determining the empirical formula of copper chloride purpose of the experiment to determine the...
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Determining the Empirical Formula of Copper Chloride
Purpose of the Experiment
To determine the empirical formula of a compound containing only
copper and chlorine.
The mass in grams of 1 mole of a compound is equal to the mass in grams of the components.
Molar Mass (Molecular Weight)
H2O MW = 2(1.008) + 15.999 g
AlCl3 MW = 26.98 + 3(35.45) g
(Note: The Significant Figures.)
So the mass in grams of 1 mole of each compound is…
The mass in grams of 1 mole of each compound is…
Molar Mass (Molecular Weight)
H2O MW = 18.015 g
AlCl3 MW= 133.33 g
Percent Composition
Defn: The percentages of a compound’s mass that are due to each of the component elements.
Mass of C = 2 x 12.011g
Mass of H = 6 x 1.008 g
Mass of O = 1 x 15.999 g
Mass of 1 mole of C2H5OH = ??
C2H5OH
Percent Composition
Defn: The percentages of a compound’s mass that are due to each of the component elements.
Mass of C = 24.022 g
Mass of H = 6.048 g
Mass of O = 15.999 g
Mass of 1 mole of C2H5OH = 24.022 + 6.048 + 15.999 = 46.069g
C2H5OH
Mass percent of C = ??
Percent Composition
Defn: The percentages of a compound’s mass that are due to each of the component elements.
Mass of C = 24.022 g
Mass of H = 6.048 g
Mass of O = 15.999 g
Mass of 1 mole of C2H5OH = 24.022 + 6.048 + 15.999 = 46.069g
C2H5OH
52.144%100%46.069g
24.022g100%
OHHC 1mol of mass
OHHC of mol 1in C of massC ofpercent Mass
52
52=×=×=
Empirical Formula
Represents the simplest whole-number ratio of the various types of atoms in a compound.
What is the Empirical Formula for the following compounds?
Scent of Bananas
Examples
P4O10
CaffeineDrying Agent
C14H28O4 C8H10N4O2
Empirical Formula
Represents the simplest whole-number ratio of the various types of atoms in a compound.
What is the Empirical Formula for the following compounds?
Scent of Bananas
Examples
P4O10
CaffeineDrying Agent
C14H28O4 C8H10N4O2
P2O5 C7H14O2 C4H5N2O
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Empirical Formula Example 1
What is the empirical formula?
1st Step:
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
What is the empirical formula?
1st Step: Because it is a percent we can choose any mass we want and multiply that mass by the percents of the components.
What is the easiest mass to choose?
Empirical Formula Example 1
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
What is the empirical formula?
What is the easiest mass to choose?
100.00 grams
1st Step: Because it is a percent we can choose any mass we want and multiply that mass by the percents of the components.
Empirical Formula Example 1
So since we chose 100.00 g of compound we have 43.64g P & 56.36 g O.
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Can we compare grams directly?
Empirical Formula Example 1
2nd Step:
43.64g P x ???
56.36 g O x ???
So since we chose 100.00 g of compound we have 43.64g P & 56.36 g O.
2nd Step: Convert grams to moles.
Can we compare grams directly?
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Empirical Formula Example 1
43.64g P x (1 mol P / 30.974g P) = 1.409 mol P
56.36 g O x (1 mol O / 15.999g O) = 3.523 mol O
So since we chose 100.00 g of compound we have 43.64g P & 56.36 g O.
2nd Step: Convert grams to moles.
Can we compare grams directly?
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Empirical Formula Example 1
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
What can we do to compare the moles?
So we now have 1.409 moles of P & 3.523 moles of O.
Empirical Formula Example 1
3rd Step:
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
What can we do to compare the moles?
So we now have 1.409 moles of P & 3.523 moles of O.
Do this by dividing both mole values by the smaller one.
O ?1.4093.523
and P ?1.4091.409
3rd Step: Convert the smaller number to one.
Empirical Formula Example 1
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
What can we do to compare the moles?
So we now have 1.409 moles of P & 3.523 moles of O.
3rd Step: Convert the smaller number to one.
Do this by dividing both mole values by the smaller one.
O 2.51.4093.523
and P 11.4091.409
Empirical Formula Example 1
So by dividing both mole values by the smaller one we deduced:
O 2.51.4093.523
and P 11.4091.409
This yields the formula PO2.5
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Is this an acceptable empirical formula?
