Lesson Four
1. Please read the material below.
2. Complete the worksheet attached to this Power School link.
Topics:
* Types of Chemical Reactions
* Reactions Involving Carbon Compounds
The Mole
Day 1
A.
Types of Chemical Reactions
There are five general types of reactions that we will study this
year in our chemistry unit. All either require heat (endothermic) or give off
heat (exothermic) but generally these heat changes are not shown in the
reaction statement.
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1.
Formation reaction
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2.
Single decomposition reaction
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3.
Complete combustion
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4.
Single replacement
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5.
Double replacement
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1)
Formation reactions
• In this reaction a compound is formed out of 2 elements. The
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compound can either be ionic (metal and nonmetal) or it can be molecular
(nonmetal and nonmetal)
Mg (s)
+
O2 (g)
---->
MgO(s)
solid magnesium +
oxygen gas ----> solid magnesium oxide
an element
+
an element ----> a compound
2)
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Single decomposition reaction
This reaction is basically the opposite of a formation reaction. To
decompose means to break down , and in this reaction a compound breaks
down into its elements.
H20 (l)
---->
H2 (g)
+
O2
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(g)
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liquid water
----> hydrogen gas
a compound
an element
+
oxygen gas
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3) Combustion
+
an element
Hydrocarbons are compounds which contain carbon and hydrogen like
methane, methanol or glucose. They burn in the presence of oxygen and
always produces carbon dioxide gas and water. Water is a vapor if the
burning occurs in air and a liquid if the reaction occurs in an organism.
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(g)
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CH4 (g)
+
O2 (g)
methane gas + oxygen gas
carbon dioxide gas
4)
----->
H2O(g)
----> water vapor
+
CO2
+
Single replacement
There is a replacement of one element for another .
• In this example a metal (Ag) which is part of a compound AgNO3 is
replaced by another metal (Cu) to form another compound Cu(NO3)2.
Cation is replaced for cation.
5)
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Double replacement
An element from one compound is switched with an element from
another compound. With the single replacement reaction, only 1
compound had elements switched in it , but this time both compounds
have elements switched so this is a double replacement reaction.
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aqueous
aqueous
aqueous
solid
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calcium chloride + sodium carbonate ---> sodium chloride +
calcium carbonate
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Read pages 103- 105 of your text and study the chart on p.
114
Some experiments that you may want to try . .
Surface tension on coins
Start with a dry penny and using an eye dropper gently add water
to its surface one drop at a time. How many drops will the penny hold
before it overflows.
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Evidence of chemical change
Many commercial processes involve chemical reactions.
Products can be purchased for cleaning copper and removing tarnish
from silver. Most of these products have been developed using an
understanding of chemistry. The products have important
applications in our lives.
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This activity takes some of the mystery out of how these
products work. It also helps to show students that inexpensive
alternatives can sometimes be found to replace expensive products
which essentially do the same thing. This helps to develop a sense of
consumer awareness to help the environment.
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Experiment #1
Compare some crystals to the products formed when chemical
reactions occur. Dissolve some epsom salts in water. Pour off some of
the solution into an evaporating dish and allow the liquid to evaporate.
Add ammonia to the remaining epsom salt solution. Allow the mixture to
settle. Decant the clear liquid and allow it to evaporate. Compare the
ammonium sulfate crystals formed with the original epsom salts.
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Repeat, replacing epsom salts with alum. Add vinegar to the
precipitate. Evaporate and compare the product to the powder
produced when an alum solution is allowed to evaporate.
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Experiment #2
Make saturated salt solutions. (This means to add salt to almost
boiling water until it will no longer dissolve and some stays in the
bottom of the pot in spite of stirring.) Use sodium chloride (table
salt), potassium chloride (salt substitute), and/or powdered alum.
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Take some of this solution and hang a weighted string in the salt
solution. As evaporation occurs,keep the level in the container constant
by adding more saturated salt solution. Once the crystals have grown
for several weeks, remove and dry them. Observe them carefully with a
hand magnifier. Draw the shape of the crystals. Are all the crystals
alike?
