Conservation of Matter: Wherever you go, There You Are
Archaeological evidence suggests that people have been using copper
for at least 11,000 years. Relatively easy to mine and refine, people
discovered methods for extracting copper from its ores at least 7,000
years ago. The Roman Empire obtained most of its copper from the
island of Cyprus, which is where copper's name originated.
Used in large amounts by the electrical industry in the form of wire, copper is second only to silver in
electrical conductance. Since it resists corrosion from the air, moisture and seawater, copper has been
widely used in coins. Although once made nearly entirely from copper, American pennies are now made
from zinc that has been coated with copper. Copper is also used to make water pipes and jewelry, as well
as other items.
Pure copper is usually too soft for most uses. People first learned about 5,000 years ago that copper can be
strengthened if it is mixed with other metals. The two most familiar alloys of copper are bronze and brass.
Bronze, the first alloy created by people, is a mix of copper that contains as much as 25% tin. Early people
used bronze to make tools, weaponry, containers and ornamental items. Brass, a mix of copper that contains
between 5% and 45% zinc, was first used about 2,500 years ago. The Romans were the first to make
extensive use of brass, using it to make such things as coins, kettles and ornamental objects. Today, brass
is also used in some musical instruments, screws and other hardware that must resist corrosion.
But…what happens when copper runs the chemical gauntlet?
A. Cu (s) + HNO3 (aq) →Cu(NO3)2 (aq) + NO2 (g) + H2O (l)
NOTE: This reaction should be carried out in a well-ventilated area (preferably a hood). Weigh
approximately 0.2 g of copper foil pieces. Place the copper pieces in a 150 mL beaker and add 2.0 mL of
concentrated nitric acid (HNO3). Carefully touch the bottom of the beaker and note any change in
temperature. Using a glass rod continuously stir the reaction mixture until the solution turns green and the
evolution of gas has stopped. Add 50 mL of distilled water to the beaker. Be careful with the nitric acid:
like other strong acids, it will sting if you get it on your skin and can damage clothing; unlike most
other acids, it will also stain the affected area yellow. You may now remove the solution from the fume
hood and take it to your lab bench.
1. Write a balanced equation for reaction [A] and indicate the type of reaction that occurred (Table 1).
2. On addition of the acid, did the contents of the beaker heat up (exothermic reaction) or cool down
(endothermic reaction)?
3. When the nitric acid was added to the copper a gas formed. Write the formula and name of the gas
formed, and indicate the color of this gas.
4. Record the color of the liquid solution you have prepared. Give the formula and name of the product
responsible for the color observed in the solution after adding the water
B. Cu(NO3)2 (aq) + NaOH (aq) →Cu(OH)2 (s) + NaNO3 (aq)
Immerse a pH probe into the reaction mixture and slowly add 12ml of 3M NaOH solution copper solution
until the reaction turns basic. Use a pH probe to monitor the change in pH. Note any chemical changes that
occur during the course of the reaction. Be careful in handling NaOH as it is a strong base which will
sting if it contacts the skin.
1. Write a balanced equation for reaction [B] and indicate the type of reaction that occurred (Table 1).
2. Give the formula and name of the copper compound that has been formed in reaction [B].
3. What is the color of this copper compound?
C. Cu(OH)2 (s) → CuO (s) + H2O (l)
Heat the contents of the beaker with constant stirring until the solution reaches a temperature just below
boiling. Remove the beaker from the hot plate and begin heating about 100mL distilled water for the wash
step. Allow the precipitate (solid) to settle. Pour off (decant) the liquid into another beaker. To the solid
that remains add approximately 80 mL of hot distilled water. Stir the resulting mixture, allow the precipitate
to settle, and again pour off the liquid. (If this liquid is clear and doesn’t contain any solid, it may be
disposed of in the sink.)
Note: Do not allow the mixture to come to a boil. Boiling makes the black CuO so fine that the filtration
step is excessively long. Heat the beaker until all the blue Cu(OH)2 disappears and is replaced by black
CuO. Separate the solid CuO using vacuum filtration. Rinse the Filter and wash the CuO. Keep the solid
on the filter paper, and discard the filtrate. Discard the filtrate after the CuO has been thoroughly rinsed.
1. Write a balanced equation for reaction [C] and indicate the type of reaction that occurred (Table 1).
2. Give the formula and name of the copper compound (precipitate) that has been formed in reaction [C].
3. What is the color of this copper compound?
4. The addition of the hot water is called a wash step. The wash step removes the water soluble chemical
impurities that stick to the surface of the precipitate, purifying it. Give the formula and name of the
product from reaction [B] that is being removed by the water washes.
