Transcript Mendelian Genetics - Shumway AST Homepage

The Mole Concept
Goal: To develop the concept of the mole
as a useful measurement and to apply this
in calculations involving mass and volume.
Objectives
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1. Define the mole in terms of Avogadro's number, formula
mass or molecular mass.
2. Given the number of moles, determine the number of
molecules of a covalent compound, formula units of an ionic
compound or atoms of an element.
3. Given the number of particles present in a sample,
determine the number of moles.
4. When given the formula, calculate the molecular mass of
a covalent compound or the formula mass of an ionic
compound.
5. Given the number of moles in a substance, determine the
mass of the substance.
6. Given the mass of a substance, determine the number of
moles of the substance.
7. When given the formula, calculate the percentage
composition of each of the elements in the compound.
Molecular and Formula Mass
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Molecular Mass is determined by the sum of the atomic
masses of each element in a compound.
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This name is incorrect when applied to an ionic
substance.
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The atomic mass of hydrogen in atomic mass units is 1, and the
atomic mass of oxygen is 16.
Therefore, the total mass of a water molecule, H2O, is 1+1+16,
or 18 amu.
E.g. Sodium chloride, NaCl, is an ionic substance which does not
exist in molecular form.
A better name for the mass of ionic substances is
formula mass.
Formula mass is determined by the sum of the atomic
masses of all atoms in the formula unit of an ionic
compound.
EXAMPLE: Formula Mass
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Find the formula mass of sodium sulfate, Na2SO4.
Solving process:
Add the atomic masses of all the atoms in the Na2SO4
unit.
2 Na atoms, 2 x 23 = 46 amu
1 S atom, 1 x 32 = 32 amu
4 O atoms, 4 x 16 = 64 amu
The total formula mass = 142 amu
EXAMPLE: Formula Mass
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Find the formula mass of calcium nitrate, Ca(NO3)2
Solving process:
Add the atomic masses of all the atoms in the Ca(NO3)2
formula unit. Remember that the subscript applies to the
entire polyatomic ion.
1 Ca atom, 1 x 40 = 40 amu
2N atoms, 2 x 14 = 28 amu
6 O atoms, 6 x 16 = 96 amu
Total formula mass is 164 amu
Groups of Atoms
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There is one problem in using the molecular and formula
masses of substances.
These masses are in atomic mass units, which is only 1.66 x
10-24 g.
The mass of a single molecule is so small that it is
impossible to measure it in the laboratory.
For everyday use in chemistry, a larger unit, such as a gram,
is needed.
One helium atom has a mass of 4 amu, and one nitrogen
atom has a mass of 14 amu. The ratio of the mass of one
helium atom to one nitrogen atom is 4 to 14, or 2 to 7.
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No matter what number of atoms we compare, equal numbers of
helium and nitrogen atoms will have a mass ratio of 2 to 7.
In other words, the numbers in the atomic mass table give
us the relative masses of the atoms of the elements.
Avagadro’s Number
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The standard laboratory unit of mass is the gram.
We would like to choose a number of atoms which would
have a mass in grams equivalent to the mass of one
atom in atomic mass units.
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Chemists have found that 6.02 x 1023 atoms of an
element have a mass in grams equivalent to the
mass of one atom in amu.
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The same number would fit all elements since equal numbers of
different atoms always have the same mass ratio.
E.g. 1 hydrogen atom has a mass of 1.0079 amu
6.02 x 1023 atoms of hydrogen have a mass of 1.0079 g.
This number, 6.02 x 1023 is called Avogadro's
Number in honor of a 19th century Italian scientist.
What is the Mole?
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6.02x1023 atoms of an element = 1 mole
of that element
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E.g. 6.02x1023 atoms of Na = 1 mole of Na
6.02x1023 molecules of a compound = 1
mole of that compound
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E.g. 6.02x1023 molecules of NaCl = 1 mole of
NaCl
Mole Trivia
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If there were a mole of rice grains, all the
land area in the whole world would be
covered with rice to a depth of about 75
meters.
