Exercise 4
Illinois Central College
Chemistry 130
Page 1
Name _________________________
Law of Definite Proportions and Types of Chemical Reactions
Objectives
Today's experiment will use the Law of Conservation of mass and the Law of Definite Proportions
to determine the percent composition of each element in the newly synthesized compound. (More
specifically, determine the % Cu and % S by weight in the compound, copper sulfide)
We will also take a close look at the four classes of chemical reactions and learn more
regarding how to write correct chemical formulas and balanced equations.
Part I: The Law of Definite Proportions
Background
A compound by definition is a substance made up of 2 or more elements combined in a fixed
proportion by weight. In lab, two methods are used in determining the composition of a pure
compound; 1) analysis and 2) synthesis. In analysis the compound is broken down into its
component elements. Whereas, in synthesis, the pure elements are chemically combined to
produce the compound.
In today's lab, synthesis will be used to determine the amounts of the different elements which
make up a compound. 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.
Two Laws that will be helpful in today's lab are:
1) Law of conservation of mass - mass can neither be created nor destroyed during a chemical
reaction
2) Law of definite proportions - a compound is always made up of elements in the same
proportion by weight.
Exercise 4
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Determination of % composition of elements by experiment
% element =
In general:
mass element
mass compound
× 100
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.
iron + oxygen →
Example:
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 =
%O=
mass iron
mass iron oxide
mass oxygen
mass iron oxide
× 100
× 100
=
=
0.597 g
0.849 g
0.252 g
0.849 g
× 100
× 100
=
=
70.3 % Fe
29.7 % O
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 × atomic weight)
×
formula weight of compound
100
Formula weight of a compound is the sum of the atomic weights of all the atoms in the compound.
The atomic weights are found on the periodic table
Exercise 4
Page 3
Example:Find the % iron and % oxygen in the possible compounds of iron oxide: FeO, FeO2,
Fe2O3
Formula weight of iron oxide compound = (atoms of Fe x atomic wt. of Fe) + (atoms of O x atomic
wt. of O)
FeO Fe : 1 × 55.85amu = 55.85amu
O : 1 × 16.00amu = 16.00amu
Formula weight = 71.85amu
1 × 55.85amu
71.85amu
% Fe =
%O=
1 × 16.00amu
71.85amu
× 100 = 77.73 % Fe
× 100 = 22.27 % O
Final Check: Add up the percentages, which should add up to 100 %.
% Fe + % O = 77.73 % + 22.27 % = 100 %.
Fe : 1 × 55.85amu = 55.85amu
O : 2 × 16.00amu = 32.00amu
Formula weight = 87.85amu
FeO2
1 × 55.85amu
87.85amu
% Fe =
%O=
2 × 16.00amu
87.85amu
× 100 = 63.57 % Fe
× 100 = 36.43 % O
Final Check: Add up the percentages, which should add up to 100 %.
% Fe + % O = 63.57 % + 36.43 % = 100 %.
Fe 2O3
Fe : 2 × 55.85amu = 111.70amu
O : 3 × 16.00amu = 48.00amu
Formula weight = 159.70amu
% Fe =
%O=
2 × 55.85amu
159.70amu
3 × 16.00amu
159.70amu
× 100 = 69.94 % Fe
× 100 = 30.06 % O
Final Check: Add up the percentages, which should add up to 100 %.
% Fe + % O = 69.94 % + 30.06 % = 100 %.
The experimental percentages found in the first example closely match the percentages found
from the chemical formula of Fe2O3. Therefore, the iron oxide compound synthesized in
example 1 is Fe2O3.
The reaction of copper and sulfur that we will conduct today is referred to as a Combination
Reation where two elements combine to form a compound.
Part II: Types of Chemical Reactions
Many chemical reactions fall into one of the following four categories:
A. Combination:
B. Decomposition:
C. Displacement:
A + B
AB
AB + C
D. Double Displacement: AB + CD
AB
A + B
CB + A
AD + CB
Exercise 4
Page 4
Note: Because you will investigate examples of each reaction type and will write balanced equations
representing each reaction you carry out, a Periodic Table of the Elements and your valence
sheet will be helpful as you write the required formulas for the compounds in the equations.
