Unit 4
COUNTING PARTICLES
Counting By Weighing
We can weigh a large number of the
objects and find the average mass.
Once we know the average mass we can
equate that to any number of the
objects.
EXAMPLE:
◦ The average mass of a book is 40.0 grams.
◦ How many books are present in a sample with a mass of 2000.0 grams?
◦ 2000.0g/40.0g = 50.0 books
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Counting By Weighing
When we know the average mass of the
atoms of an element, as that element
occurs in nature, we can calculate the
number of atoms in any given sample of
that element by weighing the sample.
The atomic mass of an element, as
found on the periodic table, allows us
to count by weighing.
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Relative Masses
•Once early chemists realized they could
determine the relative masses of molecules of
gases, they chose weighable samples of the
elements as the standard amount of each
element.
Molar Masses –
•The molar masses (formerly known as atomic weights)
we find in the Periodic Table of the elements were
originally determined relative to the mass of
hydrogen.
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Section 6.3: The Mole
The mole (mol) is known as the
“chemists dozen” and
represents
Not this type of mole!
6.022 x 1023 things (atoms,
particles, molecules, etc).
•A mole (from the Latin - “lump
of stuff”), was originally defined
as 1.0g of the lightest element
known – hydrogen.
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Mole Video
The Mole: Interesting Mole Facts
•6.022 X 1023 watermelon seeds: would be found
inside a melon slightly larger than the moon.
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6.02 X 1023 grains of sand: would be
more than all of the sand on Miami Beach.
6.022 X 1023 pennies: would make at least 7
stacks that would reach the moon.
6.02 X 1023 blood cells: would be more than the
total number of blood cells found in every human on
earth.
A mol is A LOT of particles:
602,200,000,000,000,000,000,000
The Mole
•It is VERY important to understand that the value listed as the mass
number on the periodic table really tells us TWO things:
– The mass of one atom in units of amu’s.
– The mass of one mole of atoms in grams.
–One mol of Al = 26.98 g & one atom = 26.98 amu
–One mol of Au = 196.97 g & one atom = 196.97 amu
–One mol of B = 10.81 g
& one atom = 10.81 amu
–Remember one mol = 6.022 X 1023
atoms (or particles or molecules)
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The Mole
•To summarize: a sample of any element that
weighs a number of grams equal to the molar
mass (from the periodic table) contains
6.022 x 1023 atoms (1 mol) of that element.
•Therefore, atomic weight of an element =
#g in one mol of that element
•We write that as g/mol.
6-12
Using the mole in calculations
•How many mols are in 7.0 g of N?
• 7.0 g
1
1 mol
14 g
= 0.50 mol
•How many atoms are in 7.0 g of N?
23 atoms
6.022
x
10
• (0.5mol)(
/1 mol)=
• OR
• (7.0 g)(1 mol/14.0 g)(6.022 x 10
3.011 x 1023
23 atoms
/1 mol)=3.011
atoms.
x 1023atoms
The Mole: Practice
1) Calculate the number of moles and the number
of atoms in a 25.0 g sample of calcium.
2) Calculate the number of moles and the number
of atoms in a 57.7 g sample of sulfur.
3) Calculate the number of atoms in a 23.6 mg
sample of zinc.
4) Calculate the number of atoms in a 128.3 mg
sample of silver.
6-14
1. .624 mol and 3.76 x 1023 atoms
2. 1.80 mol and 1.08 x 1024 atoms
3. 3.61 x 10-4 mol and 2.17 x 1020 atoms
4. 1.189 x 10-3 mol and 7.160 x 1020 atoms
The Mole: More Interesting Mole Facts
•A one-liter bottle of water contains 55.5 moles
H20 molecules.
•A five-pound bag of sugar contains 6.6 moles of
C12H22O11 (sucrose).
•We have 3 types of moles that live underground in
North America: Eastern Mole, Hairy-Tailed Mole
and Star-Nosed Mole
•The "Mexican" Mole is a chocolate sauce or turkey
stew. It comes from the Aztec word "molli."
6-16
Molar Mass
•A chemical compound is a collection of atoms.
•One methane (CH4) molecule contains one C atom
and four H atoms.
•It follows then that one mole of methane molecules
contains one mole of C atoms and four moles of H
atoms.
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Figure 6.3:
Various
numbers
of methane
molecules.
6-18
Molar Mass (g/mol)
•The molar mass of any substance is the
mass (in grams) of one mole of the
substance.
•The molar mass of a compound is
obtained by summing the masses of ALL
component atoms.
6-19
Molar Mass
•If we know how many atoms and how many moles
are present, we can calculate the mass of one mole
of a compound.
•This is called the “molar mass” or sometimes you
may also see it referred to as “molecular weight”.
•Since 1 mol C = 12.01 g and 4 mol H = 4(1.008)
or 4.032 g, 1 mol CH4 = 12.01 + 4.032 or 16.04 g
(sig figs).
•Remember to use mass numbers from the periodic
table.
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•Remember:
can talk about one mole of atoms or
Molar we
Mass
one mole of molecules.
•One mole of oxygen atoms (O) weighs 16.00 g.
•One mole of oxygen molecules (O2) weighs 32.00 g.
•Two moles of O atoms weigh 32.00 g.
•Two moles of O2 molecules weigh 64.00 g.
•And so on . . .
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Molar Mass: Practice
1) Calculate the following molar masses:
a) Water – H2O
b) Ammonia – NH3
c) Propane – C3H8
d) Glucose – C6H12O6
2) Calculate the mass of 1.48 mol C3H8.
3) Calculate the mass of 4.85 mol HC2H3O2.
6-22
Molar Mass: Practice
1. Calculate the number of moles of H2CO3 present
in a 7.55 g sample.
