RooneyCuSlab
MISE – Chemistry 511
Mary Anne Rooney
October 22, 2006
Lab Report – Chemical Formula of a Copper-Sulfur Compound
Mass of crucible (constant weight)
10.8800 g
Mass of crucible + copper wire
11.3902 g
Mass of Cu reacted
0.51 g
Mass of crucible + copper sulfur compound after heating 11.5188 g
Mass of copper sulfur compound
.6388 g
Calculations of empirical formula of copper sulfur compound (Show all work).
g Cu = p * AW Cu
gS
q
AW S
.51
63.546
.1288  .0080  2
32.06
.0040
1
Empirical formula of copper sulfur compound
Cu2S
Pre lab questions:
1. How do you know when all of the sulfur added (to the fixed amount of Cu) has reacted?
All of the sulfur has reacted when the flames have ceased in the covered crucible.
2. a. How do you minimize the possible reaction of the Copper with oxygen (O2) from the
air to limit the amount of CuO that could be produced as a by-product?
To limit the amount of CuO that is produced you keep the lid on the crucible.
b. Write a balanced chemical reaction for the production of CuO.
2Cu + O2 = 2CuO
3. Why is it crucial to weigh the Cu on an analytical (higher precision) balance but use a
triple beam (lower precision) balance to mass the sulfur?
Because the copper has such a low mass, it is more accurate to weigh it on an
analytical balance.
4. Why was this reaction performed under a hood?
This reaction was performed under a hood to insure safety during the reaction that was
taking place between the sulfur and the copper (as well as the air).
Post lab questions:
1. Consider an empty crucible – heated to a constant mass of 20.0605 grams. A quantity of
solid elemental silicon (Si) is placed in the initially empty crucible; whereupon the mass of
the crucible and contents is found to be 20.5117 grams. The crucible and solid silicon are
subsequently heated in a pure nitrogen (N2) environment. The nitrogen is considered to be
in excess. Following a procedure similar to the one performed in this experiment, the final
constant mass of the crucible and newly formed compound of silicon and nitrogen weighs
20.8116 grams. Determine – with explanation – the empirical formula of this resultant
compound of Si and N.
The amount of silicon used is 0.4512 grams because you needed to subtract 20.5117
(crucible + silicon) minus 20.0605 (crucible). The amount of compound SiN2 (formula
used at this point is simplified from the known symbols of silicon and nitrogen) is 0.7511
grams because you subtracted 20.8116 (crucible + compound) minus 20.0605
(crucible). The amount of nitrogen is 0.299 grams because you subtracted 0.7511
(compound) minus 0.4512 (silicon).
From this point we use the formula for law of multiple proportions:
g Si = p * AWSi
g N2 q AW N2  0.4512/0.299 = 1/1 * 28.0855/14.0067 
1.5090/2.0051 = 2.0051  AW = 0.75 = ¾  Si3N4
I should add at this point that I don’t believe that this formula would balance because
the symbol for nitrogen is N2.
2. For the purposes of this question, presume that the chemical formula of the coppersulfur compound is Cu2S. How many kilograms of this compound could maximally be
produced from a 1.00 foot length of copper wire – with a cross-sectional area of 2.00 square
millimeters (mm2) if there is more than enough sulfur available? The density of copper at
these conditions is 8.92 g/mL (i.e., 8.92g/cm3).
I used dimensional analysis to begin:
1.00 ft x 1 cm
x 1000 mm = 30.84 cm (length)
2.54 in
100 cm
2 mm2 x
1 cm x 1 cm = 0.02 cm2 (area)
100 mm
100 mm
0.02 cm2 x 30.84 cm = 0.6096 cm3 = Volume
D = m/V  8.92 g/cm3 = m / 0.6096 cm3  5.4376 g ÷ 1000 = .0054 kgCu
g Cu / g S = 2/1 x AW Cu/AWS  2/1 x 63.546/32.065 = 127.092 g/32.065g 
0.0054 kg Cu / .12709 kg Cu2S = 0.0428 kg