Download BRCC CHM 101 Chapter 5 Notes (Chapter 4 in older text versions)

BRCC
CHM 101
Chapter 5 Notes
(Chapter 4 in older text versions)
Page 1 of 8
Chemical Reactions
atomic mass - the average mass of an element expressed as atomic mass units, amu (sometimes
called atomic weight)
formula weight - the sums of the atomic weights of all the atoms in a molecular formula for a molecule
or ionic compound (sometimes called molecular weight)
How do you calculate formula weight?
ex. CO2 C has atomic weight 12.01 and there’s one C atom C = 12.01 x 1 = 12.01
O has atomic weight 16.00 and there’s two O atoms O = 16.00 x 2 = 32.00
--------Formula Weight = 44.01 amu
(NH2 )2 CO
N = 14.01 x 2 = 28.02
H = 1.01 x 4 = 4.02
C = 12.01 x 1 = 12.01
O = 16.00 x 1 = 16.00
---------Formula Weight = 60.05 amu
Ba3 (PO4 )2
Ba = 137.33 x 3 = 411.99
P = 30.97 x 2 = 61.94
O = 16.00 x 8 = 128.00
---------Formula Weight =
601.93 amu
Try the following for practice : C12 H22 O11 , NaHCO3 , CuAl6 (PO4 )4 (OH)8
Moles
mole - the formula weight of a substance expressed in grams
*
In chemistry we are interested in the relative masses of different atoms - using moles is a way to
relate these masses.
Consider the following:
H atom weighs about 1 amu
He atom weighs about 4 amu
He is 4 X heavier than H
100 He atoms would weigh 4 X as much as 100 H atoms
THEREFORE, there are the same number of atoms in 4 grams of He as there are in 1 gram of H.
 This number of atoms has been determined and given the name Avogadro’s Number
23
Avogadro’s number = 6.02 x 10
<------- a really BIG number
BRCC
CHM 101
Chapter 5 Notes
(Chapter 4 in older text versions)
Page 2 of 8
mole - Avogadro’s number of anything
1 mole of H atoms
1 mole of He atoms
1 mole of C atoms
1 mole of gummy bears
=
=
=
=
23
6.02 x 10
23
6.02 x 10
23
6.02 x 10
23
6.02 x 10
H atoms
He atoms
C atoms
gummy bears
=
=
=
=
1 gram
4 grams
12 grams
many, many grams!
 Moles are used as a practical and convenient means of measuring or weighing out a given number
of atoms or molecules.
How do we use moles?
* converting from grams to moles
If you know grams of something, then you can convert to
moles using the formula weight.
ex. convert 8.2 grams of CO 2 to moles
8.2 g CO2
x
1 mole CO2
----------------44.0 g CO2
=
0.19 moles CO2
(notice how g CO2 cancels)
Try converting these: 27.5 g SnF2 , 756.2 g ZnCl2
* converting from moles to grams
If you know the moles of something, then you can convert to
grams using the formula weight.
ex. convert 3.41 moles of C2 H5 OH to grams
first find the formula weight
3.41 moles C2 H5 OH
x
C = 12.01 x 2 = 24.02
H = 1.01 x 6 = 6.06
O = 16.00 x 1 = 16.00
--------46.08 g/mole or amu
46.08 g C2 H5 OH
----------------------- = 157 g C2 H5 OH
1 mole C2 H5 OH
Try converting 0.145 mole NaCl to grams:
Try converting 2.84 mole K2 S to grams:
BRCC
CHM 101
Chapter 5 Notes
(Chapter 4 in older text versions)
Page 3 of 8
* using the ratios between individual atoms and the molecule. The mol ratio is represented by the
subscripts in the formula.
ex. 1 mol NaHCO3 contains 1 mol Na : 1 mol H : 1 mol C : 3 mol O
In 21.4 moles of TNT, C7 H5 (NO2 )3 , how many moles of N atoms and how many moles of O atoms
are there?
* In 1 mole of TNT, there are 3 moles of N and 6 moles of O.
21.4 moles TNT
x
3 moles N
-----------------1 mole TNT
x
6 moles O
----------------1 mole TNT
21.4 moles TNT
= 64.2 moles N
= 128 moles O
Try this one: The formula for aspirin is C9 H8 O4 . How many moles of C are in 0.4375 moles of aspirin?
* convert grams to number of molecules - not usually very practical, but easy to do
ex.
A drop of water weighs 0.050 g. How many molecules of water are in the drop?
Chemical Equations
A chemical reaction has reactants and products represented by a chemical equation.
ex. 2 H2
(g)
+
O2
(g)
------------------------>
2 H2 O (l)
Balanced chemical equation - an equation where the same number of atoms of each element are on
both sides of the equation, which gives the relative numbers of molecules and relative numbers of
moles.
