Lab 11 EMPIRICAL FORMULA DETERMINATION - COPPER SULFIDE EXPERIMENT
Honors Chemistry
Objectives
1. To prepare the compound “copper sulfide”.
2. To determine the empirical formula of “copper sulfide”.
Theory
Atoms of different elements chemically combine to form different compounds following two fundamental laws.
1) Law of conservation of mass - mass can neither be created nor destroyed during a chemical reaction
2) Law of definite (or multiple) proportions - a compound is always made up of elements in the same proportion by weight.
If we analyze sample of water, we will find that the ratio of the masses of hydrogen and oxygen in each sample is the same, regardless of the source of the water.
The law of the multiple proportions states that if two elements combine to form a series of compounds, the mass ratio of the element are small whole numbers. For example, hydrogen forms two compounds with oxygen: water
(H
2
O) and hydrogen peroxide (H
2
O
2
). In the first compound, 16.0 grams of oxygen (one mole of O) combine with
2.016 grams of hydrogen (two moles of H); and in the second compound, 32.0 grams of oxygen (two moles of O) combine with 2.016 grams of hydrogen (two moles of H). The ratio of masses of oxygen that combine with the hydrogen in these compounds is 16.0 g/32.0 g or 1:2, which supports the law of multiple proportion.
The simplest whole-number ratio of the number of atoms of each element in a compound is called the empirical formula of the compound. The molecular formula shows the actual number of each type of atom present in a compound.
Types of Chemical Reactions:
Many chemical reactions fall into one of the following four categories:
A. Synthesis or Combination: A + B -> AB
B. Decomposition: AB-> A + B
C. Single Displacement: AB + C -> CB + A
D. Double Displacement: AB + CD AD + CB
When a new compound is prepared, its formula must be determined. This task is most commonly achieved from the percent composition of the compound-the percent by mass of each element relative to the total mass of the compound. It is obtained by dividing the mass of each element in the compound by the molar mass of the compound, and multiplying by 100. In one mole of water (MW = 18.02 g/mol) are two moles of H (2.016 g) and one mole of O (16.0 g). Therefore, %H = (2.016/18.016)100 = 11.19 percent and %O = (16.0/18.016)100 = 88.81 percent.
The empirical formula of the compound can be determined either by chemical analysis or by synthesis. In chemical analysis, a known mass of the compound is decomposed to obtain the masses of the elements themselves or some of their known derivatives (for carbon, the derivative is CO2 and for hydrogen, it is water).
From these masses, the percent composition of the compound is determined. In synthesis, known masses of the elements are allowed to form a compound and the mass of the compound is determined. From these data, it is possible to derive the empirical formula of the compound.

Introduction:
In this lab, you will heat a known mass of a coil of copper metal with excess powdered sulfur in a covered crucible to produce the compound “copper sulfide”. This experiment requires you to burn sulfur – producing poisonous sulfur dioxide gas. Thus this experiment MUST be completed in the FUME HOOD.
In today's lab, synthesis will be used to determine the amounts of the different elements which make up a compound. While copper forms 2 different ions – Cu 1+ and Cu 2+ , only one compound is formed – you will determine what the empirical formula of the compound you form is.
Upon heating, the excess sulfur vaporizes to form gaseous sulfur, which escapes from the crucible. When the hot sulfur gas reaches the air, it reacts with oxygen to produce gaseous oxides of sulfur (mainly sulfur dioxide, SO2).
Thus only the copper sulfide remains in the crucible. From the measured mass of the product and the mass of copper used, you may determine the mass of sulfur in the product. Then you will use the known masses of the copper and sulfur and their atomic masses to calculate the simplest formula of the copper sulfide.
The metal, copper, and the non-metal, sulfur (in the vapor state), will be chemically combined to form a compound, copper sulfide, which will have completely different properties than the copper or the sulfur.
Synthesis reaction: metal + nonmetal -> compound copper + sulfur vapor ® copper sulfide
By using a predetermined amount of copper and determining the mass of copper sulfide produced, it is possible to calculate the percentage by mass of each element in the compound. From this information the chemical formula of the compound can be determined.
Determination of % composition of elements by experiment
In general: % element = mass element x 100
mass compound
Assuming the Law of conservation of mass is true, then mass metal + mass nonmetal = mass compound
If two out of the three masses are known, the unknown mass can be calculated using this equation.
Example: iron + oxygen -> iron oxide

The mass of iron is 0.597 g. The iron metal is now burned in oxygen and forms the compound iron oxide which has a mass of 0.849 g.

