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Chapter 5: Chemical Formula Relationships – The Mole
So far, we have considered elements and compounds at the atomic
(particulate) level. However, we can’t count or weigh atoms or compounds
at the particulate level. We must be able to deal with them at the
macroscopic level, that is, work with quantities that we can see and handle,
i.e. weigh out.
Counting by Weighing
The Haber-Bosch reaction converts hydrogen and nitrogen gas into
ammonia:
H2(g) + N2(g) →
NH3(g)
Dalton showed that elements combine in fixed whole number ratios, so the
relative numbers of atoms in a chemical reaction are important. A balanced
equation tells us the relative numbers of molecules of reactants and
products in a chemical reaction, e.g
3H2 + N2 →
2NH3
I will need 3 molecules of hydrogen gas (6 H atoms) for every 1 molecule of
nitrogen gas (2 N atoms).
3H2
3
30
300
3x106
3 million
18.066x1023
3 moles
+
N2
1
10
100
1x106
1 million
6.022x1023
1 mole
→
2NH3
2
20
200
2x106
2 million
12.044x1023
2 mole
The word “mole” is the name given to a specific number, just like we give
names to many numbers:
1
Number or Count
Name
12
24
144
Ream
6
1,000,000 (1 x 10 )
1,000,000,000 (1 x 109)
1,000,000,000,000 (1 x 1012)
6.022 x 1023
Mole
We cannot physically count molecules or atoms because they are far too
small to count. So how do we measure them out?
• We weigh them – i.e. by mass measurement.
• Mass measurement substitutes for a count of the uncountable
This is how we already do it in the macro world, e.g. a carpenter needs a
known number of nails to build a house. If you know how much a nail
weighs, you can buy the nails by the pound and know pretty accurately how
many you are getting. If you wanted 1000 jelly beans, the clerk at the candy
store would not count out 1000 jelly beans. She would perhaps count out
and weight 10 jelly beans (e.g. total = 50 g), divide that weight by 10 to
determine the average weight of a single jelly bean (5.0 g). She would then
multiply the average weight of a jelly bean by 1000 to get the weight of
1000 jelly beans (5.0 g/bean x 1000 beans = 5000 g.). She would proceed to
weigh out 5000g of beans in order to get a count of a 1000 beans.
In the laboratory, even the smallest quantities we work with contain
enormous numbers of molecules or atoms, numbers so large that they are
virtually incomprehensible. So, we cannot count them out; we must weigh
them.
How do we relate the weight of an atom to a count of atoms?
7.3 The Mole Concept: Counting by Weighing
Scientists chose a standard to compare all atoms to, and that standard is
Carbon. Specifically, a mole was defined as the number of atoms contained
2
in exactly 12 g of isotopically pure 12C. The number of atoms contained in
exactly 12 g of 12C was determined to be 6.02214 x 1023 atoms. Thus, the
name given to this huge number, 6.022 x 1023, is a “mole”
So,
•
•
•
•
1 mole of 12C atoms = 6.02214 x 1023 atoms
1 mole of 16O atoms = 6.02214 x 1023 atoms
1 mole of 35Cl atoms = 6.02214 x 1023 atoms
1 mole of anything = 6.022 x 1023 anything
A mole of anything contains a count of 6.022 x 1023.
Keep in mind that a mole is not a weight (mass), it is a count, the name
given to a number.
This number was also given a second name in honor of a famous scientist,
Avogadro. Thus, Avogadro’s number is 6.02214 x 1023. Avogadro’s
number is a mole’s worth of anything; it is still a count equal to 6.022 x 1023
of whatever is being counted.
Converting Counts to Weights (Mass)
The Atomic Mass Unit (amu):
Atoms are too small to be counted. Being able to convert a count (how
many) to a mass (how much) is a fundamental concept in chemistry and one
everyone must master.
Given that 12 g of 12C = 6.022 x 1023 atoms (exactly), one could calculate
the mass of a single atom:
12 g 12C
6.02214 x 10
atoms
• 1 atom of
= 1.992647 x 10-23 g / atom
23
12
C = 1.992647 x 10-23 g
3
12
C
To avoid working with such impossibly small numbers, a new unit was
devised for this number, the atomic mass unit, abbreviated amu.
The atomic mass unit (amu) was defined as one 12th the mass of a 12C
atom. Hence, by definition:
•
•
•
•
12 amu = the mass of a single atom of 12C
12 amu = the atomic mass of 12C
12 amu = 1.992647 x 10-23 g
1 amu = 1/12th the mass of a 12C atom = 1.992647 x 10-23 g / 12
= 1.6605391 x 10-24 g
Can we calculate the mass of a proton and neutron?
