO

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Determining Formulas from Percent Composition

• The key the ratio of atoms in the formula is the
  same as the ratio of moles of those elements

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Determining Formulas from Percent Composition

• The key the ratio of atoms in the formula is the
  same as the ratio of moles of those elements
• If you determine the ratio of moles, you know the
  formula.

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Determining Formulas from Percent Composition

• The key the ratio of atoms in the formula is the
  same as the ratio of moles of those elements
• If you determine the ratio of moles, you know the
  formula.
• But! You get the empirical formula.

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Example

• A compound has Ca, S, and O.
• Ca 29.45
• S 23.55
• O 47.00
• What is the empirical formula?

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Empirical vs. Molecular Formulas

• Example Ethene is C2H4
• Percent composition tells us mol H/mol C 2
• The empirical formula is CH2.
• The molecular formula is C2H4.

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• Example
• A hydrocarbon has 82.65 C and 17.34 H
• Molar mass is 58.12 g/mol
• What are the empirical and molecular formulas?

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Hydrated Compounds

• Solids in which molecules of water are trapped
  and become part of the compound.
• Ex Gypsum CaSO4 2H2O
• is hydrated calcium sulfate
• CaSO4 is anhydrous calcium sulfate

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Determining the number of waters of hydration

• 1.023 g CuSO4 x H2O is heated to drive off the
  water. The resulting anhydrous CuSO4 has a mass
  of 0.654 g. What is the value of x?