Download CHE 110 Dr. Nicholas Bizier Office DS 337b email

CHE 110
Dr. Nicholas Bizier
Office DS 337b
email: [email protected]
What is General Chemistry about?

Chemistry = Study of properties and changes in matter

Matter= anything that takes up space

Learn to describe states of matter, both in words and
mathematically

You will learn to predict properties of matter

You will learn how to describe and predict chemical reactions that
take place.

You will learn about the energy associated with chemical reactions

You will be introduced to the periodic table.
How is Science Done?

Hypothesis- a proposed explanation

The Hypothesis is then tested.

From this type of testing you get two types of data:

Qualitative Data- nonnumerical data

Quantitative Data-obtained from measurements to produce
numeric data.

Theory-A Hypothesis that has not been disproven
Substance

Substance= matter that has a fixed composition and distinct
properties

These properties are divided into two categories:

1. Physical and Chemical

Physical properties- are properties that can be measured without
changing the basic identity of the substance. Examples of this: color,
m.p. b.p. density

Let’s first look at physical properties because they are things we
can visualize and measure
Geometric shape area and volume
formula’s you must memorize

Area of rectangle= l x w,

Volume of a cube= l x w x h

Cylinder = area of base = πr2, volume= area of base x h

Volume of a Sphere= 4/3 πr3
Greek Prefixes

Modifying the units of measurement are Greek Prefixes

The ones you will have to be comfortable with are:
mega  106
kilo  103
centi  10 2
milli  10 3
micro  10 6
Density

Density is how much matter (expressed in mass) is found in a given
volume

The mathematical expression for this:

Typically this value is expressed as
Density  mass / volumne
D  g / ml
or
D  g / cm3
Units of Measurement

A closer look at the last equations
D  g / ml
or
D  g / cm3

Weights: typically given in grams (scientific measurement)

Volume in liters

While not in the last equation: Temp in degree’s Celsius
Dimensional Analysis

Using a conversion factor-fraction whose numerator and
denominator express the same quantity given in different units.

Ex: Butane has a density of 0.579 g/mL or 1mL/0.579g

What is the volume of 5.25g of butane.
 1 ml 
5.25 g 
  9.07 ml
 0.579 g 

This type of calculation is an example of a physical property
because we can measure both volume and mass without changing
the substance.
Chemical Properties

Chemical Properties describe the ways a substance reacts to form
other substances. Ex. Flammability (oxygen combines with what you
are burning to give off heat and gas)

During a chemical reaction the substance(s) which are called the
(reactants) are transformed into different substances(s) (products).
Mixture

Combination of 2 or more substances.

Two types:

(1) heterogeneous- compositions in different pats of the sample.

(2) homogeneous-uniform compositions throughout the sample

One use various means to separate mixtures into pure substances.
Ex. Filtration or chromatography
Significant Figures

Rules:

(1) All nonzero numbers are significant: 376.8 (4 sig. figs), 921 (3 sig.
figs)

(2) Zeros between two nonzero numbers are significant: 1004 (4 sig.
fig), 20.025 (5 sig. fig).

(3)Zeros to the right of a decimal point are significant.

(4) Zeros at the end of number and to the left of a decimal point,
change to scientific notation.
200  2.00 x 102 (3 sig. figs )
2.0 x 102 (2 sig. figs )
Preforming Math with Sig. Figs

(1) Addition and Subtraction-the final answer should have as may
digits beyond the decimal point as the number with the fewest digits
beyond the decimal.
100.09  20.1  75.989  (196.179)  196.2

Multiplication and Division-the final answer should have as many sig.
figs as the number with the fewest sig. fig.
4
459
92.91

(42645.69)

4.26
x
10
 

3.58 x 104
 6.84
5231
Examples using the principles
covered thus far

How many liters are there in 10.0 gal of water?
 4 qt  0.946 L 
10.0 gal 

  37.8L
 1 gal  1 qt 

One edge of a cube is measured and found to be 13 cm. What is
the volume of the cube in m3?
 1m 
3
3
3
13 cm 

(0.13
m
)

0.002197

2.2
x
10
m

 100 cm 
Policy on Sig Figs in Class vs lab

You will follow instructions on quizzes and test. All answer will be 3 sig
figs. (You will be reminded of this on each test and quiz at the top).

You will follow your lab instructor directions regarding what you turn
in for lab. Any question regarding any assignment in lab will be
directed directly to the lab instructor. Treat this as two separate
courses. Theory in class, practicle application in lab.
Examples continued.

Alcohol has a density of 0.76 g/ml. How many grams of alcohol
would it take to fill a 2 fluid ounce shot glass?
32 oz  1 qt
1.057 qt  1 L
mL  L  qt  oz
 1L   1.057qt   32 oz 
2
1mL x 
x
x

3.4
x
10
oz
 
 

 1000 mL   1l   1 qt 
 0.78 g 

 x 2  45.9 g  46 g
2
 3.4 x oz 
Percent by Mass

This is a type of measurement which tells you what the weight of
substance out the total mass present.
 weight x 
precent by mass  
 x 100
 total weight 

Ex, what is percent by mass of sucrose in a water solution add 15 g
of sucrose to 200 grams of water.


15 g ( sucrose)
percent by weight sucrose  
 x100  7.0%
 200 g ( water )  15 g ( sucrose) 
Make up off Atoms

Atoms are made up of three different types subatomic particles:

Protons which have positive charge and are in the nucleus

Neutrons which have no charge and are also in the nucleus

Electrons which are negatively charged, which are found in diffuse
orbitals outside of the nucleolus.
Make up of atoms cont.

The number of protons in an atom is the atomic number and this
determines the identity of the atom.

The number of protons and neutrons is equal to the atomic mass.

Atoms made of same element must have the same number of
protons, but can have different numbers of neutrons

Two atoms with same number of protons, but different number of
neutrons are called isotopes of one another.
Elemental symbol

The elemental symbols used in the periodic table can have different
amounts of information on them:

Ex:

12
C
6 protons, 6 electrons, and 6 neutrons
14
6
C
6 protons, 6 electrons, and 8 neutrons
These two types of carbon atoms are isotopes of one another
Average Atomic weight:

When you look at the periodic table
You will find decimal numbers for
atomic weights.
These represent a weighted average
of the naturally occurring isotopes.
total wt. sample of x atoms
 AWaverage
total number of atoms x in sample
no. atoms isotope
%isotope 
total no. atoms
 % #1 
 %#2 
AWaverage  
xAW

isotope #1


 xAWisotope #2  ....
100
100




Osmium (Os) as an example of
Average Atomic weight

Osmium has atomic number 76 (# of protons)

It has 6 stable isotopes

Isotope

%Abundance

So the equation for the atomic weight of Osmium would be:
186Os
1.6
187Os
1.6
188Os
189Os
13.4
16.2
190Os
192Os
26.4
41.1
 1.6 
 1.6 
 13.4 
 16.2 
 26.4 
 41.1 
193.2 amu  
x
186
amu

x
187
amu

x
188
amu

x
189
amu

x
190
amu











 x 192 amu
 100 
 100 
 100 
 100 
 100 
 100 
This comcept can also be integrated with mass: 100 g of naturally
occuring Osmiun would cotain 41.1 grams of 192Os
Finding an unknown isotope

Chlorine consists of a mixture of 35Cl (34.97 amu) and 37Cl (36.97
amu). Given the average atomic weight for Cl on the periodic table
(35.45 amu), what is the relative % abundance of each isotope in an
average sample?
𝑥
100 − 𝑥
35 𝑎𝑚𝑢 +
37 𝑎𝑚𝑢
100
100
3545 𝑎𝑚𝑢 = 35 𝑥 𝑎𝑚𝑢 + 3700 − 37𝑥 𝑎𝑚𝑢
𝑥 = 37.78% 𝑤ℎ𝑖𝑙𝑒 100 − 𝑥 = 62.22 %
35.45 𝑎𝑚𝑢 =
Weights of subatomic particles

1 amu (atomic mass unit)= 1.66054 x 10-24 g

1 g =6.02 x 1023 amu

Proton = 1.0073 amu

Neutron = 1.0087 amu

Electron = 5.486-4 amu

To find the atomic mass of an element = add the number of protons
and neutrons together.
Finding the Formula
weight/Molecular weight

The molecular formula is a list of the identity and number of atoms that
make up molecule:

Ex. Ethanol: CH3CH2OH

 12.01 amu 
 16.00 amu 
 1.01 amu 
2C x 

6
H

1
O


  78.01amu


1O
 1H 
 1H



Formula weight is similar to molecular weight, however it used when
referring to ionic compounds

Ex. FeCl3

Note we use the term term formula weight, because ions are discreet
molecules, however the formula weight is how the ions add together to
form a neutral speices.
 55.85 amu 
 35.5 amu 
1 Fe x 

3
Cl


  162.4amu
 1 Fe 
 1Cl 
Concept of Moles

A mole is the amount of matter that contains as many objects as
the number of atoms in 12 g of 12C.