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Experiment #3
Add some table salt to vinegar. The solution will clean pennies.
After a number of pennies have been cleaned using the solution, add a
nail to the solution and observe what happens.
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Experiment #4
Add some baking soda to hot water and dissolve. Pour the mixture
into an aluminum tray. Put a tarnished piece of silverware into the tray
and let it stand for about an hour. Remove the silverware, rinse with
tap water, and dry it. Compare it to another tarnished piece of
silverware which did not receive this treatment.
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Experiment #5
Pour about 30 mL of vinegar into a pop bottle. Put some baking
powder in a balloon. Tap the baking powder so that it goes down to the
bottom of the balloon.
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Fit the opening of the balloon over the mouth of the pop bottle.
Lift the balloon and shake the contents into the pop bottle. When the
baking powder and the vinegar mix, a chemical reaction takes place. Gas
is released. The balloon expands. Do this on a mass balance to show
that mass is conserved during a chemical reaction.
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B
Reactions Involving Carbon Compounds
The study of carbon-containing compounds and
their propertis is called ORGANIC CHEMISTRY.A few carboncontaining compounds( such as carbon dioxide, carbon monoxide and
ionic carbonates) are not considered organic compounds. This still
leaves millions of different compounds. In fact organic compounds
far outnumber inorganic compounds.
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A HYDROCARBON is an organic compound that
contains only the elements carbon and hydrogen, People obtain
useful hydrocarbons by refining crude oil and natural gas. About
95% of these hydrocarbons are burned as fuels in exothermic
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combustion reactions. The resulting thermal energy warms homes,
businesses, and schools and provides energy for transportation. A
combustion reaction is one where there is burning and O2 is
required along with the hydrocarbon fuel. Typical products of
burning are C02 and H20
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Kinds of Combustion
1) Complete Combustion: A reaction where a fuel is burned in
an oxygen rich enevironment to produce C02 and H20 vapor and a
release of thermal energy ( exothermic) . Ity looks like this in an
equation:
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Hydrocarbon( fuel) + O2 ----> C02 and H20 + Thermal
energy
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(vapor)
(Heat)
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Some examples of hydrocarbons that are fuels are:
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Methane CH4
Ethane C2H6
Propane C3H8
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Read page 108 of your text
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2) Incomplete combustion : When hydrocarbons are burned in
a poor supply of oxygen we have incomplete burning. The reaction
is still exothermic but it does not generate as much thermal
energy( heat) as complete combustion. The products of incomplete
combustion are carbon dioxide and water as before, but also
carbon (soot) and carbn monoxide. Carbon monoxide is a colorless,
odourless, highly toxic gas.Breathing carbon dioxide interferes with
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your body’s oxygen transport system. Blood cells contain a
chemical, called hemoglobin, that combines with oxygen in your
lungs. The oxygen is later released to other cells in your body.
Carbon monoxide bonds to about 200 times more strongly than
oxygen to hemoglobin. If you breathe carbon monoxide, you reduce
the amount of oxygen that can bind to hemoglobin. As a result,
cells die. To reduce the danger of carbon monoxide poisoning,
never use fuel-burning apparatus without proper ventilation
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Read p. 109 of your text
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One sure sign that a chemical reaction has occurred is the
formation of a gas. Since gases such as hydrogen , oxygen and
carbon dioxide are clear and colourless, chemists have developed
standard tests to identify them Take careful precautions when
performing these tests. See p. 109 of your text or below
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Day 2
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C
The Mole
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When you buy a carton that says it contains 1 dozen
eggs, you know you have 12 eggs, The term “dozen” is a counting
unit. We use counting for convenience all the time. For example you
group your socks into counting units of two , called pairs. Yu might
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buy a six-pack of pop. there are lots of other examples.
The mole is one such counting unit and chemists use the
mole to represent the number of atoms, molecules or formula units
in a sample of substance, In even a small sample the number of
atoms , molecules or formula units is a mole is very large, in fact
the number of these structures in a mole is this large number. It
is
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6.02 x 10 23
So 1 mole( mol ) of any substance contains 6.03 x 10 23
atoms, molecules or formula units. Thus we can say
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* The mole( symbol mol) is defined as the amount of
substances that contains as many elementary entities( atoms,
molecules, or formula units) as exactly 12 g of carbon-12, the
most common isotope of carbon.