How do we balance equations? Only by adjusting the coefficients in front of the molecules.
Rules :
1. Cannot change the formula of any reactant or product.
2. Cannot break up a formula.
Steps for balancing equations:
1.
Write the equation using the correct formulas for all of the reactants and products.
2.
Count the number of atoms of each element on the left of the arow and write the numbers down.
Do the same for the right side.
3.
If the number of atoms of each element is not the same on both sides of the arrow, change the
coefficients until they are.
BRCC
4.
CHM 101
Chapter 5 Notes
(Chapter 4 in older text versions)
Page 4 of 8
When the equation is balanced, check the coefficients to make sure they are in the lowest possible
terms.
ex.
C3 H8
+
O2
------------------------>
CO2
C3 H8
+
O2
------------------------>
3 CO2
+
H2 O
C3 H8
+
O2
------------------------>
3 CO2
+
4 H2 O balance H
C3 H8
+ 5 O2
------------------------>
3 CO2
+
4 H2 O balance O
Try this one
C3 H6
+
O2
------------------------>
+
H2 O
CO2
+
balance C
H2 O combustion of propane
Stoichiometry - the study of weight relationships in chemical reactions
 A balanced chemical reaction gives proportions - how much starting material is needed to make so
much product.
 Coefficients in a chemical reaction represent molecules. Moles are proportional to molecules,
therefore coefficients also represent numbers of moles.
ex.
1 mole C3 H8
+ 5 mole O2
------------------------>
3 moles CO2
+
4 moles H2 O
Why is this important? Most all chemists never deal with 1 or 2 molecules of something. How can you
weigh one molecule?! You can’t. A mole is a quantity of something that you can work with.
Consider this :
2 molecules H2
+
1 molecule O2
2 moles H2
+
1 mole O2
4 grams H2
+
32 grams O2
------------------------>
------------------------>
------------------------>
2 molecules H2 O
2 moles H2 O
36 grams H2 O
What kind of problems can we solve?
With a given amount of starting material, how much product is produced?
N2
+
3 H2
------------------------>
2 NH3
How many grams of ammonia, NH3 , will be produced by reaction 100 g of N2 with excess H2 ?
formula mass NH 3 = (14.01 x 1) + (1.01 x 3) = 17.04 g/mole
formula mass N2 = (14.01 x 2) = 28.02 g/mole
100 g N2
x
2 moles NH3
1 mole N2
---------------- x -----------------28.02 g N2
1 mole N2
x
17.04 g NH3
-------------------- = 122 g NH3
1 mole NH3
Try this : How many grams of N2 would you need to produce 7.5 grams of NH3 ?
BRCC
CHM 101
Chapter 5 Notes
(Chapter 4 in older text versions)
Page 5 of 8
Rules:
1. Convert mass to moles. (If given moles, just write it down.)
2. Get proportion from the balanced equation. Write it as a ratio with moles of the compound of
unknown mass on top.
3. If the answer needs to be in mass, convert from moles to mass.
Percent Yield
 In most chemical reactions, you get less product than you expect. Why? In the real world, nothing is
perfect. ex. there are errors in technique or some reactant remains unreacted...
 To indicate how well your reaction worked, the idea of percent yield was developed.
actual yield - the yield you get
theoretical yield - the yield you expected if everything goes perfectly
actual yield
percent yield = ------------------------ x 100%
theoretical yield
ex. If your theoretical yield is 153.7 g and you obtain 124.2 g, then what is your percent yield?
124.2 g
------------- x 100%
153.7 g
=
80.81% yield
Reactions of Ions in Water
 Ionic compounds dissolve in the form of ions in water.
CaCl2
(s)
-------------->
2+
Ca
(aq)
+
-
2 Cl
(aq)
where (s) = solid and (aq) = aqueous
dissociation - when negative and positive ions are separated from each other by water molecules.
Ionic bonds are broken by the water.
What happens when 2 ionic solutions are mixed?
If negative and positive ions come together and the ionic bonds between them are too strong for the
water molecules to break, then the ions form a solid which falls out solution. The solid is called a
precipitate and the reaction is called a precipitation reaction.
ex. Mix AgNO3 solution with NaCl solution. What happens?
+
-
+
-
First the ions all separate in solution to form Ag , NO3 , Na , Cl
+
-
Next the Ag and Cl come together to form a precipitate AgCl (s)
The other ions which do not take part in the reaction are called spectator ions. Ionic equations show
only the ions which react and not spectator ions. The next ionic equation for this reaction is:
Ag
+
(aq)
+
-
Cl (aq)
-------------->
AgCl(s)
BRCC
CHM 101
Chapter 5 Notes
(Chapter 4 in older text versions)
Page 6 of 8
The following solubility rules can be used to predict whether ions will come together to form precipitates.