Empirical Formula Example 1
So by dividing both mole values by the smaller one we deduced:
O 2.51.4093.523
and P 11.4091.409
This yields the formula PO2.5
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Is this an acceptable empirical formula?
Empirical Formula Example 1
No.
What must we do to fix the formula PO2.5?
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers.
Empirical Formula Example 1
What must we do to fix the formula PO2.5?
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers.
4th Step: Convert to whole numbers.
Do this by multiplying both mole values to get whole numbers.
Empirical Formula Example 1
What must we do to fix the formula PO2.5?
Empirical Formula = P2O5
A white compound is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass.
Since compounds must contain whole numbers of atoms, the empirical formula should contain only whole numbers.
4th Step: Convert to whole numbers.
Do this by multiplying both mole values to get whole numbers.
1 P x 2 = 2 P 2.5 O x 2 = 5 O
Empirical Formula Example 1
Empirical Formula Example 2
Mg (solid – silvery white) + O2 (gas) → MgxOy (solid – white)
heat
(0.353g)limiting reagent
(0.585g)resulting mass
Atmospheric oxygen is in excess
A student burns 0.353 g of magnesium ribbon in an open crucible. A white powder forms and is found to weigh 0.585 g. What is the empirical formula?
How do we determine the empirical formula?
Empirical Formula Example 2
Mg (solid – silvery white) + O2 (gas) → MgxOy (solid – white)
heat
(0.353g)limiting reagent
(0.585g)resulting mass
Atmospheric oxygen is in excess
A student burns 0.353 g of magnesium ribbon in an open crucible. A white powder forms and is found to weigh 0.585 g. What is the empirical formula?
How do we determine the empirical formula?
1st Step: Calculate mass percents.
60.3%100585.0
353.0100
OMg of mass total
g Mg, of mass% Mg,Percent
yx=×=×=
%7.39100g 0.585
g 0.232100
OMg of mass total
O of mass% O,Percent
yx=×=×=
Mass of O = total mass of compound – mass of Mg = 0.232 g
Mg = 60.3% and O = 39.7%
Empirical Formula Example 21st Step: Calculate mass percents.
Formula masses and percent composition of three theoretical compounds of Mg and O
Empirical Formula Example 22nd Step: Formula Masses and their Percent Composition.
Mg = 24.31 g/mole O = 15.999 g/mole
Formula of Oxide MgxOy Formula Weight %Mg %O
MgO 24.31+15.999 =
MgO2 24.31 + 2(15.999) =
Mg2O 2(24.31) + 15.999 =
Mg = 60.3% and O = 39.7%
Formula masses and percent composition of three theoretical compounds of Mg and O
Empirical Formula Example 22nd Step: Formula Masses and their Percent Composition.
Mg = 24.31 g/mole O = 15.999 g/mole
Formula of Oxide MgxOy Formula Weight %Mg %O
MgO 24.31+15.999 = 40.309
MgO2 24.31 + 2(15.999) = 56.308
Mg2O 2(24.31) + 15.999 = 64.619
Mg = 60.3% and O = 39.7%
Formula masses and percent composition of three theoretical compounds of Mg and O
Empirical Formula Example 22nd Step: Formula Masses and their Percent Composition.
Mg = 24.31 g/mole O = 15.999 g/mole
Formula of Oxide MgxOy Formula Weight %Mg %O
MgO 24.31+15.999 = 40.30924.31/ 40.309
15.999/ 40.309
MgO2 24.31 + 2(15.999) = 56.30824.31/ 56.308
2 x 15.999/ 56.308
Mg2O 2(24.31) + 15.999 = 64.6192 x 24.31/
64.61915.999/ 64.619
Mg = 60.3% and O = 39.7%
Formula masses and percent composition of three theoretical compounds of Mg and O
Empirical Formula Example 22nd Step: Formula Masses and their Percent Composition.
Mg = 24.31 g/mole O = 15.999 g/mole
Formula of Oxide MgxOy Formula Weight %Mg %O
MgO 24.31+15.999 = 40.309 60.3 39.7
MgO2 24.31 + 2(15.999) = 56.308 43.2 56.8
Mg2O 2(24.31) + 15.999 = 64.619 75.2 24.8
Mg = 60.3% and O = 39.7%
Formula masses and percent composition of three theoretical compounds of Mg and O
Empirical Formula Example 22nd Step: Formula Masses and their Percent Composition.
