Transcript Emperical Formula PreLab (ZnCl2)

The Mole
A. We use a package for atoms and molecules called a mole
1. A mole =
a. the number of Carbon atoms in 12 g of 12C
b. 6.022 x 1023 units = Avogadro’s Number
c. The amount of an element equal to its atomic mass
3. 1 mole of natural C atoms weighs 12.01 g and has 6.022 x 1023 atoms
4. 1 mole of He atoms weighs 4.003 g and has 6.022 x 1023 atoms
5. 1 mole of Al atoms weighs 26.98 g and has 6.022 x 1023 atoms
6. Example: What is the mass of 6 Americium atoms?
6 Am atoms
 243 g 
-21




2
.
42
x
10
g



23
mol
 6.022 x 10 atoms 

1 mol
B. Calculating moles, mass, and atoms
1. Example: # atom / moles in 10g Al

Cu
I2

Hg
10.0 g Al 1 mol Al   0.371 mol Al
Al
 26.98 g 
 6.022 x 10 23 atoms Al 
  2.23 x 10 23 atoms Al
0.371 mol Al 

1 mol Al


S
Fe
C. A mole is the chemists “dozen”
1. A dozen marbles and a dozen peas both have 12
2. A dozen marbles might weigh 100 grams
3. A dozen peas might weigh only 15 grams
4. Example:
5.68 mg Si = ? atoms Si
 1g 
  5.68 x 10-3 g Si
5.68 mg 
 1000 mg 
mol Si 
  2.02 x 10-4 mol Si
5.68 x 10-3 g Si  128.09
g

 6.022 x 10 23 atoms Si 
  1.22 x 10 20 atoms Si
2.02 x 10 mol Si 
1 mol Si



-4

Empirical Formula = smallest whole number ratio of elements in the formula
I.
Today’s Reaction
xZn(s) + yHCl(aq) --------> ZnxCly(aq) + H2(g)
Measured Mass excess
II.
Measured Mass
Safety
1.
2.
3.
4.
6 M HCl can burn skin or put holes in clothes; wash off with water
H2(g) is flammable; do the reaction in the hood.
Use hot plate to boil off the extra water from the reaction
White fumes coming off dry solid is gaseous ZnxCly
III. Calculations (example)
1.
2.
1.013g ZnxCly solid - 0.800g Zn(s) = 0.213g Cl in the compound
Change the mass of Zn and Cl to moles
0.800 gZn
1mol 
  0.0122molZn
 65.38 g 
3.
0.213gCl 
1mol 
  0.00601molCl
 35.45 g 
Divide by the smallest number of moles; then round or multiply to whole #’s
Zn 0.0122Cl0.00601
 Zn 2.03Cl1  Zn 2 Cl1
0.00601
IV. Hints for determining empirical and molecular formulas
1. Convert mass % to grams of the element in 100 total grams of sample
Determine the Empirical Formula of an unknown if its Percent
Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen
100g 
47gC
 47gC
100g
100g 
47gO
 47gO
100g
100g 
6.0gH
 6.0gH
100g
2. Change these masses to moles by using the atomic mass of each element
1 mol C
47g C 
 3.9 mol C
12.01g
1 mol H
6.0 g H 
 6.0 mol H
1.008g
47 g O 
1 mol O
 2.9 mol O
16.00g
3. Divide each number of moles by the smallest to get a small # ratio
C3.9 H 6 O 2.9
2.9
 C1.34H 2.07O1
4. Round off to a whole # if molar #’s are close to that whole number
5. Multiply the whole ratio by a factor to get all numbers to whole numbers
C1.34H2.07O1  C1.34H2 O1 x 3  C4 H6 O3
6. Multiply empirical formula by factor needed to give the correct molar mass
a. Molar Mass of unknown = 204.2 g/mol but C4H6O3 = 102.09 g/mol
b. Correct Molecular Formula: C4H6O3 x 2 = C8H12O6
Incident: Mixing Acid with Water
Incident: Unrecognized Fume Hood Failure