Chemistry I, Honors
Mr. von Werder
The Mass of an Atom


- weighted average of the masses of the element’s isotopes
- this is why atomic masses are not whole numbers
Calculation of atomic mass requires you to know the:
- mass of each isotope (protons + neutrons)
- relative abundance of each isotope
Atomic mass
(decimal equivalent of the percentage…move decimal two spaces to the left)
average atomic mass = (isotope mass1)(rel.abund.1) + (isotope mass2)(rel.abund.2) + …


EXAMPLE:
“Vegium” and “Pennium” acitivities
Chemical Measurements
 Measuring Matter


What ways do we measure the amount of something?
COUNT
(number of students in class, number of right answers on a test...)
MASS
(your weight, pound of sugar...)
VOLUME (liters of pop, ounces of medicine...)
The Mole -- is one unit that relates to all three of these

First, what are the names of the particles in chemistry? Well, for …
Elements - call the particles atoms
- EXCEPT seven elements that exist as diatomic molecules:
H2, N2, O2, F2, Cl2, Br2, I2 (and a few molecules too, P4, S8)
Molecular compounds (nonmetal-nonmetal(s)) - call the particles molecules
Ionic compounds (metal-nonmetal(s)) - call the particles formula units
Ions themselves are particles sometimes called for.

Counting particles in chemistry is impossible -- particles are too small.
So we need a unit to represent a large number of these particles:

MOLE
■
■
from the Latin moles meaning “huge bulk”
1 mole of a substance = 6.02 x 1023 particles (Avogadro’s number)
[Just as . . .
[
[
[so then...
1 dozen
1 gross
1 ream
1 mole
=
=
=
=
12
items ]
144
items ]
500
items ]
x
6.02 1023 items ]

MOLAR MASS
■
■
Examples:
the mass of one mole of any substance (atoms, molecules, formula units)
equal to the atomic mass, or sum of atomic masses, from the
periodic table, in grams
1 mole S
= 32.07
grams S
1 mole H2O
= 18.02
grams H2O
1 mole K2Cr2O7
= 294.14 grams K2Cr2O7
MASSES TAKEN FROM
THE PERIODIC TABLE
SHOULD BE ROUNDED
TO THE SECOND
DECIMAL PLACE.
10-2 Mole Conversions
 Mole-Mass Conversions
 Molar mass can be used as a conversion factor.
For example, the relationship of:
can become conversion factors:
1 mole CO2 = 44.01 grams CO2
1 mole CO2
44.01 grams CO2

44.01 grams CO2
1 mole CO2
Mole-Particle Conversions
 The definition of a mole with Avogadro’s number is another conversion factor:
1 mol
6.02 x 1023 particles
6.02 x 1023 particles
1 mol
- You may need to calculate the specific number of atoms or ions in a substance.
- This simply requires you to create another conversion factor, like . . .
3 ions NH4+
1 form.un. (NH4)3PO4
4 atoms C
1 molecule C4H10
Concentration - measure of the amount of solute dissolved in a given amount of solution.
- concentrated - large amount of solute
These terms are too
- dilute - small amount of solute
vague, and are relative.

Molarity - most important unit of concentration in chemistry
- number of moles of solute per liter of solution (M)
2.5 moles HCl
1 L HCL sol’n
- 2.5
M HCl
Moles of solute
Liters of solution
M
or
1 L HCL sol’n
2.5 moles HCl

Converting Between Units with Moles
The mole is often an intermediate in multi-step chemical calculations.
So far, we have discussed three ways to use the mole; more will come later.
This diagram will grow to include more mole conversions.
particles
grams
Avogadro’s
number
particle
ratio
from chemical formula
particles
Avogadro’s
number
molar
mass
moles
mole
ratio
from chemical formula
molarity
liters
of solution
moles
Calculating Percent Composition
 Percent composition - identifies what elements are present and their percentages in a compound
part
 Remember that percentages are calculated as:
% = ------ x 100
whole
 Several ways to calculate percent composition of a substance.

1.
From experimental data:
- Mass of each element (“part”) in the compound is given.
- Add the mass of each element to find the “whole” mass.
2.
From the chemical formula: We assume to work with one mole.
- Mass of each element (“part”) is taken from the periodic table.
- Add the molar mass of each element to find the “whole” mass.
3.
From the combustion analysis data:
- Determine the grams of C from the grams of CO2 given.
- Determine the grams of H from the grams of H2O given.
- Determine the grams of O (or N), if needed, by subtracting
- Once you have grams of each element, determine percents.
Calculating Empirical Formulas
 Empirical formula
- lowest whole number ratio of elements
- may or may not be the same as the actual formula
- Think of the chemical formula as a ratio of moles of each element
- So, we need to find the moles of each element in the compound.
 General procedure; (but this is best understood by example)
1. Percent composition data;
or
Assume a 100 gram amount
1. Gram data;
already in grams.
1. Combustion analysis;
Calculate grams
(percents “become” grams).
2. Convert each gram amount to moles (now have a ratio of moles).
Percents to grams
Grams to moles
Divide by smallest
Multiply ‘til whole
3. Divide each mole value by the smallest number of moles.
4.If needed, multiply each part of ratio to convert to whole numbers.