It is just a collection of 6.02 x 1023 objects just as a dozen is a
collection of 12 objects.

Avogadro’s number 6.02x 1023.

A mole is the connection between nanoscale and macroscale

The Molar Mass (g)= mass of 1 mol of substance.

Ex. FeCl3 = 162.4 amu, on mole of FeCl3 weighs 162.4 grams

FeCl3= 162.4 grams/mole
Using the concept of moles

How many moles of COCl2 (Phosgene) are in 352g?
 12.01g 
 16.00 g 
 35.5 g  98.91 g
1C 

1
O

2
Cl
COCl2





mol
 mol 
 mol 
 mol 
 1 mol 
352 g x 
  3.56 mol COCl2
98.91
g



How many grams of PBr3 are in 0.789 mol?
 30.97 g 
 79.90 g  270.7 g
1P 
PBr3
  3Br 

mol
 mol 
 mol 
 270.7 g 
0.789 mol 
  217 g PBr3
 mol 
Mass→ Moles→ Molecules

How many molecules of CH4(methane) are in 48.2 g?
 12.01g 
 1.008 g  16.04 g
1C 

4
H



mol
mol
mol




 1 mol 
48.2 g 
  3.00 mol CH 4
16.04
g


 6.02 x1023 molecules 
24
3.00 mol CH 4 x 
  1.806 x 10 molecules of CH 4
mol


Molecules and Ions

There are two types of bonding, molecular and ionic

Molecular bonding involves sharing of electrons between two nonmetals

Molecular formula-indicates the actual numbers and types of atoms
in a molecule: Ex. H2O2, C2H4

Empirical formula-a relative number of atoms in a molecule

Ex. HO, CH2

Structural formula-a pictorial representation of a molecular formula

Organic compounds that contain carbon and hydrogen and usally
oxygen, nitrogen, and sulfur
Naming (non-ionic) Inorganic
compounds

For Binary compounds

(1) Element having more positive character is named first.

(2) Element having more negative character is named second, and
ide is added at the end of the element.

(3) Use prefixes (mon (1),di (2), tri (3), tetra (4), pent (5), hexa (6),
hepta (7), octa (8), nona (9), deca (10) to denoted how many
elements there are.

E.x N2O5= dinitrogen pentoxide, CO= carbon monoxide, CO2carbon dioxide
Strong Acids

There are six inorganic molecules you need to know by site and
name. These are strong acids (something we will expand upon
later).

HCl= Hydrochloric acid

HBr= Hydrobromic acid

HI = Hydroiodic acid

H2SO4= Sulfuric acid

HNO3= Nitric acid

HClO4= Perchloric acid
One weak acid

There is one weak organic acid you will need to know by site and
name:

CH3CO2H = acetic acid
Inorganic Compounds that don’t
contain metals or charged species

These are molecules that have covalent bonds but contain no
carbon-hydrogen bonds

Ex. N2O5, CO,CO2, HCl
Ionic Compounds

These compounds are made of a combination of cations (positively
charged species which have lost electrons) and anions (negatively
changed species which have gained electrons.

Most of the time they are made up of a positively charged metal
and a negatively charged non-metal, or group of non-metals which
are called poly atomic anions.

Simple Ionic compound ex: NaCl, MgBr2

Metal and Polyatomic ion examples:
NaNO3 (made of Na+ and NO3- )
CaSO4 (made of Ca+2 and SO4-2)
Na2SO3 (made of 2 Na+ and SO3-2)
What are metals, non-metals, and
metalloids.
You will be expected to know what
each one is by sight
For ionic compounds:
Metals=form cations (+)
Nonmetals=form anions (-)
Metalloids form both (you have to look
at what it is bonded to and make a
judgement call)
Charge states of certain groups
Example to try

Give an element that will react with magnesium to form an ionic
compound with general formula (Mg+2)3 (X+3)2

X= N or P
Naming inorganic compounds

Very important to do this stepwise and pay attention to what is the
cation and what is the anion

Step 1: Cation
A.
Monatomic cation: Na+ = Sodium, Ca+2 = calcium
B.
This Elements that can form more than one positive ion-transition
metals- use Roman numeral
C.
Cu+- copper(1) ion written as Cu(I), Cu+2- copper(2) written as Cu(II)
D.
Name the anion with -ide at the end: Cl-=chloride, Br-= bromide
N-3= nitride
Naming inorganic compounds

Put the two ions together: NaCl = sodium chloride,

Copper (II) Bromide Cu(II)Br2
Note figuring out what the roman numeral on is for one of the transition
metals can be accomplished by noting the charge on the anion, and
the number.
The charge on the anion multiplied by the number of atoms tells you
what the roman numeral is on the transition metal cation.
List of poly atomic ions:

NO3- = nitrate

H3O+ = hydronium

NO2- = nitrite

NH4+ = ammonium

SO42- = sulfate

CrO42- = chromate

SO32- = sulfite

Cr2O72- = dichromate

PO43- = phosphate

MnO4- = permanganate

CO32- = carbonates

BO3- = borate

HCO3- = hydrogen carbonate

SiO4- = silicate

Cl- = chloride

OH- = hydroxide
All of polyatomic ions in white will be in the place of the non-metal of a simple binary compound,
because they are anions.
The ones in red are cations, and will take the place of the metal in a simple binary compound.
Naming with Polyatomic cations

Name the simple cation or anion as you would a typical binary
cation, and then use the name of polyatomic ion in its place.

Barium hydroxide= Ba+2 + 2OH- → Ba(OH)2.

Ammonium Iodide= NH4+ + I- → NH4I

Copper(II) Sulfate= Cu+2 + SO42- → Cu(ii)SO4

Which of the following are ionic substances and which are
molecular?
(a) CaO (b) SiCl4 (c) Mg(NO3)2 (d) NOCl (e) B2H6 (f) CH3OH (g) Ag2SO4
Percent Composition

This calculation tells what percent the weight of a type of atom in a
compound or molecule is of the whole (in terms of weight)
(# of atoms ) x ( AW )
%
x 100
FW
What is the % composition of each
type of atom in glucose
(C6H12O6)?

Step 1 calculate MW
6 x C (12.01amu )  72.06amu
12 x H (1.008amu )  12.12amu
96.00amu
6 x O (16.00amu )  
180.19amu
% comp Glucose continued
(6)(12.01 amu )
%C 
x 100  40.00%
180.2 amu
(12)(1.01 amu )
%H 
x 100  6.73%
180.2 amu
6(16.00 amu )
%O 
x 100  53.28%
180.2 amu
Calculation of Empirical and
Molecular Formulas

Molecular formula-indicates the actual numbers and types of atoms
in a molecule

Empirical formula-indicates the ratio of the types of atoms to one
another
Glucose



Molecular
C6H12O6
Empirical
CH2O
Combustion Analysis

Combustion analysis is a process that determines the empirical formula of
molecules by burning them in the presence of excess oxygen.

The process involves placing a sample of know weight, in the analyzer and
recording the weight of both CO2 and H2O which is emitted from the machine.

From this information you can calculate the amount of carbon and hydrogen in
the sample. However since oxygen is in excess you must find oxygen through
indirect means (the mass comes from what is not accounted for by carbon and
hydrogen, in a sample that only contains CHO).

The information you get from this analysis provides you with the empirical
formula of the molecule.

If there are other elements present in the molecule, such nitrogen or chlorine,
you make get information by decomposition analysis (same process as
described for combustion, however a different term is used when it is not carbon
or hydrogen.
Combustion Analysis of Vitamin C

If 0.200 g of Vitamin C are subject to combustion analysis, the mass
of CO2 and water found is 0.2889 g and 0.0819 g respectively. What
is the Empirical formula.