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* The mole ( 1 mol) of a substance has been determined to
contain 6.03 x 10 23 elementary entities of the substance. This
large number is called Avogadro’s number.
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eg: there are 6.03 x 10 23 carbon atoms in 1 mol of carbon,
C. There are 6.03 x 10 23 carbon dioxide molecules in 1 mol of
carbon dioxide, CO2. There are 6.03 x 10 23 sodium chloride
formula units in 1 mol of Sodium Chloride, NaCl(s).
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The number 6.03 x 10 23 is enormous . To give you some idea
of the madnitude of Avogadro’s number read figure 3.22 and figure
3.23 on p. 116 of your text.
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How much does 1 mol of any
substance weigh?
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The molar mass of substance has been determined for each
element in the periodic table. 1 mol of each element has a mass
called the atomic molar mass and is represented as g/mol. If you
look at the periodic table you will see the molar atomic mass for
each element in the upper right corner of each square. Thus the
molar atomic mass for:
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Na = 22.99 g/mol or 1 mol of Na weighs 22.99g or
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6.02 x 10 23 atoms of Na weigh 22.9 g
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Cl= 35.5 g/mol or 1 mol of Cl weighs 35.5 g/mol or
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6.02 x 10 23 atoms of Cl weighs 35.5 g
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etc.......
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Some elements exist as molecules , not atoms. For example,
the element nitrogen exists as a molecule composed og 2 nitrogen
atoms, N2(g). Therefore 1 mol of nitrogen molecules contains 2
mol of nitrogen atoms. The molar mass of nitrogen molecules is
therefore twice the atomic molar massof nitrogen as shown in the
periodic table: 2 x 14.01 g/mol = 28.02 g/ml. In other words 1
mol of molecular nitrogen has a mass of 28.02 g.
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The Molar mass of a Compound
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The molar mass (M) with units g/mol. is used to refer to mass
of 1 mol of any pure substance. You can determine the molar mass
of a compound by using the formula for the compound. For
example, teh formula for carbon dioxide , CO2 tells you taht each
molecule contains 1 carbon atom and 2 oxygen atoms. So it looks
like this from the Periodic table
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The Molar mass of C02 is :
C= 12.01g/mol
0 = 2 x 16.00g/mol = 32.00 g/mol
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Total = 12.01g/mol + 32.00g.mol = 44.01 g/mol
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That means 1 mol of CO2 weighs 44.01 g or M of CO2 =
44.01 g./mol
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Let’s find the M of Mg( NO3)2 :
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Using the Periodic Table:
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Mg: 24.31 g/mol
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N: 2 x 14.01 g/mol = 28.02 g/mol
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0: 6 x 16.00 g.mol = 96.00g/mol
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Total + 24.31 g/mol + 28.02 g/mol + 96.00g/mol= 148.33
g/mol
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therfore the M of Mg( NO3)2 = 148.33 g/mol.
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Read p. 118-120 of your text
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Converting between Mass and Moles
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It is often useful for chemists to convert between the
amount of a sample expressed in moles and the mass of the sample
expressed in grams. To do thsi we use a method called the
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Factor Label Method. The factor label method uses
conversion factors to change units without affecting the value. You
can use conversion factors to change an amount of a sample
expressed in moles to the mass of the sample expressed in grams.
In this case, teh conversion factor is the molar mass of teh
substance.
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eg: To determine the mass of 2 mol of helium, use the molar
mass of helium ( periodic Table) , 4.00g/mol, as a conversion
factor. You can do this because 4.00g og Helium is equivalent to 1
mol of Helium.
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conversion factor - 4.00g He/ 1 mol He
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thus multiply 2 mol by the conversion factor to determine its
mass
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massof Helium= 2 mol He x 4.00g He/ 1 mol He
cancel)
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= 8.00g of He
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(mol He
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Notice that the answer has the desired units, grams. The
problem below shows how to use this method to convert from
amount to mass of a compound.
