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EmpFormula.TB.ppt.ppt
1. Determining the Empirical Formula
of Copper Chloride
Purpose of the Experiment
Determine the empirical formula of a
compound containing only copper
and chlorine
2. The molecular weight is the mass, in grams,
of one mole of a compound.
Molar Mass or Molecular Weight
H2O MWt = 18.0148 g/mol
AlCl3 MWt = 133.341 g/mol
3. Percentage Composition
The percentage composition is the percent of a
compound’s mass that results from each of its
constituent elements.
For ethanol, C2H5OH,
2 12.011 g/mol C = 24.022 g/mol of C2H5OH
6 1.008 g/mol H = 6.048 g/mol of C2H5OH
1 15.999 g/mol O = 15.999 g/mol of C2H5OH
Molecular Weight = the sum = 46.069 g/mol of C2H5OH
% C = (24.022 g/mol ÷ 46.069 g/mol) 102 = 52.14 %
% H = ( 6.048 g/mol ÷ 46.069 g/mol) 102 = 13.13 %
% O = (15.999 g/mol ÷ 46.069 g/mol) 102 = 34.73 %
Sum = 100 %
4. Empirical Formula
The empirical formula is the simplest whole-
number ratio of the atoms in a compound.
Examples: benzene, CH
phosphorus pentoxide, P2O5
Molecular Formula
The molecular formula is the simplest whole-
number ratio of the atoms in a single
molecule of a compound.
Examples: benzene, C6H6
phosphorus pentoxide, P4O10
5. Schematic diagram of the combustion device used
to analyze substances for carbon and hydrogen
CxHy + O2 ----> y/2H2O + xCO2
excess
6. A white compound has been analyzed and found to
contain 43.6 wt. % phosphorous and 56.4 wt. % oxygen.
In a separate study it molecular weight has been
found to be 284 g/mol.
What is the empirical formula of this compound?
Assume exactly 100 g of the compound, then one has
43.6 g P ÷ 30.974 g/mol P = 1.408 mol of P
56.4 g O ÷ 15.999 g/mol O = 3.525 mol of O
then the molar ratios are
1.408 mol ÷ 1.408 mol = 1.000 mol P
3.525 mol ÷ 1.408 mol = 2.503 mol O
and the empirical formula is P2O5.
Determination of the Empirical Formula
7. A white compound has been analyzed and found to
contain 43.6 wt. % phosphorous and 56.4 wt. % oxygen.
In a separate study it molecular weight has been
found to be 284 g/mol.
What is the molecular formula of this compound?
First determine the empirical weight of P2O5:
2 30.974 g/mol P = 61.948 g/mol P
5 15.999 g/mol O = 79.997 g/mol O
The empirical weight of P2O5 = the sum = 141.945 g/mol
Molecular weight ÷ Empirical weight = 284 g/mol ÷ 141.945 g/mol
= 2.001
and the molecular formula is twice the empirical formula and
the molecular formula is P4O10.
Determination of the Molecular Formula
8. Mg(s, silvery-white) + O2(g)
limiting reagent
MgxOy(s, white)
0.353 g of Mg solid
Atmospheric oxygen in excess
heat
Result: Mass of MgxOy = 0.585 g
Determining the Empirical Formula of Magnesium Oxide
The goal is to determine x and y in the unknown
magnesium oxide, MgxOy, by burning a known
mass of Mg in an excess of oxygen.
9. Determining the Empirical Formula of
Magnesium Oxide
Note: mass of O = total mass – mass Mg
= 0.585 g – 0.353 g = 0.232 g
then
% Mg = (mass Mg ÷ total mass MgxOy) 102
= (0.353 g ÷ 0.585 g) 102 = 60.3 %
% O = (mass O ÷ total mass MgxOy) 102
= (0.232 g ÷ 0.585 g) 102 = 39.7 %
Mg = 60.3 % and O = 39.7 %
10. Stoichiometry and the percentage composition
of three possible magnesium oxides
Possible Oxides
MgxOy Mg O
MgO 60.3 % 39.7 %
MgO2
43.2 % 56.8 %
Mg2O 75.2 % 24.8 %
Mg = 60.3% and O = 39.7%
and the unknown oxide is MgO
11. Today’s Experiment
Al(s, silvery white) + CuxCly(aq, blue) AlCl3(aq) + Cu(s, reddish)
known mass
25 mL
Limiting reagent
~0.3 g ea. strip
(excess) known mass
Zn(s, silvery white) + CuxCly(aq, blue) ZnCl2(aq) + Cu(s, reddish)
Original experiment
Modified experiment
13. Copper is a transition metal with partially filled d orbitals.
A transition metal may exhibit multiple oxidation states, such
as +1, +2, or +3, oxidation states which are not easily
predicted by its position in the periodic table.
Because of the partially filled d orbitals, a transition metal ion
in aqueous solution frequently is brightly colored, e.g.,
copper ions are blue in aqueous solution.
Zinc and aluminum are both stronger reducing agents than
copper, see their redox potentials, so either will yield metallic
copper from a solution of a copper salt.
14. These potentials indicate the relative thermodynamic
tendency for the indicated half-reaction to occur.
15. Zn(s) + 2 HCl(aq) ----> ZnCl2(aq) + H2(g)
2 Al(s) + 6 HCl(aq) ----> 2 AlCl3(aq) + 3 H2(g)
Cu(s) + n HCl(aq) --x--> no reaction
Note: Cu is below H (0.00 V ref) in table, so
It will not react with acids H+(aq) to form H2(g)
Metals above H in the table will react with acids.
Other Reactions in the Procedure:
Removal of Excess Reducing Agent
16. Reagents in Lab
CuClx solution in 4 L spigot jugs, use ~25 ml for
each analysis
Record data: 0.08067 g CuClx/ml, d = 1.074 g/ml
10 % HCl in 1 L wash bottles, use ~5 ml
Use solid NaHCO3 on acid spills
Checkout
2 piece Al foil, ~0.3 g ea
1 pr Beaker Tongs
17. 25 mL copper chloride, weigh and
use exact density to get mass of CuClx
Add Al foil
Stir (takes about 5 min)
Add 5-10 drops of 10 % HCl and stir
(HCl will dissolve excess Al)
Decant the supernatant liquid
Cu
Flow Chart for Procedure
waste
18. CAUTION:
Do not overheat
to avoid oxidation
Wash with distilled water to
remove aluminum chloride
Transfer Cu residue to a
pre-heated and pre-weighed
casserole
heat
Determine the mass of Cu
Cu
Flow Chart for Procedure
waste
waste
19. Procedure Notes
Record all weights to 0.001g
Weigh 25 ml of CuClx solution, use exact density to calculate exact
volume, then calculate the mass of CuClx
Do not use metal forceps or spatulas
Add Al foil until blue color is gone, allow excess foil to dissolve
Allow container to cool before weighing
Speed up cooling by placing flask in front of hood sash raised 4-6”
The second beaker does not have to be 150 mL
A casserole will also work as an evaporating dish
20. Hazards
10 % HCl is a corrosive strong acid
CuClx solution-heavy metal, irritant
Hot surfaces - hotplates, glassware
Waste
Liquid waste: Al+3 solution and HCl
Cu solids