D. CuO (s) + H2SO4 (aq) →CuSO4 (aq) + H2O (l)
Add 12 mL of 6 M sulfuric acid (H2SO4) to the copper compound purified in reaction [C]. Questions
Reaction [D]
1. Write a balanced equation for reaction [D] and indicate the type of reaction that occurred (table 1).
2. Give the formula, color, and name of the copper compound formed in reaction [D].
E. CuSO4 (aq) + Zn (s) →Cu (s) + ZnSO4 (aq)
BE SURE TO CARRY OUT THIS REACTION IN THE HOOD! Return your reaction [D] solution to
the hood and slowly add about 0.8 g of zinc powder, stirring the solution constantly until the evolution of
gas ceases. If blue color remains in the liquid, return your reaction mixture to the hood, add another 0.2g of
zinc and repeat the process. Using vacuum filtration, separate the resulting solid from the solution. Discard
the solution and dry the solid in an oven. Be sure to mass the filter paper before drying. Once the drying
process is complete, mass the material and the filter paper. Record the mass of the dried solid.
Two reactions are taking place when zinc powder is added to the reaction [D] solution. (Hint: both reaction
involve zinc metal, one with H2SO4 and the other with CuSO4.
1. Write a balanced equation for the reaction of zinc with H2SO4 and indicate the type of reaction
that occurred (Table 1).
2. Give the formula, color, and name of the gas that has been formed in reaction E.
3. Write a balanced equation for the reaction of zinc with CuSO4 and indicate the type of reaction that
occurred (Table 1).
4. Why was it necessary to add more zinc if the solution remained blue in the last step?
5. At this point is copper present as a compound, element, or ion?
6. If you originally started with 1.50 grams of copper and you were able to recover 1.20 grams of
copper, what is your percent recovery (percent yield)?
Now, remember when you determined the mass of the copper you used in reaction A? How much copper
should you have recovered from reaction E?
A can of lentils says you can’t prove it!
OK…alle meine kinder…for you I have a hint.
In reactions carried out under normal laboratory conditions, matter is neither created nor destroyed, and
elements are not transformed into other elements. We’ve already established that equations depicting
reactions must be balanced; that is, the same number of atoms of each kind must appear on opposite sides
of the equation.
Chemists ordinarily work with weighable quantities of elements and compounds. For example, in an ironsulfur equation (Fe + S →FeS) the symbol Fe represents 55.845 grams of iron, S represents 32.066 grams
of sulfur, and FeS represents 87.911 grams of iron sulfide (ed note: where do these values come from?).
Because matter is not created or destroyed in a chemical reaction, the total mass of reactants is the same as
the total mass of products. If some other amount of iron is used, say, one-tenth as much (5.585 grams), only
one-tenth as much sulfur can be consumed (3.207 grams), and only one-tenth as much iron sulfide is
produced (8.791 grams). If 32.066 grams of sulfur were initially present with 5.585 grams of iron, then
28.859 grams of sulfur would be left over when the reaction was complete (what is a limiting reagent?).
The reaction of methane (CH4, a major component of natural gas) with molecular oxygen (O2) to
produce carbon dioxide (CO2) and water can be depicted by the chemical equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Here another feature of chemical equations appears. The number 2 preceding O2 and H2O is a stoichiometric
factor…Huh?! What’s that? This indicates that one molecule of methane reacts with two molecules of
oxygen to produce one molecule of carbon dioxide and two molecules of water (a recipe). The equation is
balanced because the same number of atoms of each element appears on both sides of the equation (here
one carbon, four hydrogen, and four oxygen atoms). Analogously with the iron-sulfur example, we can say
that 16 grams of methane and 64 grams of oxygen will produce 44 grams of carbon dioxide and 36 grams
of water. That is, 80 grams of reactants will lead to 80 grams of products.
The ratio of reactants and products in a chemical reaction is called chemical stoichiometry. Stoichiometry
depends on the fact that matter is conserved in chemical processes, and calculations giving mass
relationships are based on the concept of the mole. One mole of any element or compound contains the
same number of atoms or molecules, respectively, as one mole of any other element or compound. By
international agreement, one mole of the most common isotope of carbon (carbon-12) has a mass of exactly
12 grams (this is called the molar mass) and represents 6.02214179 × 1023 atoms (Avogadro’s number).
One mole of iron contains 55.847 grams; one mole of methane contains 16.043 grams; one mole of
molecular oxygen is equivalent to 31.999 grams; and one mole of water is 18.015 grams. Each of these
masses represents 6.0221 × 1023 molecules
What this tells is that the mole can be described several different ways. Moles can be defined
in terms of particles…..The key to moving back and forth between these different ways of thinking
about substances is to envision the mole as a middle man and by using a diagram such as the one you
see below to plot your moves.