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Alternatively you nmay use the following equation to conmvert
between mass and moles:
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n= m/M
n= amount ( mol)
m= mass(g)
M = Molar mass(g/mol) from periodic table
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eg: What is the mass of 7.50 mol of H2O.
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!st step: Find the M of H20 ( using periodic table)
H= 2 x 1.01 g/mol = 2.02 g/mol
0= 1 x 16.00 g/mol = 16.00g/mol
M of H20
= 18.02 g/mol
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using the formula n= m/M amd solving for m we get
m= nM
= 7.50 mol of water x 18.02 g/mol (mols
cancel)
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= 135.15g or
= 1.35 x 10 2
g is the mass of 7.50 mol of
water
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Another Example: The mass of an iron bar is 16.8g. What
amount ( in mol) of iron is in the sample.
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Iron is Fe and using the periodic table you see that it’s M is
155.85 g/mol
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Using the formula: n= m/M we can find the number of moles
of Fe
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= 16.8 g / 155.85 g/mol
( g’s
cancel)
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= 0.301 mol
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Therefore you can say that 16.8 g of iron contains 0.301 mol
of iron.
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Read p. 120-121 of your text
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D
Mass
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The Mole and the Law of Conservation of
Using the mole concept , you can relate the coefficients in
balanced chemical equations( the numbers in front of the formulas)
to the mass of the substances involved. Start by thinking about
how the coefficients relate to moles of substances, For example,
let’s look at this balanced equation
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2H2(g) + O2(g)----------> 2H20(l)
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2 molecules of H2 + 1 molecule of O2-----> 2 molecules of
H20
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2 moles of H2 + 1 mole of O2 --------> 2 moles of
water
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4 atoms of H
+ 2 atoms of O --------> 4 atoms of H
and 2
atoms of
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O
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2 x 6.03 x 10 23 + 1 x 6.03 x 10 23 -----> 2 x 6.03 x 10
23
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molecules
molecules
molecules
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Notice the # of each kind of atom is balanced on each side
of the equation so mass is conserved
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I can multiply everything by 2 and repeat the above and
repeat the proceedure above so that mass will still be conserved.
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4H2(g) + 2O2(g)----------> 4H20(l)
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4 molecules of H2 + 2 molecule of O2-----> 4 molecules of
H20
4 moles of H2 + 2 mole of O2 --------> 4 moles of
water
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8 atoms of H
+ 4 atoms of O --------> 8 atoms of H
and 4
atoms of
O
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4 x 6.03 x 10 23 + 2 x 6.03 x 10 23 -----> 4 x 6.03 x 10
23
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molecules
molecules
molecules
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What this all shows is that in a balanced equation the total
mass of the product is equal to the total mass of the reactants
OR the total mass of the substances on the reactants side is equal
to the total mass of the substances on the products side. The
table below shows that balanced chemical equations obey the law
of conservation of mass when they refer to moles
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Read p. 124 of your text
Science 10
Unit A
Chemical Reactions
Assignment Sc 10.04
Topics:
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* Types of Chemical Reactions
* Reactions Involving Carbon Compounds
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The Mole
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Sc 10.04 assignment
(48 marks )
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Instructions: Please complete each section as
required
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DAY 1
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Part A. Complete questions 2, 3, 6 and 7 on page
114-115 of your text. ( 36 marks)
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2.( 8 marks - 2 marks for each question)
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a.
b.
c.
d.
3. ( 8 marks- 2 marks for each question)
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a.
b.
c.
d.
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6.( 8 marks - 2 marks for each question)
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a.
b.
c.
d.
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7. ( 12 marks - 2 marks for each question)
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a.
b.
c.
d.
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e.
f.
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Day 2
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Part B: Complete the following Practice problem
on p. 123 of your text ( 12 marks)
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17. ( 4 marks)
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a.
b.
c.
d.
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18. ( 4 marks)
a.
b.
c.
d.
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19. ( 4 marks)
a.
b.
c.
d.