Calculating Molecular Formulas
 Molecular formula - actual formula, or true formula, of a compound
- is a whole number multiple(n) of the empirical formula
- may be the same as the empirical formula (whole number multiple is 1)
- Examples:
CH2O
is the empirical formula
C6H12O6
is the molecular formula (multiple, n, is 6)
CO2 is the empirical formula
CO2 is the molecular formula (multiple, n, is 1)
 The key to finding the molecular formula is to find the multiple, n:
The multiple should be a whole number, or very nearly so.
n=
molar mass of compound (given)
molar mass of empirical formula
Then you simply multiply the subscripts in the EF by “n” to get the subscripts in the MF
molecular formula = (empirical formula) x n
Chemical Equations
 We describe chemical reactions, the rearranging of atoms, by writing chemical equations,
the “sentences” of the chemical language.
 You already know the “words” - they are the formulas you learned to write in Chapter 2.
 What we observe in a beaker or test tube or elsewhere can be communicated with an
equation, which in turn can be interpreted at the microscopic level.

Word Equations
- The reactants and products are written by their names.

Formula Equations
- The reactants and products are written by their formulas.
- Symbols used in equations:

+
(s)






(l)
(g)
(aq)
MnO2
Pt
Catalyst - helps speed up a chemical
reaction, but is not used up itself.
Balancing Chemical Equations
Equations need to be consistent with the Law of Conservation of Mass - atoms are not
created or destroyed, they are rearranged.
Needs to be exactly the same number of each type of atom on each side of the reaction (of
the arrow).
We do this using COEFFICIENTS which, similar to math, are numbers placed in front of a
chemical formula to indicate how many of that substance we need.
- A coefficient of 1 is understood and need not be written.
- Coefficients need to be whole numbers, but using a fraction as
an intermediate step is OK - as long as you clear the fraction.
Some tips:
- NEVER change the formula of substance to balance an equation.
- If a polyatomic ion (SO42-, CO32-, PO43-…) is present on both sides
of the reaction unchanged, balance the whole ion as one piece.
- In a complicated looking equation, it is helpful to leave an element sitting
on its own as the last thing to balance - because changing that
element does not affect the others you have already worked with.
- It can be helpful at times to write water as HOH - allowing you to balance
the OH as a polyatomic ion.
- Reduce coefficients to their lowest whole number ratio.
- Practice, Practice, Practice
What is Stoichiometry?
Stoichiometry
- study of the quantitative relationships in chemical reactions
- involves the relationships between reactants and products
- will be used in many topics yet to come in chemistry

Interpreting Balanced Chemical Equations
Consider the reaction:
Mg
+
2 HCl
--->
1 atom
+
1000 atoms
6.02 x 1023 atoms
1 mole
+
2 molecules
--->
MgCl2
+
H2
1 formula unit
+
1 molecule
+
1000 molecules
2000 molecules ---> 1000 formula units
+ 12.04 x 1023 molecules ---> 6.02 x 1023 formula units + 6.02 x 1023 molecules
+
2 moles
--->
1 mole
+
1 mole
So, we go from interpreting chemical equations based on the coefficients in terms
of particles (e.g. ...) to interpreting them in terms of moles.
Interpreting equations in terms of moles allows us to relate chemical equations to
actual measurable amounts in grams and liters.

Mole-Mole Problems
The coefficients in equations are most useful in allowing use to see the relative amounts of
reactants and products.
For example:
2 H2O ---> 2 H2 + O2
2 moles H2O
2 moles H2O
2 moles H2
2 mole H2
1 mole O
1 mole O
2 mole H2
1 mole O
1 mole O
2 moles H2O
2 moles H2O
2 moles H2
These “molar ratios” are now useful in converting moles of one substance to another.
(We need only write the molar ratio that is needed as a conversion factor for a given situation)
Examples:

Verifying the Law of Conservation of Matter
Again, consider the reaction:
Mg
1 mole Mg
+
+
2 HCl
2 moles HCl
1 mole Mg x 24.31 g Mg
2 mole HCl x 36.46 g HCl
1 mole Mg
1 mole HCl
= 24.31 g Mg
= 72.92 g HCl
97.23 g reactants
--->
--->
MgCl2
1 mole MgCl2
1 mole MgCl2 x 95.21 g MgCl2
+
+
H2
1 mole H2
1 mole H2 x 2.02 g H2
1 mole MgCl2
= 95.21 g MgCl2
1 mole H2
= 2.02 g H2
97.23 g products
Notice the mass of the reactants = the mass of the products
Verification of the Law of Conservation of Matter.