Soluble Compounds:
All nitrates (NO3 -) and acetates (C2 H3 O2 -) are soluble.
All chlorides (Cl-) are soluble, EXCEPT: AgCl, Hg2 Cl2 , and PbCl2 .
Most sulfates (SO4 2-) are soluble. Exceptions include BaSO4 , SrSO4 , CaSO4 , and PbSO4 .
Compounds of Group 1 elements and ammonium ion (NH4 + ) are soluble.
Insoluble Compounds:
All carbonates (CO3 2-) and phosphates (PO4 3-) are insoluble, EXCEPT those combining with the
Group 1 elements or ammonium ion.
All hydroxides (OH -) are insoluble, EXCEPT the hydroxides of the group 1 elements, ammonium ion,
Sr(OH)2 , and Ba(OH)2 which are soluble.
All sulfides (S2 -) are insoluble, EXCEPT those combining with the Group 1 elements, the Group 2
elements, and ammonium ion which are soluble.
ex. A solution of Ba(NO3 )2 is mixed with a solution of K2 SO4 .
Ba
2+
Ba
(aq)
2+
(aq)
+
+
2 NO3
SO4
(aq)
+
2(aq)
+
2K
(aq)
+
-------------->
SO4
2(aq)
BaSO 4
(s)
What happens?
-------------->
????
net ionic equation
oxidation-reduction reactions:
 oxidation-reduction reactions or redox reactions are an important class of chemical reactions which
include combustion, respiration, rusting, bleaching, and the chemistry of batteries.
Redox rxns were discovered over a 100 years ago and the following definitions were given:
oxidation - gain of oxygen or loss of hydrogen
reduction - loss of oxygen or gain of hydrogen
“
Redox rxns do not always involve oxygen or hydrogen. The modern definitions are:
oxidation - the process of losing electrons.
reduction - the process of gaining electrons.
“ This makes the definitions of oxidation and reduction confusing when reducing means gaining
electrons.
“ Oxidation and reduction always occur together!
This means that when one species loses electrons, another species must gain electrons. The total
reaction may be broken into 2 parts called half-reactions. One half-reaction involves the oxidation
and the other half-reaction involves the reduction.
BRCC
CHM 101
Chapter 5 Notes
(Chapter 4 in older text versions)
2+
2+
ex. Consider the reaction of copper ion, Cu , with aluminum wire, Al. The Cu
2+
and is therefore reduced; the half-reaction for the reduction of Cu ion is:
Cu
!
-
2+
(aq)
+ 2e
Page 7 of 8
ion gains electrons
Cu (s)
The aluminum loses electrons, therefore is oxidized; the half-reaction for the oxidation of Al is:
Al (s)
!
3+
Al
-
(aq)
+ 3e
“ Cu2+ ion takes electrons from Al metal. Therefore, Cu2+ ion is the agent that causes the Al metal to
2+
be oxidized. Cu
ion is called the oxidizing agent, or oxidant, in this reaction.
“ Al metal gives up electrons to Cu2+ ion so that it can be reduced. The Al metal is called the
reducing agent, or reductant.
‘ The oxidizing agent always gets reduced.
‘ The reducing agent always gets oxidized.
“ Oxygen, O2 , is always an oxidizing agent.
ex. Consider the following combustion reaction:
CH4
+
2 O2
!
CO2
+
CH4 is the reducing agent.
O2 is the oxidizing agent.
2 H2 O
Heat of Reaction
heat of reaction - the heat gained or lost in a chemical reaction.
exothermic - gives off heat to the environment - it loses heat
endothermic - absorbs heat from the environment - it gains heat
ex. water evaporates - it takes up heat to change phase from liquid to gas - endothermic
Energy Value of Food:
What nutritionists call Calories are actually kilocalories, kcals.
carbohydrate =
protein
=
fat
=
4.0 kcal/g
4.0 kcal/g
9.0 kcal/g
alcohol
=
7.0 kcal/g
BRCC
Example:
CHM 101
Chapter 5 Notes
(Chapter 4 in older text versions)
Consider Chef Boyardee Spaghetti and Meatballs.
Calories per serving
grams of protein
grams of carbohydrate
grams of fat
:
:
:
:
250 kcal (2 servings per can)
9g
32 g
10 g
Calculate the percent of Calories (kcals) from fat.
10 g fat
9.0 kcal
X ---------g fat
= 90 kcal
90 kcal
----------- X 100% = 35 %
250 kcal
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