The Mole

1
The Mole
2
Mass Relationships in Reactions

• The Mole
• All reactions occur at the molecular level, with
  individual atoms, ions, or molecules doing the
  reacting. Since it is impossible to count out
  these individual units, we need a unit represent
  visible quantities of these.
• The MOLE is defined as the number of atoms in
  exactly 12.0000 grams of C-12
• Mole value 6.02 x 1023 particles of anything
• Counting number like dozen or gross
• Called Avogadros number
• Mole of any atom atomic weight in grams of that
  atom
• Mole of any substance sum of weights of all of
  the atoms in its formula

3
(No Transcript)
4
The Mole

• The number equal to the number of carbon atoms
  in exactly 12 grams of pure 12C.
• 1 mole of anything 6.022 ? 1023 units of
• that thing

5
Molar Mass

• A substances molar mass (molecular weight) is
  the mass in grams of one mole of the compound.
• CO2 44.01 grams per mole

6
Formula Weight and the Mole

• Sum of the atomic weights rounded to the nearest
  0.1 unit
• Distribute into parentheses correctly
• One mole of a substance sum of weights
  of all of the atoms in its formula

7
Calculating Molar Masses
8
Mole T

• PARTICLES??MOLES??MASS (grams)
• LITERS of GAS

9
Examples
p. 311,12
10
Calculations with Moles

• Moles x MM grams of substance
• 1mole 6.02x1023 particles MM in grams
• 23.5 g NaCl ? Mol
• 23.5 g / 58.5 MM 0.402 mol
• 2.4 mol CaCO3 ? Grams
• 2.4 mol x 100.0 MM 240 grams
• Use dimensional analysis bridge

11
Mole Calculations
12
Homework 11a

• p. 311 1-3 all
• p. 312 4, 5, 6, 10
• p. 322 25, 26
• p. 323 27-29 all
• p. 324 30
• p. 326 31-35 all

13
Percent Composition

• Mass percent of an element
• For iron in iron (III) oxide, (Fe2O3)

14
Sample Composition

• C10H14O --carvone-from carraway seeds
• C 10 x 12.0 120g --gt 120/150 80
• H 14 x 1.0 14g --gt 14/150 9.3
• O 1 x 16.0 16g --gt 16/15010.7
• Total 150 g

15
Another Composition

• C14H20N2SO4 --pennicillin F
• C 14 x 12.0 168 H 20 x 1.0 20.0
• N 2 x 14.0 28.0 S 1 x 32.1 32.1
• O 4 x 16.0 64.0 Total 312.1
• C168/312.153.8 H20/312.16.4
• N28.0/312.19.0 S32.1/312.110.3
• O64.0/312.120.5

16
Formulas

• molecular formula (empirical formula)n
• n integer
• molecular formula C6H6 (CH)6
• empirical formula CH

17
Empirical Formula Determination From Percents

• 1. Base calculation on 100 grams of compound.
• 2. Determine moles of each element in 100 grams
  of compound.
• 3. Divide each value of moles by the smallest of
  the values.
• 4. Multiply each number by an integer to obtain
  all whole numbers.

18
Sample Empirical Formula

• 43.64 P 56.36 O
• 43.64g 1mol 1.41 56.36g 1mol 3.52
• 31.0g
  16.0g
• 1.41/1.411.0 3.52/1.412.50
• Multiply by 2 to fix the Oxygen
• P2 O5 therefore P2O5 is the formula

19
From Grams Found in Experiment

• Cl 71.65/ 35.5 2.02 H 4.07 / 1.0 4.07
• C 24.27 / 12.0 2.02
• C 2.02/2.02 1 Cl 2.02/2.02 1
• H 4.07/2.022
• Therefore H2CCl is the formula

20
Molecular Formula

• If molar mass is given, empirical can be adjusted
  to make molecular formula
• From a previous slide, Cl71.65 C24.27
• H4.07 Empirical formula was H2CCl
• If MM given as 98.96, divide by Empirical mass
  (EM) to get the Multiplier
• EM 2.012.035.5 49.5
• MM/EM 98.96/49.5 2
• Therefore formula is 2 x H2CCl H4C2Cl2

21
Examples
p. 332, 334
22
Homework 11b

• p. 333 46-50 all
• p. 335 51-53 all
• p. 337 54-57 all

23
Moles of Atoms in Formulas

• If the fundamental particle of a substance has
  more than one atom of an element, then the number
  of moles of that element are multiplied by the
  subscript of that element
• 3 moles H2O? 3 x 2 moles H
• 2.36 moles Fe4(Fe(CN)6)3? 2.36 x 7 moles Fe

24
Homework 11c

• p. 331 42, 43
• p 346ff 90, 98, 111, 112, 116, 121, 136, 139,
  142, 144, 146, 148, 150