Note that 1 amu expressed in grams can be shown, as a first approximation,
to be essentially the mass of a single proton or neutron. For example:
• The mass of an atom is the sum of protons + neutrons
• Proton mass and neutron mass are essentially equal
• 12C has 6 protons plus 6 neutrons for a total of 12
• 12C has 12 P+N, and weighs 12 amu
Therefore:
• 1 proton = 1 amu
• 1 neutron = 1 amu
• (1.6605391 x 10-24 g/amu) x 12 amu/atom = 1.992647 x 10-23 g/atom
Note the use of the word “essentially” above. The mass of the proton,
neutron, and electron are actually known to a high degree of accuracy, and
should noted below that the mass of the proton and neutron are not exactly 1
amu, although they are extremely close. :
Particle
Mass, g
Mass, amu
-24
Proton
1.673 x 10
1.007276
-24
Neutron
1.675 x 10
1.008665
-24
Electron
0.0009 x 10
0.0005486
Thus, the mass of an atom calculated simply by adding up the actual mass of
number of protons and electrons turns out to be greater than what is actually
found.
Why is the actual atomic mass smaller that the sum of its parts?
4
This difference, albeit extremely small, is highly important in understanding
atoms. The mass difference is due to the nuclear binding energy required to
overcome the repellent forces of the positive charges (proton) which are
forced to come into close proximity in forming the nucleus. Where does this
energy come from?
Consider Einstein’s famous equation relating mass and energy:
E = mc2
Where E = energy; m = mass; c = a constant, the speed of light
Simply put, some of the particle mass is converted to nuclear energy, the
energy required to form a stable nucleus, and hold everything together. This
is the same nuclear energy which is released in nuclear reactions such as
nuclear fission when the nuclei of atoms are broken apart.
Note that this nuclear binding energy mass is not a fixed fraction of the
atom’s total mass; rather the nuclear binding energy is unique to each
element, and the amount of particle mass converted to energy is also unique.
Hence, the mass of each element is not a simple multiple compared to
each other element. The mass for each element must be experimentally
determined. Thus all elements have their atomic masses experimentally
determined by comparison to carbon – 12 using a technique called mass
spectrometry.
Atomic Mass
There is one additional complication to the number underneath the element’s
symbol for atomic mass. If you look at the number underneath the symbol
for carbon in the periodic table, that number is 12 (actually 12.01, but more
about that shortly). This number is called the atomic mass number or
shortened to atomic mass. It refers to the mass of a single atom of the
element in atomic mass units (amu).
For comparison, consider the element magnesium, specifically the isotope
magnesium – 24. We know that the mass number of 12 amu for the carbon
isotope carbon – 12 is the sum of the mass of the protons and neutrons.
Since magnesium – 24 has twice as many particles as carbon – 12, the mass
number for magnesium – 24 is 24 amu.
Likewise, the mass in amu (atomic mass units) of a single atom of any
element in the Periodic table is called the atomic mass of the element. That
5
number is the subscript number written underneath the symbol of each
element (see Table 8.1 below).
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8–4
How do we use this information?
Example:
Compute the number of moles and number of atoms in 10.0 g of Al
7.4 Molar Mass vs Atomic Mass
Now look again at the Periodic Table. What is the number underneath the
element Carbon? We said that that number (12) represents the mass (in
amu) of a single atom of carbon. That number can be referenced in a second
way to give it an additional, more practical value.
We know that by definition that a mole of 12C atoms weighs exactly 12
grams. Therefore, to measure out a mole’s worth of 12C atoms, you would
weigh out exactly 12.00 g of carbon-12. We refer to this number as the
molar mass, i.e. the mass of a mole of atoms.
To summarize:
• 1 atom of 12C = 12 amu
• 1 mole of 12C atoms = 12 g
• 12 g of 12C = 6.022 x 1023 atoms
6
• Atomic masses allow us to convert weights into numbers of atoms
Example:
If our sample of carbon weighs 3.00 x 1020 amu, we will have 2.50 x 1019
atoms of carbon:
1 C atom
= 2.50 x 1019 C atoms
12.01 amu
___________________________________
3.00 x 10 20 amu x
Look at other elements in the Periodic table. What do you see?
A mole is always the same number, but a mole of different substances will
always have different masses because of different numbers of protons and
neutrons.
Copyright © Houghton Mifflin Company.All rights reserved.
8–7
7
A $1000 in dollar bills and $1000 in dollar coins both are 1000 of each
(paper and coin), however, the mass of $1000 in bills compared to coins will
be dramatically different. It’s the same for a mole of different elements.
The number of protons + neutrons is different for each element, therefore,
the identical number of atoms of each element will have a different mass.
To summarize then, the number underneath the element in the periodic table
gives you the mass of a single atom in amu, or the mass of a mole’s worth of
atoms in grams. This number is known as both the atomic mass of an
element, and also the molar mass of an element.
Why aren’t the mass numbers whole numbers?
Because of isotopes (see chapter on atomic theory / nuclear structure).
Isotopes:
Consider: if one mole of 12C weighs exactly 12 g, why is the subscript
number 12.01 rather than exactly 12.00?