You are reminded that the free (elemental) forms of hydrogen, nitrogen, oxygen,
fluorine, chlorine, bromine, and iodine are diatomic; that is, they are symbolized as: H2,
N2, F2 O2, C12, Br2, and I 2 respectively.
A thorough knowledge in balancing chemical reactions is required and helpful in predicting the
outcome of a chemical reaction. Complete the prelab exercise at the end of this section before
you come to your laboratory class. An answer sheet is also provided at the end of this exercise
for you to complete during the lab.
Procedure: Part I Law of Definite Proportions
A. Determining percent composition from experimental results:
Use the laboratory balance to determine the mass of a small strip of copper. Then bend the copper
strip into a "u" shape. In the hood there are flasks containing boiling sulfur, so that reddish-brown
sulfur vapors almost fill the flasks. Using a glass rod with a bend at one end, place the copper strip
loosely on the glass hanger and carefully lower it into the sulfur vapor. Try to hold the copper in
the center of the flask, keeping it from touching the sides of the flask or the boiling liquid at the
bottom. The reaction between the copper and the sulfur vapor will cause the copper strip to glow.
The glow will start at the bottom of the copper strip and then will travel up to the bend at the top. As
soon as the top of the strip stops glowing, the reaction is complete. (The reaction will take only about
5 seconds.)
Carefully remove the hanger with the suspended copper sulfide from the flask, but continue to hold it
well back in the hood until the excess sulfur has burned away.
When the burning of the sulfur has ceased, remove the copper sulfide to a 50 mL beaker and allow
it to cool. If the copper sulfide tends to stick to the glass hanger, it can be pulled away with forceps.
The copper sulfide is brittle and it may break. If it does, be careful not to lose any of the pieces in
transferring the copper sulfide to the beaker.
If the copper sulfide has remained in one piece, simply measure its mass after it has cooled. If it
is in pieces, determine its mass in the beaker, and then determine the mass of the empty beaker and
subtract to determine the mass of the copper sulfide. The mass of the sulfur in the copper sulfide
compound is the difference in mass between the copper and the copper sulfide. Calculate the
percent copper and the percent sulfur in the compound produced in this experiment.
B. Determining percent composition from a chemical formula:
Calculate the percent copper and the percent sulfur in CuS, CuS 2 and Cu2S.
C. Discovering the chemical formula of a synthesized compound.
Compare the percent composition of the compound prepared experimentally with those calculated
from the chemical formulas CuS, CuS 2 and Cu2S.
Exercise 4
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Procedure: Part II Types of Chemical Reactions
A. Combination (or Addition) Reaction:
An example:
Ca + S
CaS
In the above example two elements combine to form a compound. The combination of two smaller
compounds to form a larger one also meets this criterion.
1. The products of simple combustion reactions results in the addition of oxygen in the air to form
the oxides of each of the elements in the compound, carbon dioxide and water. Write a
balanced equation for the burning of propane, C3H8.
2. Using tongs hold a small piece of magnesium metal in the flame. The magnesium reacts with the
oxygen in the air to form magnesium oxide. Describe what happens. Write a balanced
equation for this reaction.
3. The burning of magnesium occurs at such a high temperature that some nitrogen in the air can
react with the magnesium to form some magnesium nitride. Write a balanced equation for
this reaction.
B. Decomposition Reaction:
An example:
2 KClO 3
2 KCl + 3 O2
The starting material, usually a single compound, is broken into its component elements or simpler
compounds.
1. When mercury(II)oxide is heated, it decomposes to metallic mercury and oxygen gas. Write a
balanced equation for this simple de composition reaction.
2. In the exercise on physical/chemical changes you heated ammonium chloride. Recall or repeat
the experiment where approximately one gram of ammonium chloride is gently heated in a large
test tube over a flame. Carefully smell the vapor. Ammonia, NH3 , is detected along with
another gas. Write the balanced equation for this reaction.
Exercise 4
Page 6
C. Displacement Reaction:
An example:
CuCl2 + Zn
Cu + ZnCl2
An element reacts with a compound in this type of reaction. The free element, which is chemically more
active, forces an element to leave and takes its place.
