determining the empirical formula of copper chloride purpose of the experiment to determine the...

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.