Mg = 24.31 g/mole O = 15.999 g/mole
Formula of Oxide MgxOy Formula Weight %Mg %O
**MgO 24.31+15.999 = 40.309 60.3 39.7
MgO2 24.31 + 2(15.999) = 56.308 43.2 56.8
Mg2O 2(24.31) + 15.999 = 64.619 75.2 24.8
Mg = 60.3% and O = 39.7%
Today’s Experiment
Al(s) + CuxCly(aq) AlCl3(aq) + Cu(s)
known massof ~25 ml
Limiting reagent(excess) known mass
of product
(silvery white) (reddish)(blue soln) (gray soln)
Ground state electron configuration: [Ar].3d10.4s1Shell structure: 2.8.18.1
Transition metals may exhibit multiple oxidation states(+1, +2, +3, etc…). These are not easily predicted by position in the periodic table.
Transition metals ions in aqueous solutions frequently are brightly colored, also due to d orbitals (e.g. Cu ions are blue).
Copper is a Transition Metal.
Last week we worked with Zinc. Zn and Al are both stronger reducing agents than copper. (Note: the redox potentials on next slide.)
Because of this either one would work to produce metallic copper from a solution of a copper salt. We chose aluminum to work with because the reaction between Zinc and Copper Chloride is quite exothermic and extreme caution would have had to have been observed to avoid burns.
**
*
The oxidation of aluminum is more likely than the oxidation of zinc.
These potentials indicate the relative thermodynamic tendency for the indicated half-reaction to occur.
Al <--> Al+3 + 3 e– E = -0.166 voltsZn <--> Zn+2 + 2 e– E = -0.763 voltsCu <--> Cu+2 + 2 e– E = +0.34 volts
2Al(s) + 6HCl(aq) --> 2AlCl3(aq) + 3H2(g)
Cu(s) + n HCl(aq) -x-> No Reaction
Removal of Excess Reducing Agent
Once you have ensured that the aluminum is in excess (some aluminum will be floating on the surface), then it is time to proceed with the next step: removal of the aluminum. This is accomplished by adding HCl in excess.
The HCl reacts with the aluminum and not the copper.
Reagents in LabCuClx solution in 4L spigot jugs
- Take ~25 ml for each run (using 100 ml Grad Cylinder)
Checkout2 pc Al foil – Add in excess. Do not fold.
1 - pair of Beaker Tongs
Record data: (0.08067 g CuClx / ml, d=1.074 g/ml)
10% HCl in 1L wash bottles - Take ~5 ml for each run (using 10 ml Grad Cylinder)
(N.B. solid NaHCO3 is to be used for acid spills.)
25 mL copper chloride, weigh and use exact density to get mass of CuClx
Add Al foil
Stir gently (takes about 5 min)
Add 2-5 ml of 10% HCl and stir gently. ( HCl will dissolve excess Al.
If it does not – DECANT and add more HCl.)
Decant the supernatant liquid
Cu
Flow Chart for Procedure
waste
Do not overheat. *Overheating will cause oxidation.
Wash with distilled water to remove aluminum chloride
Transfer Cu residue to a pre-weighed casserole.
heat
Determine the mass of Cu
Cu
Flow Chart for Procedure
waste
waste
*Hot plates should be set on 3. If they are not please notify TA. Anything higher than 3 may cause product to overheat.
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 when stirring solution.
Add Al foil until blue color is gone, allow excess foil to dissolve also.
Allow container to cool before weighing.
Speed up cooling by placing in front of hood sash raised 4-6”.
Hazards10% HCl-strong acid, corrosive
CuxCly solution-heavy metal, irritant
Hot surfaces - hotplates, glassware (Be Careful: Hot glassware looks just like cold glassware.)
WasteLiquid Waste: Al+3 / HCl
Used Solids: Cu
Results (calculations)Collected data
Mass of CuClxMass of Cu
Mass percent of CuMass of ClMass percent of ClEmpirical formula
Summary of Data & Calculations*
*Note that on the bottom of page 79 it states that the calculations which need to be shown on a separate sheet of paper are indicated by asterisks (*).
Homework Assignment - Due September 22-25.
Read: Dimensional Analysis (pp.22-26)
Do: Problem Set #2 (p.23) 2.1-2.4 & Set #3 (26) 3.1-3.2
(On a separate sheet of paper, show your calculations.)
For September 22-25Read: Separating the Components of a
Ternary Mixture p. 83.
Turn-In: EF pp. 79-82; Calculations Page
(explained on p.79) & DA Homework.