Step 1: First determine the mass of carbon in 0.2998g of CO2, by
converting to mol C. Then to grams of C.
1 mol CO2
1 molC
0.2998 mol CO2 x
x
 0.006812 mol C
44.01 g CO2 1 mol CO2
12.011 g C
0.006812 g C x
 0.0812 g C
1 mol C
Determining H in Vitamin C

Similar fashion for finding H2O.
1 mol H 2O
2 mol H
0.819 g x H 2O x
x
 0.00909 mol H
18.02 g H 2O 1 mol H 2O
1.008 g H
0.00909 mol H x
 0.00916 g H
1 mol H
Determining O in Vitamin C and
percent composition of elements

Weight of original sample minus the weight of carbon and hydrogen

(0.200g –(0.08182 g C + 0.00916 g H) = 0.1090 g O
0.08182 g C
x 100 %  40.91% C
0.2000 g sample
0.00916 g H
%H 
x 100%  4.58% H
0.2000 g sample
0.1090 g O
%O 
x 100 %  54.50 % O
0.2000 g sample
1 mol O
? mol O  0.1009 g O x
 0.006813 mol O
15.999 g O
1
C0.006812 H 0.0909O0.006813 x
 CH1.33O  C3 H 4O3
0.006812
%C 
Combustion and Decomposition

While combustion analysis only gives information about carbon
hydrogen and oxygen indirectly, you can get information about
other elements from decomposition analysis.

In these types of problems you will have combustion analysis data
from one sample with a specific weight, and decomposition data
from another sample with a different weight.
Lysine Example

Lysine is an amino acid which has the following elemental
composition: C, H, O, N. In one experiment, 2.175 g of lysine was
combusted to produce 3.94 g of CO2 and 1.89 g H2O. In a separate
experiment, 1.873 g of lysine was burned to produce 0.436 g of NH2.
The molar mass of lysine is 150 g/mol. Determine the empirical and
molecular formula of lysine.

Step One: determine the mass of each element present.

Carbon: 3.94 g x (12.011 / 44.0098) = 1.0753 g
Hydrogen: 1.89 g x (2.016 / 18.0152) = 0.2115 g
Nitrogen: 0.436 g x (14.007 / 16.023) = 0.38114 g
Oxygen: cannot yet be done

Why can't the oxygen be determined yet? It is because our C, H,
and N data come from TWO different sources.
Lysine continued

Step Two: Convert mass of each element to percentages.

Carbon: 1.0753 g ÷ 2.175 g = 49.44 %
Hydrogen: 0.2115 g ÷ 2.175 g = 9.72 %
Nitrogen: 0.38114 g ÷ 1.873 g = 19.17 %
Oxygen: 100 - (49.44 + 9.7 + 19.17) = 21.67 %

Step Three: Determine the moles of each element present in 100 g
of lysine.

Carbon: 49.44 g ÷ 12.011 g/mol = 4.116 mol
Hydrogen: 9.72 g ÷ 1.008 g/mol = 9.643 mol
Nitrogen: 19.17 g ÷ 14.007 g/mol = 1.3686 mol
Oxygen: 21.67 g ÷ 15.9994 g/mol = 1.3544 mol
Lysine continued

Step Four: find the ratio of molar amounts, expressed in smallest,
whole numbers.

Carbon: 4.115 mol ÷ 1.3544 mol = 3.04
Hydrogen: 9.643 mol ÷ 1.3544 mol = 7.12
Nitrogen: 1.3686 mol ÷ 1.3544 mol = 1.01
Oxygen: 1.3544 mol ÷ 1.3544 mol = 1

The empirical formula is C3H7NO.

In order to determine the molecular formula, we need to know the
"empirical formula weight." This value is 73.1.

We see that the approximate molecular weight is just about double
this value, leading to the molecular formula of C6H14N2O2
Molarity

Unit of concentration.

M is used to show this value.

Molarity = moles of solute/volume of soln in liters(l)

This unit of measurement is useful when you are given a volume and
concentration of substance in a solvent (this solvent will be assumed
to be water unless specified).
Example calculations

Calculate the molarity of the solution prepared by dissolving 62.3 g
of sucrose C12H22O11 in enough water to form 0.500l of solution.

Grams → moles → molarity
 1 mol sucrose   1 
62.3 g sucrose x 
 x
  0.364M sucrose
 324.34 g sucorse   0.500 L 
How would you prepare, in grams,
250ml of 0.600M aq. KBr solution?
 1l   0.600 mol KBr   119.00 g KBr 
250 ml x 
 x
 x
  18.0 g KBr
1l
 1000 ml  
  1 mol KBr 
Examples continued.

Calculate the molarity of chloride ions in 7.68g of CaCl2 which was
dissolved in water to make a 250ml solution.

CaCl2(aq) → Ca2+(aq) + 2Cl-(aq)

G CaCl2 → moles CaCl2 → moles Cl- → molarity of Cl-
 1 mole CaCl2   2 moles Cl    1 
1
7.68 g CaCl2 x 
x
x

0.554
M
Cl
 
 

 110.98 g CaCl2   1 mole CaCl2   0.250 l 
Dilution

Taking a concentrated solution to a diluted solution:

M1V1= M2V2

How much 5.00M HCl would be need to make 250ml of 0.350M HCl?
Vconc HCl
(0.350 M HCl ) x0.250 ml

 18.0 ml HCl
5.00 M HCl
Stoichiometry

Stoichiometry-calculation of the quantifies of elements or
compounds involved in a chemical reaction

Atoms are neither created nor destroyed during a chemical
reaction

Chemical reaction

H2(g)+ O2(g)→ H2O(l)

States g=gas, l=liquid, s-solid, aq=aqueous substance dissolved in
water

Balanced equation

2H2(g)+ O2(g)→ 2H2O(l)
Rules for balancing chemical
equations

(1) The chemical formulas of the reactants and products are fixed
and can’t be altered by changing the subscripts. Only change the
coefficients in front of the formulas.

(2) Balance elements contained in only one substance first

(3) Balance H and O atoms last.

Ex. H2SO4(aq) + KOH(aq)→ K2SO4(aq) + H2O(l)
H2SO4(aq) + 2KOH(aq)→ K2SO4(aq) + 2H2O(l)

Ex. AgClO3(aq)+ CaBr2(aq) → AgBr(s) + Ca(ClO3)2(aq)
2AgClO3(aq)+ CaBr2(aq) → 2AgBr(s) + Ca(ClO3)2(aq)
Molecular reactions are balanced
the same way

Ex. C3H8(g) + O2(g)→ CO(g)+ H2O(g)

2C3H8(g) + 7O2(g)→ 6CO(g)+ 8H2O(g)
Types of chemical reactions








(1)Combination reactions=two or more substances react to form one
product.
Ex. 2H2(g) + O2(g)→ 2H2O(l)
(2) Decomposition reactions= substance react to form two or more
different products.
Ex. BaCO3(s) → BaO(s) + CO2(g)
(3)Displacement reactions= Single element reacts with a compound to
form a new product and new element.
Ex. 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)
(4) Exchange reactions= exchanging the partners between two
compounds. The driving force is the removal of a solid, molecule or gas
from solution.
Ex Na2SO4(aq) + PbCl2(aq)→ 2NaCl(aq) + PbSO4(s)
Coefficients

Coefficients in a chemical eq. can be interpreted as numbers of
molecules at a nanoscale level.

Ex. 3H2(g) + N2(g)→ 2NH3(g)

3 molecules of H2 reacts with 1 molecule of N2 to produce 2
molecules of NH3

When then can use Avagadro’s number to bring it to macroscale:

3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3
Use of mass (in grams) of reactants
and chemical equations

If you are given grams of substance you can use moles to predict
the moles of product and ultimately the grams of the product

Grams of reactant → moles of reactant →use coefficient multipliers
→ moles of product → grams of products

Consider the following equation:

3H2(g) + N2(g) → 2NH3(g)

How many moles of NH3(g) are produced from 10.0 moles of N2(g)?
 2 mol NH 3 
10.0 mol N 2 x 
  20.0 mol NH 3
 1 mol N 2 
Stoichiometry cont.

3H2(g) + N2(g) → 2NH3(g)

(2) How many grams of NH3 are produced from 280 g of N2?