2. How many water molecules are present in 10.0 g
of water? (Hint: find moles first)
3. How many molecules of sucrose (C12H22O11) are
present in a five pound bag of sugar?
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Representative
Particles=
atoms, molecules,
Items, etc
Percent Composition of Compounds
•Sometimes it is not enough to know a compound’s composition in terms
of numbers of atoms; it may also be useful to know its composition in
terms of the masses of its elements.
•We can calculate the mass fraction by dividing the mass of a given
element in one mole of a compound by the mass of one mole of the
compound.
6-25
Percent Composition of Compounds
•Once we know the mass fraction we can multiply by 100 to get the
percent.
•Remember: percent = part/whole X 100.
6-26
In one mole of methane (CH4), there is one mole of C
and four moles of H:
1 mol C = 12.01 g
4 mol H = 4(1.008) = 4.032 g
1 mol CH4 = 12.01 g + 4.032 g = 16.042 g
%C= mass of 1mol C/ mass of 1 mol CH4 x 100
%C = 12.01/16.042 X 100 = 74.87% C
%H= mass of 4 mol H/mass of 1 mol CH4 x 100
%H = 4.032/16.042 X 100 = 25.13% H
Percent Composition Practice
Determine the mass percent of each
element in the following:
◦ H2SO4 (sulfuric acid)
◦ C3H7OH (isopropyl alcohol)
◦ C6H12O6 (glucose)
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Formulas of Compounds
•The object of this section is to do the opposite of the previous section.
Instead of getting the mass from the formula, we will determine the
formula from the mass.
•To do this, the mass must be converted to moles using each element’s
mass number. How can we convert mass to moles?
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Formulas of Compounds
ex) An unknown compound with a mass of 0.2015 g is is found to
contain:
◦ 0.0806 g C
◦ 0.01353 g H
◦ 0.1074 g O
It must contain:
0.0806g (1 mol C/12.01 g C) = 0.00671 mol C
0.01353g(1 mol H/1.008 g H) = 0.01342 mol H
0.1074g(1 mol/16.00 g O) = 0.00671 mol O
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The numbers from the previous slide allow us to determine the C:H:O
ratio.
0.00671 (C) : 0.01342 (H) : 0.00671 (O)
If we divide each number by the smallest number we get 1:2:1 for the
C:H:O ratio.
This leads us to a formula of C1H2O1 or CH2O.
This is not necessarily the TRUE formula of the
compound, but represents the RELATIVE numbers
of atoms.
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This represents the lowest whole number ratio of the compound.
The actual formula could be CH2O, C2H4O2, C3H6O3, C4H8O4, C5H10O5,
C6H12O6, . . .
Any formula with a C:H:O ratio of 1:2:1 is possible (in theory, an
infinite number).
C1H2O1 represents the simplest possible
formula or the EMPIRICAL FORMULA.
The multiples represent possible
MOLECULAR FORMULAS.
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Formulas of Compounds: Practice
Determine the empirical formula from each of the
following molecular formulas:
◦ H2O2 (hydrogen peroxide)
◦ C4H10 (butane)
◦ CCl4 (name?)
◦ HC2H3O2 (acetic acid)
◦ C6H12O6 (glucose)
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Calculation of Empirical Formulas
There are four steps to determine the
empirical formula of a compound:
1. Obtain the mass of each element present (in
grams).
2. Determine the number of moles of each type
of atom present (use atomic mass).
3. Divide each number by the smallest number.
4. Multiply all numbers by the smallest integer
that will make them all integers
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Empirical Formulas: Practice
1) A 1.500 g sample of a compound containing only
carbon and hydrogen is found to contain 1.198 g
of carbon. Determine the empirical formula for
this compound.
2) A 3.450 g sample of nitrogen reacts with 1.970
g of oxygen. Determine the empirical formula
for this compound.
3) When a 2.000 g sample of iron metal is heated
in air, it reacts with oxygen to achieve a final
mass of 2.573g. Determine the empirical
formula for this compound.
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If the relative amounts of elements are presented as percentages,
assume we are starting with a 100 g sample (100%). Then each
percentage simply becomes a mass (in grams).
For example if 15% of a compound is carbon, we just assume it is 15 g
of a 100 g sample; from there we convert to moles.
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1)
2)
Empirical Formulas: More Practice
The simplest amino acid, glycine, has the following mass percents:
32.00% carbon, 6.714% hydrogen, 42.63% oxygen, and 18.66%
nitrogen. Determine the empirical formula for glycine.
A compound has been analyzed and found to have the following mass
percent composition: 66.75% copper, 10.84% phosphorous, and
22.41% oxygen. Determine the empirical formula for this compound.
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Calculation of Molecular Formulas
If we know the empirical formula AND the molar mass, we can calculate
a compound’s molecular formula.
Note: without the molar mass, the best you can find is the empirical
formula.
Once the molar mass is known, one must ALWAYS find the empirical
formula before one can calculate the molecular formula. It is
impossible to do the reverse.
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Calculation of Molecular Formulas
It is also important to note that the molecular formula is ALWAYS an
integer multiple of the empirical formula. We can represent the
molecular formula in terms of the empirical formula:
(Empirical Formula)n = molecular formula
It should also be noted when n = 1, the empirical and molecular
formulas are identical to each other.
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Molecular Formulas: Practice
1)
A compound containing carbon, hydrogen, and oxygen is found to be
40.00% carbon and 6.700% hydrogen by mass. The molar mass of
this compound is between 115 and 125 g/mol. Determine the
molecular formula for this compound.
2)
Caffeine is composed of 49.47% C, 5.191% H, 28.86% N, and
16.48% O. The molar mass is about 194 g/mol. Determine the
molecular formula for caffeine.
6-40