1. Place a few mL of dilute hydrochloric acid in a test tube. Drop a small piece of aluminum metal
into the acid. Aluminum is displacing the hydrogen in the hydrogen chloride. The hydrogen
bubbles out as a gas as the aluminum dissolves.
Describe what happens.Write a
balanced equation for the reaction.
2. Place a few mL of water in a small beaker. Add a small calcium turning. Calcium is very active
and even displaces one of the hydrogens in water. Describe what happens.
Write a
balanced equation for the reaction.
D. Double Displacement (Metathesis) Reaction:
An example:
NiCI2 + CaS
CaCl2 + NiS(s)
This is a sort of "partner exchange" reaction. It frequently occurs when two ionic substances in aqueous
solution are mixed together. If there is no interaction among the ions, no reaction occurs. However, if
the ions can combine to form an insoluble solid, a gas, or a non-ionized product, such as water, then
the double displacement reaction takes place.
1. To 1 mL of potassium chloride solution add 1 mL of silver nitrate solution. An insoluble
precipitate of silver chloride appears. Describe what happens.
Write a balanced equation for this reaction.
2. To 1 mL of sodium carbonate add 1 mL of dilute hydrogen chloride. In this case one of the
products of the ion partner exchange reaction is unstable and decomposes to form water and
carbon dioxide gas. Describe what happens.
Write a balanced equation for this reaction.
3. To 1 mL of barium hydroxide add 1 mL of dilute hydrogen sulfate (sulfuric acid). Notice both
the heat released due to the formation of water and observe the formation of the precipitate.
Describe what happens.
Write a balanced equation for this reaction.
Exercise 4
Illinois Central College
Chemistry 130
Name _________________________
REPORT SHEET
Law of Definite Proportions and Types of Chemical Reactions
Part 1: PERCENT COMPOSITION OF A COMPOUND
A. Determining percent composition from experimental results:
Trial #l
Trial #2
Mass of copper
_______________
_______________
Mass of copper sulfide
_______________
_______________
Mass of sulfur
_______________
_______________
Percent Copper
_______________
_______________
Percent Sulfur
_______________
_______________
Show Calculations:
B. Determining percent composition from chemical formula:
CuS
CuS2
Cu2S
% Copper
__________
__________
___________
% Sulfur
__________
__________
___________
Show All Calculations:
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Exercise 4
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C. Questions
1. Comparing your experimental results with the %Cu and %S determined from the chemical
formulas, give the chemical formula for copper sulfide formed by this experiment
2. Which chemical law was illustrated in this experiment?
3. Which chemical law was used in this experiment?
4. a) Is the reaction exothermic or endothermic?
b) How do you know?
5. Calculate the percent composition of the following compounds from their formulas.
Show all calculations.
CaSO4
%Ca _________
%S _________
%O __________
NH4NO3
%N __________
%H __________
%O __________
Exercise 4
Part II: Types of Chemical Reactions
A. Combination Reactions
1. Equation:
2. Result:
Equation:
3. Equation:
B. Decomposition Reactions
1. Equation:
2. Equation:
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Exercise 4
Page 10
C. Displacement Reactions
1. Result:
Equation:
2. Result:
Equation:
D. Double Displacement Reactions
1. Result:
Equation:
2. Result:
Equation:
3. Result:
Equation:
Exercise 4
Illinois Central College
Chemistry 130
Page 11
Name _________________________
PRELAB: Exp.4 Types of Chemical Reactions
Balance the following chemical reactions after determining the correct formulas for each of the elements
or compounds.
1.
Aluminum + Oxygen
Aluminum oxide
2.
Hydrogen sulfate + Ammonium hydroxide
3.
Butane (C4H10) + oxygen
4.
Zinc + Hydrogen chloride
5.
Barium acetate + Sodium phosphate
Water + Ammonium sulfate
Carbon Dioxide + Water
Zinc chloride + Hydrogen(g)
Barium phosphate + Sodium acetate
Balance the following net ionic equation:
6.
______Ba +2 + ______PO4-3
7.
______H2CO3 + _____OH-
______ Ba3(PO4)2(S)
______H2O + ______CO3-2
Exercise 4
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