Grams of reactant → moles of reactant →use coefficient multipliers
→ moles of product → grams of products
 1 mol N 2   2 mol NH 3   17.04 g NH 3 
280 g N 2 x 
  340 g NH 3
 x
 x
 28.02 g N 2   1 mol N 2   1 mol NH 3 
Stoichiometry cont.

Consider the combustion of propane in oxygen

Chemical equation is: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

How many grams of O2 are required to burn 75.0 g of C3H8?
g C3H8(g) → moles C3H8(g) → moles O2(g) → grams O2(g)
 1 mol C3 H 8   5 mol O2   32.00 g O2 
75.0 g x 
 x
 x
  272 g O2
 44.11 g C3 H 8   1 mol C3 H 8   1 mol O2 
Stoichiometry cont.

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

How many grams of H2O(l) and CO2(g) are produced?
 4molH 2O   18.02 gH 2O 
1.70molC3 H 8 x 
 x
  122 gH 2O
 1molC3 H 8   1molH 2O 
 3molCO2   44.01gCO2 
1.70molC3 H 8 x 
 x
  224 gCO2
 1molC3 H 8   1molCO2 
Limiting reagents

The limiting reagent is the reactant which is consumed completely
and therefore limits the amount of product formed. (other reactants
are in excess taking into account stoichiometry)

How many moles of magnesium oxide are produced by the
reaction 3.82g of magnesium nitride with 7.73g of water?
Mg3N2(s) + 3 H2O(l) → 2NH3(g) + 3MgO(s)
Step 1: find moles of each reactant
 1 mol Mg3 N 2 
3.82 g Mg3 N 2 x 
  0.0378 mol Mg 3 N 3
 100.9 g Mg3 N 2 
 1 mol H 2O 
7.73g H 2O x 
  0.429 mol H 2O
 18.02 g H 2O 
Limiting reagent

Step 2: Determine # of moles of product produced from each.
 3 mols MgO 
0.0378 mol Mg3 N 2 x 
  0.113 mol MgO
 1 mol Mg3 N 2 
 3 mols MgO 
0.429 mol H 2O x 
  0.429 mol MgO
 3 mols H 2O 
Step 3: Which ever reactant produces the smallest # of moles of
product is the limiting reagent.
 40.3g MgO 
0.113 mol MgO x 
  4.55 g MgO
 1 mol MgO 
Limiting reagent (what’s left over)

How much excess reactant is left in the reaction from the previous
slide?

Step 1: Find moles of excess reagent used:
 3 mols H 2O 
0.133 mol MgO x 
  0.113 mols H 2O
 2 mols MgO 
0.429 moles H 2O( started with)  0.113 mols H 2O (used )  0.316 mols H 2O (leftover )
More examples


A 2.00g sample of Fe3O4(s) is reacted with 7.50 g of O2(g) to produce
Fe2O3(s). Calculate the number of grams of Fe2O3(s) produced.
4Fe3O4(s) + O2(g)→ 6Fe2O3(s)
 1 mols Fe3O4 
3
2.00 g Fe3O4 x 
  8.64 x10 mols Fe3O4
 231.54 g Fe3O4 
 1 mols O2 
7.50 g O2 x 
  0.234 mols O2
32.00
g
O

2 
 6 mols Fe2O3 
8.64 x103 mols Fe3O4 x 
  0.0130 mols Fe2O3
4
mols
Fe
O
3 4 

 6 mols Fe2O3 
0.234 mols O2 x 
  1.40 moles O2
1
mols
O

2

 159.7 g Fe2O3 
0.0130 mols Fe2O3 x 
  2.08 g Fe2O3
1
mol
Fe
O
2 3 

Iron example cont.

4Fe3O4(s) + O2(g)→ 6Fe2O3(s)

How much excess reagent is left over?
 1 mols O2 
3
0.0130 mols Fe2O3 x 

2.17
x
10
mols O2 used

 6 mols Fe2O3 
0.234 mols O2 ( started with)  2.17 x103 mols O2 (used )  0.23 mols O2 excess
More examples

A mixture is prepared from 25.0 g of aluminum and 85.0 g of Fe2O3(s).
How much iron is produced?

Fe2O3(s) + 2Al(s)→ Al2O3(s) + 2Fe(l)

How much iron is produced in the reaction in grams?
 1 mol Al 
25.0 g Al x 
  0.927 mols Al
26.98
g
Al


 1 mol Fe2O3 
85.0 g Fe2O3 x 
  0.532 mol Fe2O3
159.7
g
Fe
O
2
3 

 2 mol Fe 
0.927 mol Al x 
  0.927 mol Fe
 2 mol Al 
 2 mol Fe 
0.523 mol Fe2O3 x 
  1.06 mol Fe
1
mol
Fe
O
2
3 

 55.87 g Fe 
0.927 mol Fe x 
  51.8 g Fe
 1 mol Fe 
Al plus Fe cont.

How much excess reagent is there?

Fe2O3(s) + 2Al(s)→ Al2O3(s) + 2Fe(l)
 1 mol Fe2O3 
0.927 mol Fe x 
  0.463 mol Fe2O3 used
 2 mol Fe 
0.532 mol ( started with)  0.463 mol (used )  0.069 excess Fe2O3
Theoretical Yield/Actual Yield

Theoretical Yield= calculated amount of product to form when all of
the limiting reagent reacts

Actual Yield= amount of product that is actually obtained
 actual yield 
% yield  
 x100
 theoretical yield 
Sample Yield problem

If the reaction of 3.50 g of magnesium nitride with 7.00 g of water
produced 3.60g of magnesium oxide, what is the percent yield of
the reaction?

Mg3N2(s) + 3 H2O(l) → 2NH3(g) + 3MgO(s)
 1 mol Mg3 N 2 
3.50 g Mg3 N 2 x 
  0.0347 mol Mg3 N 2
 100.92 g Mg3 N 2 
 1 mol H 2O 
7.00 g H 2O x 
  0.388 mol H 2O
18.02
g
H
O

2

 3 mol MgO 
0.0347 mol Mg3 N 2 x 
  0.104 mol MgO
 1 mol Mg3 N 2 
 3 mol MgO 
0.388 mol H 2O x 
  0.388 mol MgO
 3 mol H 2O 
Yield continued
 40.30 g MgO 
0.104 mol MgO x 
  4.19 g MgO  theoretical yield
 1 mol MgO 
actual
3.60 g MgO
x 100 
x 100  85.4%
theoretical
4.19 g MgO
Net Ionic equations

A solution is homogenous mixture of 2 or more substances.

An aqueous solution is a mixture where water is the solvent

Solvent-component of a solution present in the greatest quantity

Solute-all other components

Electrolytes-solutes that exist as ions in solution and conduct
electricity.

NaCl(s)→Na+(aq) + Cl-(aq)

Strong electrolytes-completely ionize in solution (salts). Ex. NaCl

Weak electrolytes-only partially ionize in solution.

HC2H3O2(aq)
H+(aq) + C2H3O2-
(aq)
Non-Electrolytes

Nonelectrolytes-molecular substances that do not form ions in
solution. Ex. Sugar C12H22O11
Precipitation reactions

Positive and negative ions exchange partners. Driving force is the
removal of ions from solution. There are three general types of these
reactions

(1) Formation of a precipitate (solid)

(2) Formation of a weak or nonelectrolyte (Acid-base reactions fall
in this category

(3) Formation of a gas
Solubility rules
Examples

NaCl(aq) + AgNO3(aq)

NaCl(aq) + AgNO3(aq)→ NaNO3(aq) + AqCl(s)

Pb(NO3)2(aq)+ 2KBr(aq)

Pb(NO3)2(aq)+ 2KBr(aq) → PbBr2(s) + 2KNO3(aq)
Writing total ionic and Net ionic
equations

A total ionic equation shows all molecules and ions (drawn
separately as cation and anions) on both sides of the equation

A net ionic equation leaves out the spectator ions=ions that appear
in the identical forms on both side of the equation.