For example: Consider the following question a student research team wishes to know how many
grams are in 2.15 mole of calcium hydroxide.
2.15 mole Ca(OH)2
74 grams
1 mole Ca(OH)2
158 grams
A somewhat more involved question might ask the following: Assuming you hade 1.56 grams of Ca(OH) 2.
How many formula units of Ca(OH)2 make up that mass. Now, in this case the strategy would look like
this.
Once again, this is simply a matter of following the plan. Start with the mass, pivot through the
molar mass, continue through Avogadro's number and you're done.
1.56 g
Ca(OH)2
1 mole Ca(OH)2
6.02 x 1023 f.u.
74 grams
1 mole Ca(OH)2
1.27 x 1022 f.u. Ca(OH)2
STOICHIOMETRY
Now, stoichiometry ( http://www.shodor.org/UNChem/basic/stoic/) is a more complex kind of
operation, but it still follows the same kind of logical progression....what ideas do I begin with, what
is my ultimate goal, and what conversion factors do I use to 'pivot' from one idea to another. The
difference is that you are asking questions like, "if I use 1.56 of one substance to produce another
substance, how much product am I going to get." Another question I could ask, might be, "if I want
to do , say, a metathesis reaction with Ca(OH)2 how much of another reactant, for example,
(NH4)3PO4 am I going to need for a complete reaction.
For this kind of operation, the pathway for solving the problem is a little more involved, but again, if
you follow the steps, it's not that big a deal. What is different about the operation is that it now
takes into account the mole ratios between the various reagents and products you are working
with. Consider the reagents I just described. If we react them together, this precipitation
metathesis will look like this:
3Ca(OH)2(aq) + 2(NH4)3PO4(aq)  Ca3(PO4)2(s) + 6NH4OH(aq)
There are a number of mole ratios we can identify from this balanced question. For example, 3 moles
of Ca(OH)2(aq) need to be reacted with 2 moles of (NH4)3PO4(aq). Here's another one. 3 moles
of Ca(OH)2(aq) will produce 1 mole of Ca3(PO4)2. There are others of course, but the point is that
you choose the mole ratios that take you from where you are beginning the operation to where you
want to go.
So, lets consider the examples I just described. Say that I want to find out how much Ca3(PO4)2 is
produced from 1.56 grams of Ca(OH)2...
HERE IS MY PLAN:
I'm going to start with 1.56 grams of Ca(OH)2, pivot through the molar mass to moles and use the
3:1 molar ratio that exists between the Ca(OH)2 and the Ca3(PO4)2
1.56 g Ca(OH)2
1 mol Ca(OH)2
74 g Ca(OH)2
1 mol Ca3(PO4)2
3 mol Ca(OH)2
310 g Ca3(PO4)2
2.18 g Ca3(PO4)2
1 mol Ca3(PO4)2
SOOOOOOOOOOOOOOOO:
This leads us back to Einstein’s little challenge. Can you prove how much Copper ought to be
recovered from reaction E?
Table 1:
Unbalanced Reactions
Reaction type
Cu (s) + HNO3 (aq) →Cu(NO3)2 (aq) + NO2 (g) + H2O (l) Oxidation-Reduction Reaction
Cu(NO3)2 (aq) + NaOH (aq) →Cu(OH)2 (s) + NaNO3 (aq) Double Replacement Reaction
Cu(OH)2 (s) → CuO (s) + H2O (l)
Decomposition Reaction
CuO (s) + H2SO4 (aq) →CuSO4 (aq) + H2O (l)
Double Replacement Reaction
CuSO4 (aq) + Zn (s) →Cu (s) + ZnSO4 (aq)
Single Replacement Reaction
What makes these series of reactions interesting is that we can ‘track’ the movement of copper from atom
(Cu) to ion (Cu2+), from metal, to aqueous solution to precipitate, to solution and finally back to metal. The
differences are easy to follow; several are quite striking. And the general sense that if 0.2 grams of copper
goes into the reaction, certainly given what we have heard about the Law of Conservation of Matter, 0.2
grams of copper ought to be recovered.
Stoichiometry allows us to determine for ourselves that the mathematics associated with mole ratios and
molar masses provides us a vehicle for ‘tagging’ the copper and follow it through all five reactions.
This leaves us with the final two goals.
1.
Use stoichiometry to show that if 0.2 grams of Copper reacted with nitric acid in reaction A, then
0.2 grams of Copper must be recovered in reaction E.
2. Calculate percent error…since if you did not recover 0.2 grams you have no one to blame but
yourself.