What is the percent by weight of iron and oxygen in the compound iron oxide?
To find the mass of oxygen that reacted with the iron, the law of conservation of mass is used. mass iron + mass oxygen = mass iron oxide
0.597 g + mass oxygen = 0.849g mass oxygen = 0.849 g – 0.597 g = 0.252 g
 To find the % by weight of each element in the compound
% Fe = mass iron x 100 = 0.597 g x 100 = 70.3% Fe
mass iron oxide 0.849 g
% O = mass oxygen x 100 = 0.252 g x 100 = 29.7% O
mass iron oxide 0.849 g

Determination of % composition of elements using chemical formulas - There are some compounds that contain the same elements but in different proportions. In order to verify which compound was made, we can compare our experimental percentages (calculated in the last example) with the percentages calculated from the chemical formula (this example).
% element = (no. of atoms x atomic weight) x 100
formula weight of compound
The atomic weights are found on the periodic table.
Formula weight of a compound is the sum of the atomic weights of all the atoms in the compound.
Formula weight of iron oxide compound, FeO = (atoms of Fe x atomic wt. of Fe) + (atoms of O x atomic wt. of O)
Fe : 1 x 55.85amu = 55.85amu
O : 1 x 16.00amu = 16.00amu
Formula weight = 71.85amu
% Fe = 1 x 55.85amu x 100 = 77.73 % Fe % O = 1 x 16.00amu x 100 = 22.27 % O
71.85amu 71.85amu
Final Check: Add up the percentages, which should add up to 100 %.
% Fe + % O = 77.73 % + 22.27 % = 100 %.
Prelab:
1.
What is the difference between an empirical formula and a molecular formula?
2.
What ions does copper form?
3.
Based on your answer above, what are the two compounds that copper can make when combined with sulfur?
4.
What is the Law of conservation of mass?
5.
What is the Law of definite (or multiple) proportions?
6.
Sample calculations: In an experiment, a mixture of copper and sulfur was heated to produce a sample of copper sulfide. The following data were obtained: a.
Weight of the empty crucible 2.077 g b.
Weight of the crucible + copper 2.289 g c.
Weight of the crucible + copper sulfide 2.396 g
Using the above data, calculate the percentages of copper and sulfur in copper sulfide and derive its empirical formula. d.
Copper mass (b-a) = e.
Copper sulfide mass (c-a)= f.
Sulfur mass (e-d)= g.
Using the following formula determine the % Copper in copper sulfide =
% composition of elements by experiment = % element = mass element x 100
mass compound h.
% Sulfur in copper sulfide = i.
Start with the grams of Cu you calculated in d and find the Moles of Cu using dimensional analysis j.
Start with the grams of S you calculated in f and find the Moles of S using dimensional analysis k.
Use the moles of Cu and moles of S found in i and j to write the Empirical formula

7.
Find the % iron and % oxygen in the possible compounds of iron oxide: FeO, FeO
2
, Fe
2
O
3.
The first one has been done for you:
Formula weight of iron oxide compound, FeO = (atoms of Fe x atomic wt. of Fe) + (atoms of O x atomic wt. of O)
Fe : 1 x 55.85amu = 55.85amu
O : 1 x 16.00amu = 16.00amu
Formula weight = 71.85amu
% Fe = 1 x 55.85amu x 100 = 77.73 % Fe % O = 1 x 16.00amu x 100 = 22.27 % O
71.85amu 71.85amu
Final Check: Add up the percentages, which should add up to 100 %.
% Fe + % O = 77.73 % + 22.27 % = 100 %.
MATERIALS
Copper wire
Meker Burner
Sulfur
Crucible
Ring Stand with Ring
Wire Mesh
SAFETY
1. Wear your goggles and aprons.
Clay Triangle
Matches
Spoon
Pencil
2. Sulfur and sulfur dioxide are irritants and may be toxic, especially the sulfur dioxide. Sulfur dioxide gas becomes sulfurous acid in your lungs. Thus this experiment MUST be completed in the FUME HOOD.
PROCEDURES
1. Obtain a clean, dry crucible. To ensure the crucible is dry, heat it for 1 to 2 minutes with a Meker Burner.
Once you have heated the crucible use crucible tongs to handle the crucible
2. Allow your crucible to cool (4 to 5 minutes) on your ceramic pad. Test the heat from the crucible with the back of your hand.
3. Weigh the cool, empty crucible. Record your results in the data table.
4. Take a 15 cm piece of copper wire and use steel wool to clean it. Wrap the wire around a pencil to produce a
loose coil that will fit into the bottom of your crucible.
5. Place the wire coil in the crucible (flat against the bottom) and weigh the crucible and wire coil. Record your results.
6. Add enough powdered sulfur to cover the copper coil in the crucible.
7. Bring your crucible, crucible tons and ceramic pad to one of the stations under the hood. MAKE SURE THE
FUME HOOD IS TURNED ON.
8. Do not begin heating until you are told to do so.
9.Light your burner and place it under the crucible. The bottom of the crucible should be just above the inner cone of the burner flame. PULL DOWN THE FUME HOOD GLASS!