HCN(aq)+ KOH(aq)→KCN(aq) +H2O(l)

Total: HCN(aq)+ K+(aq) + OH-(aq) → K+(aq) + CN-(aq) + H2O(l)

Net Ionic: HCN(aq)+ OH-(aq) → CN-(aq) + H2O(l)
Gas Formation

H2CO3(aq) → H2O(l) +CO2(g)

FeCO3(s) + 2HCl(aq) → H2CO3(aq) + FeCl2(aq), (H2O(l) + CO2(g) from the
breakdown of H2CO3(aq))

Total: FeCO3(s) + 2H+(aq) +2Cl-(aq)→ H2O(l) +CO2(g) + Fe2+(aq) + 2Cl-(aq)

Net: FeCO3(s) + 2H+(aq) → H2O(l) +CO2(g) + Fe2+(aq)
Oxidation/Reduction Reactions










Oxidation=loss of e-1 by a substance. Charge on the atom increase
Reduction=grain of e-1 by a substance. Charge on the atom decreases.
Oxidation Numbers: looks at the charge on an atom if it wasn’t in a
compound.
Rules for assigning oxidation numbers:
(1) The oxidation of an atom of a pure element is zero.
(2) The oxidation number of a monoatomic ion equals its charge.
(3) Some elements have the same oxidation numbers in all of their
compounds:
(a) hydrogen is +1 except when it’s combined with a metal, then it’s -1.
(b) fluorine is -1.
(c) oxygen is -2 expect when it’s in a peroxide like H2O2 when it is -1.
Oxidation numbers cont.

The sum of the oxidation numbers in a neutral compound is 0.

The sum of the charges in a polyatomic ion equals the charge on
the ion.

2H2(g) + O2(g) → 2H2O(g)

2H2(g) → 0 to +1 losing electrons-oxidized (reducing agent)

O2 → 0 to -2 gaining electrons-reduced (oxidizing agent)
0
0
2(+1) -2
Assigning oxidation states cont.

Fe2S3 + 12HNO3 → 2Fe(NO3)3 + S3 + 6NO2 + 6H2O
2(+3) 3(-2)
(+1),(+5),3(-2)
(+3),(+5),3(-2)X3
0
+4,2(-2)
2(+1), -2
Balancing redox reaction in a step
wise manner using half reactions

(1) Identify elements changing oxidation state

(2) separate oxidation from reduction into two half reactions

(a) balance the non-H,O atoms

(b) Determine the electrons lost or gained per atom in each halfreaction

(3) Determine total electron change for element by Multiplying no.
atoms changing times electron change per atoms, then write total
half reaction

(4) Add half reactions together and use whole number multipliers as
need to make sure total gain of electrons equals total loss.
Redox steps cont.

(5) After addition the electrons cancel, add up total charge on
reactant side and compare to total charge on product side. If it is
not equal:
(a) If acidic add appropriate number of H+ ions to whichever side
need to balance the charges.
(b) If basic, add appropriate number of OH- ions to whichever
side needed to balance the charges.
(6) After charges balanced, add H2O as needed to balance H’s.
Once the H’s are balanced, the O’s should also be balanced. If not,
you made a mistake in steps (1-3).
Balancing redox reactions
examples

Cr2O7-2 + CH3OH → HCO2H + Cr+3 (acidic)
2(+6),7(-2)
4(+1), (-2),(-2)
2(+1),+2,2(-2)
+3

Reduction half reaction:

6e- +Cr2O7-2 → 2Cr+3

Oxidation half reaction:

CH3OH → HCO2H + 4e-

Addition of reaction:

2Cr2O7-2 + 3CH3OH → 3HCO2H + 2Cr+3

Charge on reactant side is = -4

Charge on product side is = +6
X2
X3
Final Steps example 1 redox

Add H+ to balance charge

16H+ + 2Cr2O7-2 + 3CH3OH → 3HCO2H + 2Cr+3

Hydrogens on reactant: 28H

Hydrogen on Product: 6H

Add H2O:

16H+ + 2Cr2O7-2 + 3CH3OH → 3HCO2H + 2Cr+3 + 11H2O
Example two redox

HClO2 + MnO2 → Cl2 + MnO4- (acidic)
(+1),(+3),2(-2)
(+4),2(-2)
0
(+7),4(-2)

Reduction:

6e- + 2HClO2 → Cl2

Oxidation

MnO2 → MnO4- + 3e-

Addition of reaction:

2HClO2 + 2 MnO2 → Cl2 + 2 MnO4-
x1
x2
Example 2 cont.

Add H+ to balance charge
2HClO2 + 2 MnO2 → Cl2 + 2 MnO4- +2H
Hydrogens are already equal at this point, so no water needed!
Basic example

CrO2- + ClO- → CrO4-2 +Cl2 (basic)
+3,2(-2)
(+1),(-2)
(+6),4(-2)
0

Reduction:

2e-1 + 2ClO- → Cl2

Oxidation:

CrO2- → CrO4-2 + 3e-1

2CrO2- + 6ClO- → 2CrO4-2 +3Cl2
X3
X2

2CrO2- + 6ClO- → 2CrO4-2 +3Cl2

Balance charge with OH-

2CrO2- + 6ClO- → 2CrO4-2 +3Cl2 + 4OH-

Balance H with water:

2H2O + 2CrO2- + 6ClO- → 2CrO4-2 +3Cl2 + 4OH-
Additional Notes about redox

Once you have the final equation you can use this equation just like
any chemical equation you have before.

However, you need to be ready for spectator ions again:

Take the previous equation we looked at:

2H2O + 2CrO2- + 6ClO- → 2CrO4-2 +3Cl2 + 4OH-

A question might ask how many grams of NaClO would it take to
produce 3 grams of Cl2. In this case the Na really isn’t part of the
calculations other then final weight of reactant. (it is part of question
because you can’t physically add a anion alone, it has to have a
cation with it).
Gases

Gases can be monoatomic, diatomic (two atoms which the same),
or polyatomic

Gases expand to fill the container they are in

If you apply pressure to a gas, the volume of a gas decreases

If you release pressure it increases the volume.

If you have two or more gases they form homogeneous mixtures

Gases molecules are far aprat
Pressure

Pressure = Force/area

F=mass x accerlation

F=Kgm/s2= 1 newton (N)

A=m2

P= N/m2 = 1 Pascal (Pa)

There are number of ways pressure is can be reported:

1 atm = 760mmHg= 760torr=1.01335x105 Pa
Pressure example

A lead cube with edges 25.0 cm in length rest on a flat surface. The
density of lead is 11.34 g/cm3. Calculate the pressure in Pascals
exerted by the cube on the surface.
F
P
A
F  ma  9.807m / s 2
 25.0cm   11.34 g   1kg   9.807m 
3
2
F 
x
x

1.74
x
10
Kgms


 x



3
2
 1   cm   1000kg   1sec 
3
2
1m 

2
Areaofsurface   25.00cmx

0.625
m

100cm 

 1.74 x103 Kgms 2 
1 2
3
P

2784
Kgm
s

2.78
x
10
Pa

2
0.625m


Simple Gas laws

Boyles Law (pressure and volume, temp and amount of gas kept
constant

PV=constant

P1V1= P2V2

Charles law-volume varies directly temp (when pressure and the
number of moles is kept constant
V
 cons
T
V1 V2

T1 T2
Simple Gas Laws cont

Avogradro’s Law-volume varies directly with the number of moles
when pressure and temperature remain constant.
V
 cons
n
V1 V2

n1 n2
Ideal Gas Law

PV=nRT

P=pressure (usually in atm’s)

V=volume (L)

T= temp (always in Kelvin)

n= # of moles

R= gas constant (0.8206 L atm mol-1 K-1)

Standard tempt= 0⁰ C, (273 K)

Standard pressure 1 atm
Ideal gas examples

The pressure of oxygen in inhaled air is 157 torr. The total volume of
the average adult lung when expanded is about 6.00 L, and body
temperature is 37⁰ C. Calculate the mass of O2.
 1 atm 
P  157 torr x 
  0.207 atm
760
torr


V  6.00 L
T  37 o C  273  310 K
n?
atm L
R  0.8206
mol K


 0.207 atm x 6.00 L   32.00 g O 
2
 x
n
  0.156 g O2
atm
L
 0.8206
x310 K   1 mol O2 

mol K


Ideal gas examples


How many additional moles of gas would have to be added to a
flask containing 2.00 moles of gas at 25⁰ C and 1 atom in order to
increase the pressure to 1.60 atm under conditions of constant temp
and volume?
Initail
Final

P = 1.00 atm
1.60 atm

V = Constant
Constant

n = 2 moles

T= Constant
?
Constant
PV  nRT
P RT

n V
P1 P2

n1 n2
1.00atm
1.60atm

 3.2molesgas
2.00moles
N2
2.00moles  1.2molesgas
Gas Laws examples cont.

