Mole Calculations

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Mole Calculations
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The Mole

• Mole measurement of the amount of a substance.
• We know the amount of different substances in one
  mole of that substance.

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Atomic Mass Unit

• Mass of 1 mole of compound
• Found by adding the atomic weights of each atom
  of each element that makes up a compound.
• H2O
• There are 2 atoms of hydrogen and 1 atom of
  oxygen.

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AMU

• H 2 x 1.01 2.02
• O 1 x 16 16
• _____
• 18.02 amu
• 1 mole of any substance the amu of that
  substance

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AMU

• Molar mass the sum of the molar masses of atoms
  of the elements in the formula
• Calculated the same way

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The Mole

• One mole of any substance has Avogadros number.
• 6.022 x 1023 atoms, molecules, ions

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The Mole

• For gases only 1 mole of a gas occupies 22.4 L

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Converting Between Units

• amu ? 1 mole ? 6.022 x 1023 atoms, mlc, ions
• (expressed
• in grams)

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Conversions

• Change 5.0 grams of sodium chloride to moles of
  sodium chloride.

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• Change 0.45 moles of barium chlorate to grams of
  barium chlorate.

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• Determine the number of atoms in 15 grams of
  water.

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Empirical Formulas
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Empirical Formulas

• The simplest whole number ratio of moles of each
  element in the compound.
• H2O
• NaCl

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Empirical Formulas

• Usually given in percentages of each element in
  the compound.
• Based on 100 of the compound.
• Can be compared to a 100 gram sample of the
  substance.
• 11.2 Hydrogen
• 88.8 Oxygen

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• 1. Change percents to grams
• 11.2 H 11.2 g H
• 88.8 O 88.8 g O

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• 2. Convert grams to moles. (We know an
  empirical formula is the lowest whole number
  ratio of moles)
• 11.2 g H 1 mol 11.089 mol
• 1.01 g
• 88.8g O 1 mol 5.550 mol
• 16 g

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• Cannot have decimals, the numbers must be whole
  numbers. Divide by the lowest.
• 11.2 g H 1 mol 11.089 mol / 5.550 2
• 1.01 g
• 88.8g O 1 mol 5.550 mol / 5.550 1
• 16 g
• H2O empirical formula

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• 36.84 N
• 63.16 O

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• 35.98 Al
• 64.02 S

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Molecular Formula
21

• Specifies the actual number of atoms of each
  element in one molecule or formula unit of the
  substance.

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• The molar mass of acetylene is 26.04 g/mol and
  the mass of the empirical formula, CH, is 13.02
  g/mol

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• 1. The problem will give you a molar mass of the
  compound.
• 2. Calculate the empirical formula as usual.

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• 40.68 carbon
• 5.08 hydrogen
• 54.24 oxygen
• Molar mass 118.1 g/mol

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• Change to grams
• 40.68 g C 1 mol 3.387 mol
• 12.01 g
• 5.08 g H 1 mol 5.030 mol
• 1.01 g
• 54.24 g O 1 mol 3.390 mol
• 16 g

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• 3.387 mol / 3.387 mol 1 x2 2
• 5.030 mol / 3.387 mol 1.5 x2 3
• 3.390 mol / 3.387 mol 1 x2 2
• Empirical Formula C2H3O2

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• To find the molecular formula
• (EFamu)x MM (molar mass)

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• C2H3O2 59.05 amu
• (59.05)x 118.1
• X 2
• (EF)x2 (C2H3O2)x2 C4H6O4

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• 65.45 C
• 5.45 H
• 29.09 O
• MM 110.0 g/mol

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• 49.98 g C
• 10.47 g H
• MM 58.12 g/mol

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• 46.68 N
• 53.32 O
• MM 60.01 g/mol