One mole of rice grains is more grains
than all the grain that has been grown
since the beginning of time. (1)
One mole of rice would occupy a cube
about 120 miles on an edge! (1)
Mole Trivia
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A mole of marshmallows would cover the
United States to a depth of 600 miles (3)
In order to put a mole of rain drops in a
30 meter (about 100 feet) diameter tank,
the sides of the tank would have to be
280 times the distance from the Earth to
the Sun. (4)
A mole of hockey pucks would be equal to
the mass of the Moon.
Mole Trivia
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Assuming that each human being has 60
trillion body cells (6.0 x 1013) and the
Earth's population is 6 billion (6 x 109), the
total number of living human body cells on
the Earth at the present time is 3.6 x 1023
or a little over half of a mole.
Mole Trivia
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If one mole of pennies were divided up
among the Earth's population, each
person would receive 1 x 1014 pennies.
Personal spending at the rate of one
million dollars a day would use up each
persons wealth in about three thousand
years.
Life would not be comfortable because the
surface of the Earth would be covered in
copper coins to a depth of at least 400
meters.
Mole Trivia
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If you had a mole of pennies and wanted to buy kite
string at the rate of a million dollars per inch, you would
get your money's worth.
After stretching your string around the Earth one million
times, and to the Moon and back twenty-five times, you
would have enough string left over to sell back at a
dollar an inch (a decided loss) to gain enough money to
buy every man, woman and child in the US a $50,000
automobile and enough gasoline to run it at 55 mph for
a year.
After those purchases, you would still have enough
money left over to give every man, woman, and child in
the whole world about $5000.
Mole Trivia
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Basis for calculations:
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Earth's circumference = 25,000
miles
Distance to moon = 240,000 miles
Cost of gasoline = $2.50 per gallon
Gasoline mileage = 20 miles per gallon
U.S. population =220,000,000
World population = 6,000,000,000
Calculating Molecular Mass
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1. Calculate the molecular or formula
masses of the following compounds, all in
amu (g/mol):
a. C2H6
Amu=30
b. SiCl4
Amu=170
c. MgCO3
Amu=84
d. Ca3(P04)2
Amu=310
Calculating Molecular Mass
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Calculate the molecular or formula masses
of the following compounds, all in amu
(g/mol):
K2S
Amu=110
CH2CHCH2OH
Amu=58
Pb3(As04)2
Amu=899
C12H22011
Amu=342
Conversions
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Make the following conversions:
1.00 x 1026 molecules of SnCl2 to moles.
Ans: 1.66 X 102 mol.
0.400 moles of H2O to molecules.
Ans: 2.41 X 1023 molecules.
76.0 grams CaBr2 to moles.
Ans: 0.380 mol. Or 3.80 X 10-1 mol.
18.0 grams HBr to moles.
Ans: 0.222 mol. Or 2.22 X 10-1 mol.
Conversions
Make the following conversions:
 9.30 moles SiH4 to molecules.
 Ans: 5.60 X 1024 molecules.
 Find the mass of one atom of Na.
 Ans: 3.82 X 10-23g.
 Find the mass of one molecule of
H2SO4.
 Ans: 1.63 X 10-22 g.
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Conversions
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Make the following conversions:
9.30 moles SiH4 to grams.
Ans: 298 g.
Find the mass of 1.405 mole of Na2SO4.
Ans: 199.5 g.
Find the mass of one molecule of H2SO4.
Ans: 1.63 X 10-22 g.
Percent Composition
1 mole of NaCl is comprised of equal parts Na and Cl
 But Na and Cl have 23 g/mol and 35.45 g/mol, respectively!
 So they have differing percent composition
 The percent composition of Na is
(molar mass Na) x 100
molar mass NaCl
Molar mass NaCl= 23+35.45= 58 g/mol
%Na= (23) x 100
(58)
%Na= 40%
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%Cl= (35) x 100
(58)
%Cl=60%
 So, while NaCl is equal parts NaCl by molar ratios, by mass, it’s a
60/40 split.
Percent Composition
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Find the percentage composition of
the following:
CsF
 Ans: 87.5%; 12.5%.
 Bi203
 Ans: 89.7%, 10.3%.
 BaH2
 Ans: 98.6%, 1.44%.
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