If all of the propane in a tank (C3H8) at 20 C and a pressure of 300
atm is burned in a combustion reaction, where the O2 is also at 20 C,
but 756 mmHg. How many liters of O2 would be required to
completely burn the propane?

5O2 + C3H8 → 3CO2 + 4H2O

Gas law for tank → take into account Stoic → gas for oxygen
Propane tank cont.
Tank
P  300atm
V  30 L
n?
T  293K
R  0.8206
atm L
mol K
 5O2 


1
C
H
 3 8
Oxygen
756 mmHg
P
 0.995 atm
760 mmHg
V ?
n  value from tank x 5
atm L
R  0.8206
mol K
T  293 atm
Propane continued

Propane tank → to moles of Oxygen
PV
RT


 300 atm x 30.00 L 
  374 moles C3 H 8
n
 0.8206 atm L x 293 K 


mol K


 5O2 
374 moles C3 H 8 x 
  1870 moles O2
 1C3 H 8 
n
Volume of Oxygen
nRT
P
atm L


1870
mol
O
x
0.8206
x
293
K
2


mol K
  45200 L O2
V 
0.995 atm






V
MW of a gas can be found using
gas laws

PV=nRT, n can also me expressed as grams/MW

Acetaldehyde is a common liquid that vaporizes readily. If the
pressure of acetaldehyde is observed to be 331 mmHg in a 125 mL
flask a 0 C and the density of the gas is 0.0855g/L what is the molar
mass of acetaldehyde?

Denistry→grams, and Gas Law to get moles
 1atm 
P  331mmHgx 
  0.436atm
760
mmHg


V  0.125 L
n  ?( g / mw)
atm L
R  0.8206
mol K
T  273K
Dg/L
 0.855 g 
g 
 x  0.125L   0.107 g
 1L 
Gas Law portion
 1atm 
P  331mmHgx 
  0.436atm
760
mmHg


V  0.125 L
n  ?( g / mw)
atm L
R  0.8206
mol K
T  273K
0.436atmx0.125 L
n
 2.43x103 moles acetlaldehyde
atm L
0.8206
x 273K
mol K
0.107 grams
MW 
 44.0
2.43 x103 moles
Dalton’s law of partial Pressure

Two or more gases will mix together to total occupy the volume they
are contained in.

Total pressure is equal to the sum of the individual pressures exerted
by the various gases in a mixture

PT= P1 + P2 + P3……

Total number of moles of gas in a mixture is equal to the sum of the
number of moles of each gas in a mixture.

nT= n1 + n2 + n3……

PTV=nTRT
Example of Partial Pressure

What is the total pressure exerted by a mixture of 3.00g of CH4 and
10.50g N2 at 301K in a 25.0L flask?
PT  PCH 4  PN 2
V  25.0 L
 1 mol CH 4 
 1 mol N 2 
nt  nCH 4  nN2   3.00 g CH 4  x 

10.50
g
N

  0.563 moles gas
2
16.0
g
CH
28.0
g
N

4 

2 
atm L
R  0.08206
mol K
T  301 K
atm L
0.563 moles gas x0.08206
x 301K
mol K
PT 
 0.556 atm gas
25.0 L
Collection of Gas over water

We can use Dalton’s law of Partial Pressure to measure the amount
of gas given off from a reaction

This is done by subtracting the amount of water vapor that would
normally be in flask at a given temperature.
Collection of Gas over water
picture.
Collection of Gas over water
problem

A sample of KClO3 is partially decomposed and the gas collected
over water.

2KClO3(s) → 2KCl(s) + 3O2(g)

(a) How many moles of O2 are collected?

(b) How many grams of KClO3 decomposed?

© When dry, what volume would the collected O2 occupy at the
same temp. and pressure?
Part A

Find partial pressure of O2 by subtracting the pressure of water from
the total.
PV  nRT
 1 atm 
PT  756 torr x 
  1.07 atmT
760
torr


 1 atm 
PH 2O  23.8 mmHgx 
  0.0313 atmO2
760
mmHg


PO2  107 atmT  0.0313 atmO2  1.04 atmo2
V  0.250 L
n?
R  0.08206
atm L
mol K
T  298 K
1.04atmo2 x0.250 L
n
 0.0106 moles O2
atm L
0.08206
x 298 K
mol K
Part B and C
Part B
 2 moles KClO3   122.55 g KClO3 
0.0106 moles O2 x 
  0.866 g KClO3
 x
 3 moles O2   1 mole KClO3 
Part C
P  1.07 atm
V ?
n  0.0106 mole O2
R  0.08206
atm L
mol K
T  293K
V
0.0106 mole O2 x0.08206
1.07 atm
atm L
x 293 K
mol K
 0.242 L O2
Energy and Chemical Reactions

Thermodynamics=study of energy and its transformations

C6H12O6(s)→6O2(g) → 6CO2(g) + 6H2O(g) + energy

Energy-the capacity to do work or transfer heat

Two types:

(1)kinetic-energy of motion

a. macroscale-moving ball-mechanical energy
b. nanoscale-moving atoms, molecules,-etc. thermal energy
C. Moving electrons-electrical energy
Energy continued

(2) Potential energy-stored energy-energy possessed by an object
by virtue of its position relative to other objects.

a. energy an object has in your hand due to the force of gravitygravitational energy

b. attraction or repulsion of charge particles-electrostatic energy

c. chemical potential energy-stored in foods and fuels
Calculated in different ways depending on type.
Energy continued

Total energy= KE + PE

Energy Units: Joule
2
2
1 2 1
m
Kgm
 
Ek  mv  ( Kg ) x   
2
2
s2
s

Calorie=amount of energy required to raise the temp of 1g of water
by 1⁰ C.

Work on Surroundings?

∆E=q+w
Energy is a State function

A State function= values which do not depend on the pathway,
only on the initial and final states. Energy is a state function.

So 100g of water going from 90⁰ C to 15 ⁰ C, will give off the same
amount of energy, if you cooled the water down to 50 ⁰ C, and then
heated to 75 ⁰ C then cooled it again.

Heat and work are not state functions!

When talking about chemical reactions, most of the energy change
is in the form of heat. Work is negligible.
Heat Capacity

Heat Capacity=amount of energy required to increase the temp of
that substance by one degree= Cp

Cp=J/(Kor ⁰C)

Specific heat=amount of energy needed to increase the temp of
one gram of substance by one degree-like density, it’s
characteristic of specific substance

Cp/g=J/(Kor ⁰C)g

Molar heat capacity=

Cp/mole=J/(Kor ⁰C)mole
Specific heat example

The Specific heat capacity of copper metal is 0.380 J/⁰Cg. Assume
you had a 75.0g cube of copper at 25.0g. What would be the final
temperature of the copper if it absorbed 150 J of heat.
q  C p x mcopper x T
T  T2  T1
150 J 
0.380 J
x75.0 g x T
o
Cg
T  5.26o C
5.26o C  T2  25o C  30.3o C
Molar heat capacity example

If a 1.00 Kg block of Al (molar heat capacity of 24.2 J/⁰C mole) at
600 ⁰C is placed in contact with 1.00Kg. Block of copper (molar heat
capacity of 24.2 J/⁰C mole), calculate the final temperature of the
blocks.

qal=qcu
 1moleAl 
J
J
1000 gAlx 
x
24.2

896

o
o
Cmole
C
 26.98 gAl 
 1moleCu 
J
J
1000 gCux 
x
25.2

385

o
o
Cmole
C
 63.55 gCu 
C pAl  C pCu
J
896 o (600 oC  T f )  385(T f  100 oC )
C
J
J
J
J
5.38 x105 o  896 o T f   385 o T f  38500 o
C
C
C
C
T f  450 oC
Calorimetry

Calorimetry-measuring the heat flow into or out of a substance,

1) if temp is increasing= Then heat is flowing from the reaction to the
surrounding solution.

2)if temp decreasing=The heat is flowing from the solution into the
reaction.

