Examples of Dimensional Analysis
Converting 12 .7 lbs. into kilograms:
12.7 lb
1
x
1 kg
2.205 lb.
= 5.76 kg
Converting 76.9 gallons into liters:
76.9 gal x 3.79 L = 291 L
1
1 gal
Converting 60.0 miles/hour into feet/second
60.0 miles x 5280 feet x 1 hour
hour
1 mile
3600 sec
= 88 feet/ sec
Chapter 7 – Chemical Quantities: The Mole
Atoms: The building blocks of matter
Subatomic Particle: Particles which compose the atoms
Particle
Location
Charge
Mass
Proton
nucleus
positive
1 amu
Neutron
nucleus
neutral
1 amu
Electron
electron cloud
negative
0.00055 amu
Carbon:
6 protons
6 neutrons
6 electrons
An atom is electrically neutral; it has the same number of protons (+) and
electrons (-).
Hydrogen: 1 proton
0 neutrons
1 electron
1
Mass of 1 proton:
0.000 000 000 000 000 000 000 001 66 g
1.66 x 10-24 grams
Mass of 1 proton: 1 atomic mass unit
1 a.m.u.
1
Mass of 1 carbon atom:
12 a.m.u
(6 protons + 6 neutrons)
1 Atomic Mass Unit is defined as 1/12 the mass of a carbon-12 atom
1 amu = 1.66 x 10-24 grams
Atomic number (Z) : The number of protons in the nucleus of an element. It
determines the type of element and does not vary.
e.g.
Z carbon = 6
Z chlorine = 17
The periodic table is arranged in order of increasing atomic numbers.
Mass Number: The mass of a particular atom.
It is determined by the total number of protons and neutrons in the nucleus.
(Electrons are not counted because of their insignificant mass.)
e.g.
Mass # for a carbon atom = 12 amu
(6 protons + 6 neutrons = 12 amu)
12
6C
Mass # for a uranium atom = 238 amu
(92 protons + 146 neutrons)
238
92 U
2
Question: How many protons does it take to make a mass of 1.0 gram?
Mass of 1 proton = 1 amu = 1.66 x 10-24 g
1gram ÷ 1.66 x 10-24 g/ proton =
In other words, it requires 6.02 x 1023 protons to make 1 gram of protons.
Mole = 6.02 x 1023
1 dozen = 12
1 gross = 144
1mole = 6.02 x 1023
Avogadro’s Constant
Abbreviation for the mole: mol
A mole is the number of atoms required in order to have the mass of a given
element in grams equal to its atomic mass in atomic mass units.
e.g. 1 atom of aluminum weighs 27 amu.
(atomic mass = 27)
Therefore a mole of aluminum weighs 27 grams.
And, there are 6.02 x 1023 atoms of aluminum in that 27 grams.
Remember:
The quantity of a mole of anything is always 6.02 x 1023.
The mass of moles of different things varies. It depends upon the mass of the
object.
1 mole of bowling balls = 1 mole of ping pong balls
mass of 1 mol bowling balls >> mass of 1 mol ping pong balls
1 mole of lead atoms = 1 mole of oxygen atoms
mass of 1 mole lead atoms > mass of 1 mole oxygen atoms
207.2 g  16.0 g
3
Mass
Formula
Mass
6.02 x 1023
Mole
# Particles
22.4 L @ STP
Volume
Percent Composition: The percent (%) by mass of each element in the compound.
H2O
2H
1O
H2O has 2 atoms of H for 1 atom of O
However, the mass of the O in the molecule is greater
than the H because of oxygen’s greater atomic mass.
Calculating the Percent Mass:
1) Calculate the total mass of the compound.
e.g. H2O
2) Divide the mass of each individual element by the total mass and multiply
by 100.
3) You can also calculate the amount of a substance in a given sample.
For example:
Given a 220.0 g sample of H2O, how much (g)
of it is O?
Empirical Formula: The lowest terms whole-number ratio of the elements in a
compound
Molecular Formula: Actual number of atoms of each element in a molecule of the
compound
Molecular Formula
H2O2
H2O
C6H12O6
C12H22O11
Empirical. Formula
HO
H2O
CH2O
C12H22O11
4
CH is the empirical formula for all of the following :
C2H2
C3H3
C50H50
C109H109
CgoogleHgoogle
Calculating Empirical Formula:
1) Determine the number of moles of each element.
e.g. 79.8 % C
Assume 100.g of compound
20.2% H
2) Find the ratio of moles of all the elements in the compound by dividing
each value by the element with the smallest number of moles.
3) Write the formula using the ratios from step #2 putting the least
electronegative element first.
Another example:
67.6% Hg
10.8 % S
21.6 % O
4) If any of the ratios from step #2 turn out to be fractions of atoms (e.g. 1.5 ,
2.33, 1.67) then multiply the whole formula in step#3 by the number which
will clear out the fraction and form a whole number.
An example of that:
1.843 g H
9.398 g C
5