Solving Stoichiometry Problems
Mass-Mass Problems
Mass-Volume
Volume-Volume
Other combinations: mole-mass, mass-mole, etc.
Mole Map
GIVEN
FIND
particle
s
grams
particle
ratio
from chemical formula
Avogadro’s
number
particle
s
molar
mass
joules
(or calories)
H of
reaction
molar
mass
moles
molar
ratio
H of
reaction
moles
molarity
liters
grams
Avogadro’s
number
molarity
molar
volume
from balanced equation
molar
volume
of solution
liters
of solution
liters
of a gas @ STP
volume
ratio
liters
of a gas @ STP
joules
(or calories)
11-3 Limiting Reactants and Percent Yield
Types of Chemical Reactions
Metathesis ( meh tath isis )
Double replacement (exchange)
Acid base
Redox
Synthesis
Decomposition
Single replacement
Combustion
Others…

COMBINATION (synthesis)
- two or more reactants combine to form a single product
R
+
S
--->
RS
Examples:
S
H2
Na
+
+
+
O2
O2
Cl2
--->
--->
--->
SO2
H2O
balance. . .
NaCl
- in predicting the product it is sometimes helpful to consult the ion
list to determine a possible combination of ions.
- nonmetal oxides (molecular oxides) and water usually produce acids
Example:
SO3(g) + H2O(l) ---> H2SO4(aq)
- metal oxides (ionic oxides) and water usually produce bases (hydroxides)
Example:
CaO(s) + H2O(l) ---> Ca(OH)2(aq)

DECOMPOSITION {DEC}
- a single reactant breaks into two or more products
RS
--->
R
+
S
H2O(l) --->
H2(g) +
O2(g)
- most decomposition reactions require energy (heat, light, electricity)
- predicting decomposition products is more difficult, so you will usually be
given one or all of the products involved.

SINGLE-REPLACEMENT {SR}
- a metal element will replace a positive ion of a compound or
a nonmetal element will replace a negative ion of a compound
T
+
U


+
RS
--->
TS
+
R
RS
or
--->
RU
+
S
- in predicting if a metal will cause single-replacement we need to
consult the ACTIVITY SERIES of metals
a metal will replace (any metal ion or H+) below it on the series
still need to be aware of the charges of the ions and write a
correct formula for the product
Example: Al + H2SO4 --->
Al + H2SO4 --->
Al2(SO4)3 + H2
balance . . .
- in predicting if a nonmetal will replace another, use Group 7 (group 17)
of the periodic table as the series
F
most active
(we will not worry about other
Cl
nonmetals at this point)
Br
I
least active
Example:
Cl2 + KI --->
Cl2 + KI ---> KCl + I2

balance . . .
- if, based on the activity series, no reaction is expected, write
“No reaction”, or just NR, and a brief explanation.
Examples:
Cu + FeSO4 ---> NR, Cu is less active than Fe
Br2 + LiF ---> NR, Br is less active than F
DOUBLE-REPLACEMENT {DR}
- two compounds exchange their ions
RS
R+S-
+
TU
+
T+U-
--->
--->
RU
R+U-
+
TS
+
T+S-
- these reactions usually take place in aqueous solutions
- Predicting the products is relatively easy - again, remember to get
the correct formulas for the products
- Predicting IF THE REACTION WILL OCCUR there must be a driving
force on the product side of the possible reaction equation:
1)
insoluble (or slightly soluble) product - a precipitate
Refer to a solubility chart. (page 574)
2)
gaseous product
3)
molecular product (H2O, CO2 , etc.)
- Example:
Pb(NO3)2 + Na2S ---> PbS
+ NaNO3 balance . . .

COMBUSTION of hydrocarbons
- complete combustion of a hydrocarbon (contains C, H, and sometimes O)
produces carbon dioxide (CO2) and water (H2O)
* incomplete combustion can produce carbon, and carbon monoxide
* usually a lot of heat and light is released - important energy reactions
CxHy
Example:
C2H6
+
+
O2
O2
--->
--->
CO2
CO2
+
+
H2O
H2O
balance . . .