There are two types of calorimetry: “coffee cup” and Bomb
“Coffee Cup” vs. Bomb
Law of Conservation of Energy

In interactions between a system and its surrondings, the total
energy remains constant-energy is neither created nor destroyed.
Applied to the change of heat this means.

qsystem + qsurrondings = 0

If this often conveint to rearrange this equation to:

qsystem = - qsurrondings
Example of calorimetry

When 4.25g of solid ammonium nitrate dissolves in 60.0 g of water in
a calorimetry, the temp drops from 22.0⁰C to 16.9 ⁰C. Calculate ∆H (in
kJ/mol of NH4NO3) for the solution process.
NH 4 NO3( s )  NH 4  ( aq )  NO3 ( aq )
Assume the specific heat of the solution is the same as that of pure water:
qreaction  qwater ( surrondings )  0
qwater ( surrondings ) 
4.184 J
o
3
x
(60.0
g

4.25
g
)
x

5.1
C


1.37
x
10
kJ
g ,o C
qreaction  (1.37 x 103 kJ )  1.37 x103 kJ
 1 mole NH 4 NO3( s ) 
4.25 g NH 4 NO3( s ) x 
 0.531 moles NH 4 NO3( s )
 80.1 g NH NO 
4
3( s ) 



1
kJ
3
1.37 x10 kJ x 
 25.8

 0.531 moles NH NO 
mol
4
3( s ) 

Enthalpy-heat content of the
system (∆H)

∆H= q

∆H⁰=enthalpy under standard conditions: 1 atm, 25 ⁰C

Enthalpy is a state function:

∆H=Hfinal-Hinitial
H  H final  H initial (endothermic )
H  H final  H initial (exothermic )
Enthalpy of Reaction

Enthalpy of reaction or heat of reaction-enthalpy change that
accompanies a reaction

∆H=Hproducts-Hreactants
Enthalpy of formation

The Enthalpy of formation (Heat of formation ∆Hf)- the enthalpy for the
reaction forming the substance from its pure elements

Standard State-the most stable form of a substance at a particular
temp (25 C) and standard atmospheric pressure (1 atm).

Standard enthalpy of formation (∆H⁰f)= change in enthalpy that occurs
with the formation of 1 mole of a substance from its elements in their
standard state.

(∆H⁰f)= 0 for C (graphite), H2(g), O2(g)

Ex. C(s) + O2(g) → CO2(g) ∆H⁰f = -393.5kJ

H2(g) + 1/2O2(g) → H2O(g)
∆H⁰f = -393.5kJ

∆H⁰rxn = ∑(x)∆H⁰f Products - ∑(y) ∆H⁰f reactants

X and Y are coefficents
Hess’s Law continued

The enthalpy of a reaction is equal to the sum of the heats of
formation of the product minus the heats of formation of the
reactants.
H o rxn  H o f products  H o f reac tan ts
Ex.rxn : 2C2 H 2( g )  5O2( g )  4CO2( g )  2 H 2O( l )
H o rxn  (4H o f CO2  2H o f H 2O(l ) )  (2H o f C2 H 2( g )  5H o f O2( g ) )
4(395kJ )  2(285.8kJ )  2(226.7 kJ )  5(0 KJ )
 2599.5kJ

Calculate the ∆H⁰rxn for the dissolution of 0.8327 g of H3AsO4 in water
give the following:
 1mole H 3 AsO4( s ) 
 5.866 x103 mol H 3 AsO4( s )
0.8327 g H 3 AsO4( s ) x 

 141.95 g H AsO 
4( s ) 
3

H 3 AsO4( s ) H o f  900.4kJ / mol
H 3 AsO4( aq ) H o f  899.7kJ / mol
H 3 AsO4( s )  H 3 AsO4( aq )
H o rxn  H o fH3 AsO4 ( aq )  H o fH3 AsO4 ( s )
 (899.7 kJ )  (900.4 kJ )  0.7 kJ / mol
0.7 kJ / mol x 5.866 x103 mol  4.0 x103 kJ
Application of Hess’s Law
Indirect Determination of Hess’s
Law

∆H is an Extensive Property (the amount matters)
N 2( g )  O2( g )  2 NO( g )  H o  180.5kJ
1
1
N 2( g )  O2( g )  NO( g )  H o  90.25kJ
2
2

∆H Changes sign when a process is reversed.
NO( g )
1
1
 N 2( g )  O2( g )  90.25kJ
2
2
Hess’s Law of Constant Heat
Summation

If a process occurs in stages or steps (even if hypothetically), the
enthalpy change for the overall process is the sum of the enthalpy
changes for the individual steps.

Example:
1
1
N 2( g )  O2( g )  NO( g )  O2( g ) H o  90.25kJ
2
2
1
NO( g )  O2( g )  NO2( g ) H o  57.07kJ
2

1
N 2( g )  O2( g )  NO2( g ) H o  33.18kJ
2
A second example

Use the following equation to determine ∆H⁰rxn for the following
reaction. 3C( graphite )  4 H 2( g )  C3 H 8( g ) H o  ?
1)C3 H 8( g )  5O2( g )  3CO2( g )  4 H 2O( l ) H o  2219.8 kJ
2)C( graphite )  O2( g )  CO2( g )
H o  393.5 kJ
1
3) H 2( g )  O2( g )  H 2O( l )
2
H o  285 kJ
3CO2( g )  4 H 2O( l )  C3 H 8( g )  5O2( g )
H o  2219.8 kJ
3C( graphite )  3O2( g )  3CO2( g )
H o  1181kJ
4 H 2( g )  2O2( g )  4 H 2O( l )
H o  1143 kJ

3C( graphite )  4 H 2( g )  C3 H 8( g )
H o  104kJ
Electron Configuration

Electron structure: arrangement of electrons

Electromagnetic radiation carries energy through space (ex. visible
light, ultraviolet, infrared, X-rays). All of these move through a
vacuum at 3.00x108(speed of light).

They all move as waves:
Electromagnetic Spectrum

Wavelength (λ)-distance between successive crests or troughs(units
of length).

Frequency (ν)-number of cycles that pass a given point in one
second (cycles/s-Hertz(Hz)).

λ ν=c (speed of light)

High frequency = high energy (gamma rays)

Low frequency = low energy (radio)
Max Plank (1900)

He assumed that radiation could only be emitted in little packets
(quanta). The energy associated with these quanta is proportional
to the frequency of the reaction.

E=hv


E=energy of a quantum, h=Plank’s constant (6.63 x-34 J,s),
v=frequency of radiation
Energy is always emitted in whole numbers-multiples of hv (hv, 2hv,
3hv, etc)
Example

Cosmic rays are a penetrating electromagnetic radiation produced
by certain catastrophic events. Calculate the energy associated
with cosmic radiation on a per mole basis and compare your result
with the energy of radio waves. The average wavelength of a radio
wave is 500nm and that of a cosmic ray is 1x10-14m.
Ecos mic  hv
Eradio  hv
v  c
cos mic  1x1014 m
v  c
radio  500m
22
3.8 x108 m 
1
 3.00 x10
v
x

14
sec
sec
 1x10 m 
3.8 x108 m  1  6.00 x105
v
x

sec
sec
 500 m 
Ecos mic
3.00 x1022
6.02 x1023 1.2 x1013 J
34

x6.63x10 J sec x

sec
1 mole
mole
Eradio
6.00 x105
6.02 x1023 2.4 x 104 J
34

x 6.63 x 10 J sec x

sec
1 mole
mole
Photoelectric Effect-Albert Einstein

If you shine light on a metal, electrons are emitted. There is a
minimum energy of light that can be show where no electrons are
emitted. One metal might require red light, whereas another might
require yellow light. Radiation strikes the metal in a stream of tiny
energy packets (massless).

Photons - energy particles. Light can have particle-like properties as
well as wavelengths. Duel nature of light.

E=hv - energy of a photon.

Monochromatic light- made up of a single wavelength

Continuous spectrum (all wave lengths)
Einstein Continued

Apply high voltage to an element (gaseous)-atoms absorb energy
and then emit energy as electromagnetic radiation. Ex. Neon-redorange light. Which wavelength of light is emitted is a characteristic
of an element.

Line spectrum - spectrum containing radiation of specific
wavelengths.
Niels Bohr - Bohr’s Model of the
Hydrogen Atom

En = energy of an electron in a hydrogen atom

RH = Rydberg constant (2.18 x 10-18 J)

n = principle quantum number- different allowed orbits for the
electron

n = 1 if electron is in first orbit, n= 2 if electron is in the second orbit,
etc.