10. Record your observations. As the sulfur burns it will turn reddish-black. Continue heating until ALL the sulfur, yellowish-black in color, has disappeared from the crucible. Also, make sure that there is no yellowish-brown gas in the crucible. You are done when only a grayish-black coil of “copper sulfide” remains.
11. Turn off your burner and let the crucible cool on your ceramic pad for at least 5 minutes. Do not attempt to move it.
12. After the crucible has cooled for at least 5 minutes, weigh the crucible and its “copper sulfide” contents.
Record your results.
13. IF TIME PERMITS, cover the “copper sulfide” coil with additional powdered sulfur and burn again until all the sulfur is gone. Cool and reweigh the crucible and contents. If the new mass is higher, record the new mass.
(Make sure that the mass increase is NOT because if it did you didn’t burn all of the excess sulfur. The purpose of this additional step is to insure that the copper has completely reacted. Without this step, the center of the coil may still contain some unreacted copper. In practice, you would continue to add sulfur and heat until the mass of the product became constant. This procedure is called “heating to constant mass”, and is very common in analytical chemistry.
14. Discard the “copper sulfide” in the trash can. Clean and dry your crucible.
15. Wash your HANDS!

DATA TABLE
MASS OF CRUCIBLE (g)
MASS OF CRUCIBLE AND COPPER COIL
(g) –
MASS OF COPPER COIL (g)
(THE SAME FOR BOTH TRIALS)
MASS OF COPPER SULFIDE and crucible
(g) after heating
Mass of COPPER SUFLIDE (g)
You Calculate this
MASS OF SULFUR (g)
You calculate this
MOLES OF COPPER(show your calculations)
MOLES OF SULFUR (show your calculations
MOLES OF COPPER:MOLES OF SULFUR
RATIO (this is a ratio rounded to the nearest whole numbers)
TRIAL I TRIAL II
Same as trial 1
NA same

EMPIRICAL FORMULA
% COPPER IN your “copper sulfide” compound
% COPPER IN the real “copper sulfide compound – see Mrs. S or Mrs. B for the THEORETICAL % Cu
Calculate your percent error
OBSERVATIONS:
DATA ANALYSIS – SHOW ALL OF YOUR WORK IN ORDER TO RECEIVE FULL CREDIT!
1. Determine the mass of copper wire used in the experiment.
2. Determine the number of moles of copper wire used in the experiment.
3. Determine the mass of copper sulfide formed.
4. Determine the mass of sulfur that combined with the copper.
5. Determine the number of moles of sulfur used.
6. Calculate the copper/sulfur mole ratio. Round off to the nearest whole number(s).
7. Give (calculate) the empirical formula for copper sulfide.
8. Calculate the percent error between your experimental percentage of copper and the theoretical percentage of copper provided by your instructor.
9. What are the major sources of error in this experiment? Explain how the copper:sulfur ratio would be affected by each error you identify. If some of the excess sulfur was not burned off during the heating, how would the copper/sulfur ratio have been affected?
10. Examine the results obtained by other lab groups. Is there good agreement among these results? What does the class data tell you about the empirical formula of a compound?

Example :In an experiment, a mixture of copper and sulfur was heated to produce a sample of copper sulfide. The following data were obtained:
Weight of the empty crucible 2.077 g
Weight of the crucible + copper 2.289 g
Weight of the crucible + copper sulfide 2.396 g
Calculate the percentages of copper and sulfur in copper sulfide and derive its empirical formula.
Answer
Copper mass = 2.289- 2.077 = 0.212 g
Copper sulfide mass = 2.396- 2.077 = 0.319 g
Sulfur mass = 2.396- 2.289 = 0.107 g
%Copper in copper sulfide = (0.212/0.319)100 = 66.5%
% Sulfur in copper sulfide = (0.107/0.319)100 = 33.5%
Moles of Cu = 66.5 g (1 mol/63.55 g) = 1.05 mol Cu
Moles of S = 33.5 g (1 mol/32.0 g) = 1.05 mol S
Empirical formula = CuS