The Mole

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The Mole

• 1 atom or 1 molecule is a very small entity not
  convenient to operate with
• The masses we usually encounter in chemical
  experiments vary from milligrams to kilograms
• Just like one dozen 12 things
• One mole 6.022 x 1023 things
• Avogadros number

NA 6.022 x 1023
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The Mole
NA 6.022 x 1023

• Why 6.022 x 1023 ?
• This is the number of carbon atoms found in 12 g
  of the carbon-12 isotope
• Molar mass mass of one mole of atoms,
  molecules, ions, etc.
• Numerically equal to the atomic or molecular
  weight of the substance in grams
• m (1 mole H2) Mr(H2)
• m (1 mole Fe) Mr(Fe)

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The Mole Example 1

• Example Calculate the mass of a single Mg atom
  in grams to 3 significant figures.

4
The Mole Example 2

• Example How many C6H14 molecules are contained
  in 55 ml of hexane (d 0.78 g/ml).

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The Mole Example 3

• Example Calculate the number of O atoms in 26.5
  g of lithium carbonate, Li2CO3.

• Solution
• Calculate the molar mass of Li2CO3
• Mr(Li2CO3) 73.89 g/mol
• Calculate the amount of Li2CO3 in moles
• n (Li2CO3) 0.359 mol
• Calculate the amount of O atoms in moles
• n (O) (0.359x3) mol 1.077 mol
• Multiply by the Avogadros number to obtain the
  answer
• N (O) 1.077 mol x 6.0221023 atom/mol
• 6.501023 atoms

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Percent Composition and Formulas of Compounds

• If the formula of a compound is known, its
  chemical composition can be expressed as the mass
  percent of each element in the compound (percent
  composition), and vice versa.
• When solving this kind of problems, we can use
  masses expressed in a.m.u. or in g/mol

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Percent Composition Example 1

• What is the percent composition of each element
  in sodium chloride, NaCl?

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Percent Composition Example 2

• Calculate the percent composition of iron(III)
  sulfate, Fe2(SO4)3, to 3 significant figures

• Solution
• Calculate the molar mass of Fe2(SO4)3
• Mr(Fe2(SO4)3) 399.88 g/mol
• Calculate the percent composition for the
  elements

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Simplest (Empirical) Formula

• The smallest whole-number ratio of atoms present
  in the compound
• Molecular formula, on the other hand, indicates
  the actual number of atoms present in a molecule
  of the compound

water
hydrogen peroxide
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Empirical Formula Example 1

• The first high-temperature superconductor,
  prepared by Bednorz and Müller in 1986, contained
  68.54 lanthanum, 15.68 copper, and 15.79
  oxygen by mass. What was the simplest formula of
  this compound?

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Empirical Formula Example 2

• A sample of a compound contains 6.541g of Co and
  2.368g of O. What is its empirical formula?

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Elemental Composition

• A combustion train for carbon-hydrogen analysis
• percent composition is determined experimentally

magnesium perchlorate
sodium hydroxide
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Empirical Formula Example 3

• 0.1172 g of a pure hydrocarbon was burned in a
  C-H combustion train to produce 0.3509 g of CO2
  and 0.1915 g of H2O. Determine the masses of C
  and H in the sample, the percentage of these
  elements in this hydrocarbon, and the empirical
  formula of the compound.

• Solution
• First we calculate the molar weights of CO2 and
  H2O
• Mr(CO2) 44.01 g/mol Mr(H2O) 18.02 g/mol
• Now we can calculate the amount of both CO2 and
  H2O in moles

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Example 3 (continued)

• Notice that 1 mole of CO2 molecules contains 1
  mole of C atoms and 1 mole of H2O
  molecules contain 2 moles of H atoms
• Therefore,
• n (C) 7.97310-3 mol n (H) 2 x 1.06310-2
  mol 2.12610-2 mol
• Now we can answer the first question determine
  the masses of carbon and hydrogen in the starting
  compound using the formula

m (C) 7.97310-3 mol x 12.01 g/mol 0.09576
g m (H) 2.12610-2 mol x 1.008 g/mol
0.02143 g Now, if you add these masses, they
should give you the mass of the starting
compound 0.1172 g
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Example 3 (continued)

• Now we can easily find the mass percentage of the
  elements in the hydrocarbon by dividing the mass
  of the element by the total mass of the
  hydrocarbon and multiplying by 100
• We can also find the percentage of hydrogen by
  subtracting the percentage of carbon from 100
• The slight difference between two results in
  caused by rounding errors.

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Example 3 (continued)

• Finally, to find the empirical formula we use the
  amounts of C and H in moles found in step 3 and
  divide them by the smallest of both numbers
• We know that the fraction like .666 is in fact
  2/3. Therefore, to convert both results to whole
  numbers, we need to multiply them by 3
• C 3.000 H 7.998
• These result tells us that the empirical
  formula for the hydrocarbon in question is
• C3H8

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Empirical Formula Example 4

• 0.1014 g sample of purified glucose was burned in
  a C-H combustion train to produce 0.1486 g of CO2
  and 0.0609 g of H2O. An elemental analysis showed
  that glucose contains only carbon, hydrogen, and
  oxygen. Determine the empirical formula of the
  compound.

This example is worked out in your textbook
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Molecular Formula

• Indicates the actual number of atoms present in a
  molecule of the compound
• To determine the molecular formula for a
  molecular compound, both its empirical formula
  and its molecular weight must be known
• The molecular formula for a compound is either
  the same as, or an integer of, the empirical
  formula

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Molecular Formula Example

• A compound is found to contain 85.63 C and
  14.37 H by mass. In another experiment its
  molar mass is found to be 56.1 g/mol. What is
  its molecular formula?

Ill work on this example in the beginning of the
next lecture.
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Other Examples

• What mass of ammonium phosphate, (NH4)3PO4, would
  contain 15.0 g of N?

You do it !
The answer is 53.2 g
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Reading Assignment

• Go through Lecture 3 notes and learn again the
  problems we solved in class
• Read Chapter 2 to the end
• Learn Key Terms (p. 82)
• Take a look at Lecture 4 notes (already posted on
  the web)
• If you have time, read Chapter 3
• Homework 1 due by 9/13