All energies are negative- the more negative, the more stable an
atom will be.
Bohr Model Continued

Ground state-lowest energy state; n = 1

Excited state- higher energy state; n = 2

n gets larger and larger until the electron completely breaks away
from the nucleus.

Zero-energy state-when an electron is removed from the nucleus.
Highest energy state: n= infinity
Bohr Continued

Energy is absorbed when it moves to a higher n.

Energy is emitted when it moves to a lower n.

+∆E when nf › ni – radiant energy is absorbed

-∆E when nf ‹ ni – radiant energy is emitted.
de Broglie

If light can have both wave-like properties and particle-like
properties, then why can’t matter (like electrons) have both? Both
radiation (light) and matter obey the following equation.

λ= de Broglie wavelength (m)

m= mass of the particle (kg)

v= speed (m/s)

h=Plank’s constant
Can Look at Electrons, Baseballs,
Molecules with the Equation.

Calculate the de Brogile wavelength of an electron traveling at
1.00% the speed of light
h  6.62 x1034 J s
m  9.11x1031 kg
v  0.01(3.00 x108 m / s )  3.00 x106 m / s
 h 

 mv 


6.62 x1034 J s
Js
10
 

2.43
x
10

31
6
m
9.11
x
10
kg
x
3.00
x
10
m
/
s


kg
s
 
Particle Wave Duality

Since electron exhibit both wave and particle like properties, one
can tell exactly how fast an electron is moving around an atom at a
given time.

Heisenberg Uncertainty Principle- It is impossible for us to know both
the exact location of an electron and its speed at the same time.
Only one or the other can be determined.

Quantum Mechanics - Schrodinger - We can’t predict the exact
speed and location of an electron, but we can determine a
probability.
Quantum Mechanics

Wave functions (ψ)- ψ2 – probability density- probability of an
electron’s location in an allowed state. Higher probability of finding
an electron in an area of high density.

Orbitals - not the same things as an orbit - In quantum mechanics n
measure the most probable distance of an electron from the
nucleus-not the radius of the a defined orbit.
Quantum numbers:

(1)n- principle quantum number-shell- a collection of orbitals with
the same quantum number n. Specifies the size and extent of the
orbital. Determines the energy of an electron in a hydrogen atom.
 1 
En  2.18 x1018 J x  2 
n 

(2)azimuthal quantum number-specifies the shape of the orbital.
Restricted to 0,1….n-1. Designated by letters: s,p,d,f which
correspond to numbers
Azimuthal quantum number
continued

l = 0 –s-sharp

l = 1 –p-principle

l = 2 –d- diffuse

l = 3- f-fundamental

Defines subshells
n  1,
n  2,
n  3,
n  4,
l  0,
s
l  0,1,
s and p
l  0,1, 2
s, p, d
l  0,1, 2,3, 4 s, p, d , f
1s
2 s, 2 p
3s3 p,3d
4 s , 4 p, 4 d , 4 f
Magnetic Quantum Number

Determines the spatial orientation of an orbital.

Has values between l and –l including zero. Defines the # of orbitals
in each subshell.
n  1 l  0 (1s ) ml  0
1 orbital in subshell
n  2 l  0 (2 s ) ml  0
1 orbital in subshell
n  2 l  0 (2 p) ml  1, 0,1
3 orbitals in subshell
n  3 l  0 (3s ) ml  0
1 orbital in subshell
n  3 l  0 (3 p) ml  1, 0,1
3 orbitals in subshell
n  3 l  0 (3d ) ml  2, 1, 0,1, 2
5 orbitals in subshell
Spin Quantum Number

(4) ms-electron spin quantum number-electrons spin in 2 opposite
directions, -1/2 or + ½

Each orbital can hold a maximum of two electrons

No two electrons in an atom can have the same two quantum
numbers- Pauli Exclusion Principle
Electron #1
n 1
l 0
ml  0
Electron # 2
n 1
l 0
ml  0
ms  1/ 2
ms  - 1/ 2
Using Arrows to Show the Pauli
Exclusion Principle

1 electron
1s

2 electrons
1s
can not be:
 
or
1s
1s
How Many Electrons are in the n=4
Shell?
l
ml
ms
0
0
 1/ 2 or  1/ 2
1
 1, 0,1
 1/ 2 or  1/ 2
2  2, 1, 0,1, 2
 1/ 2 or  1/ 2
3  3, 2, 1, 0,1, 2  1/ 2 or  1/ 2
# orbitals
1
3
5
7
# electrons
2
6
10
14
32 e-
Shapes of Orbitals
Electron Configurations

Arrangement of electrons in the orbitals. Orbitals fill up in order of
increasing energy
He  2 electrons

1s 2
1s
B  5electrons

1s

2s

2p
1s 2 2 s 2 2 p
Li 

1s 2
1s
C  6 electrons

1s

2s
 
2p
1s 2 2 s 2 2 p 2
Hund’s Rule

The ground state configuration is obtained by placing the electrons
in different orbitals with parallel spins. No orbital in the subshell
contains two electrons until each on contains one electrons.
O   7 electrons

1s

2s
   2 2 3
1s 2 s 2 p
2p
N   8 electrons

1s

2s
   2 2 4
1s 2 s 2 p
2p
Noble Gas Configurations
Fr  87 electrons

Valance electrons-outer shell electrons

Core electrons-inner shell electrons

Representing the inner shell electrons as a noble gas
[ Rn]7 s1
K  19 electrons
[ Ar ]4 s1
Na  11electrons
[ Ne]3s1
Mn  25 electrons
[ Ar ]4 s 2 3d 5
Pt  78 electrons
[ Ar ]6 s 2 4 f 14 5d 8
Removing Electrons to Form Ions

Do not remove electrons in the order in which they are filled. Always
remove from valence shell (outermost energy level).
Zn [ Ar ]4s 2 3d 10
Zn 2 [ Ar ]3d 10
Se [ Ar ]4 s 2 3d 10 4 p 4
Se
3
2
10
1
[ Ar ]4 s 3d 4 p
Two cases when electrons shift from
their ground state configuration.
1 chromium(Cr)-24 electrons
[ Ar ]4 s 2 3d 4
but really [ Ar ]4s1 3d 5

  
4s
3d
2 copper(Cu)-29 electrons
[ Ar ]4 s 2 3d 9

4s
but really [ Ar ]4s1 3d 10
    
3d
Periodic Trends

Atomic radius-electrons distribute themselves about a nucleus in a
diffuse, cloud-like manner-no sharp boundaries.

(1)atomic radius increases going down the PT-adding a whole new
shell.

(2)atomic radius decreases going across the PT- increased positive
charge in the nucleus. Therefore, pull on the electrons is greater,
making the radius smaller.
Ionic Radius

(1) Positive or negative ions of elements in the same group increase
in size moving down the PT.

(2) Radius of cation is always smaller then the neutral atom.

(3) Radius of anion is always larger than the neutral atom.
Ionization Energy

Amount of energy it takes to remove an electron completely from an
atom (making a cation)

First ionization energy (I1) =amount of energy associated with removing
one electron from a neutral atom, forming a positive ion, A+.
A g   A ( g )  e 

Second ionization energy (I2) = amount of energy associated with
removing one electron from the A+ ion to produce A2+ ion
A


g
A
2
(g)
e

The higher the ionization energy, the harder it is to remove an electron.
Ionization Energy Cont.
Na  11electrons
1s 2 2 s 2 2 p 6 3s1  1s 2 2s 2 2 p 6
I1  496kJ / mole
Na   10electrons
1s 2 2 s 2 2 p 6  1s 2 2 s 2 2 p 5

I 2  4560kJ / mole
Generally,

(1)ionization energy increases moving across the PT.

(2)ionization energy increases going down the PT.
Electron Affinity

The energy change when an atom gains an electron in the gaseous
state. Generally, for neutral and positively charged species, energy
is released when an electron is added.

Generally, for an anion, energy must be added for an electron to
be added.
A  e   A
Cl  e   Cl 

EA  349kJ / mole
Generally, the trend is moving from left to right, and up the periodic
table (excluding the noble gases).
Electronegativity

This is a value only used when talking about atoms bound in
covalent bonds.

This represents the pull that the single bond places on the shared
electrons.

Trend increases going across and going up, making Fluorine the
most electronegative .
Periodic Trends