The answer is: Because the element in the Periodic table is a mixture of
the isotopes 12C, 13C, and 14C. The number 12.01 is a weighted average of
all three isotopes of Carbon. This is the number we will work with in most
chemistry problems.
Other examples: look at the subscript for nitrogen; look at hydrogen, look at
chlorine. The number under each element refers to the mass (in amu) of a
single atom of that element, or the mass (in grams) of a moles worth of
atoms.
Working with Moles:
How do we use this information? Converting between moles (how many)
and mass (how much).
Example:
Compute the number of moles and number of atoms in 10.0 g of Al
• Use the Periodic Table to determine the mass of 1 mole of Al
1 mole Al = 26.98 g
8
• Use this as a conversion factor for grams-to-moles
10.0
g Al x
1 mol
26.98
Al
g
= 0.371
mol
Al
• Use Avogadro’s Number to determine the number of atoms in 1 mole
1 mole Al = 6.02 x 1023 atoms
• Use this as a conversion factor for moles-to-atoms
6.02 x 10 23 atoms
0.371 mol Al x
= 2.23 x 10 23 Al atoms
1 mol Al
Problems: Mole Calculation Worksheet.doc , problems 1 – 8
7.5 Conversion Among Mass, Number of Moles, and Number
of Units
Molar Mass of Molecules
What is the molar mass of a molecule, e.g. methane (CH4)?
If :
• The mass of a mole of carbon atoms is 12.01 g.
• The mass of a mole of hydrogen atoms is 1.008 g.
9
Figure 8.3: Various numbers of methane
molecules showing their constituent atoms.
Copyright © Houghton Mifflin Company.All rights reserved.
8–11
The mass of a mole of CH4 (the molar mass) would be:
12.01 g. carbon + (1.008 g / H atom x 4 H atoms) = 16. 042 g / mole CH4
Therefore:
• 1 mole of CH4 molecules contains 1 mole of C (12.01g.) and 4 moles
of H atoms (4.032 g.)
• 1 mole of CH4 molecules weighs 16.042 g
We call this the molar mass – the mass of a mole’s worth of molecules.
This is also called the molecular weight.
Problems: Mole Calculation Worksheet.doc, problems 9 – 23
Problems: Molar Mass Calculations Worksheet.doc
Calculating Moles from Mass
Questions:
• What is the molar mass for C10H603?
10
• 1.56 g of C10H603 = ______________ moles of C10H603?
Problems: Grams Moles Worksheet.doc
Molar mass vs Molecular mass vs Formula Mass
Molecular Mass would be the mass of a single molecule of CH4 in amu.
Thus, the molecular mass would be the same number as the molar mass, but
the units are different: CH4 = 16.042 amu / molecule
Formula Mass
The molecular mass and formula mass mean exactly the same and are
sometimes used interchangeably for molecular compounds, e.g.
• the molecular formula or formula unit for methane = CH4
• the molecular mass = 16.042 amu / molecule
• the formula mass = 16.042 amu / formula unit
• the gram formula mass = 16.042 g / mole of formula units
However, for ionic compounds, only the term formula mass is employed,
e.g.
• the formula unit for sodium chloride is NaCl
• the formula mass for NaCl = 58.5 amu / formula unit
• the gram formula mass for NaCl = 58.5 g / mole of formula units
• the molar mass for NaCl = 58.5 g / mole of formula units
This is because ionic compounds are a collection of ions arranged in an
enormous array of alternating positive ions and negative ions. Ionic
compounds are represented by the lowest common denominator ratio of the
ions
Molar Mass
The molar mass (also called the molecular weight or gram formula weight)
is the mass in grams of a mole of the atoms or molecules.
• The molar mass for the nitrogen atom,
N = 14.0067 g
• The molar mass for nitrogen molecule,
N2 (g) = 28. 0134 g
• The molar mass for ammonium molecule, NH3(g) = 17.0304g
11
Questions:
• What is the formula mass for Na2SO4?
• 300.0 g Na2SO4 of = ____________ moles of Na2SO4 ?
Practice in calculating formula mass. See interactive tutorial:
http://chemistry.alanearhart.org/Tutorials/Stoichiometry/index.html
Calculating Molecules from Moles
Question: Isopentyl acetate formula = C7H12O2
• 1 g C7H12O2 = ______________ molecules of C7H12O2?
• 1 g C7H12O2 = ______________ moles of C7H12O2?
How do we use all of this?
Consider the following reaction:
2H2 (g)
+
O2 (g)
# of Molecules
Multiply by
Avogadro’s #
# of Moles
# of grams
12
→
2H2O (l)
Returning to the Haber-Bosch reaction:
3H2 + N2 →
2NH3
If we want to make 2 moles of NH3, how much H2 and N2 must we weigh
out? From the balanced equation, we see:
Need 3 moles of H2
• 1 atom of H = 1.008 amu
• 1 molecule of H2 = 2.016 amu
• 1 mol of H2 molecules = 2.016 g
• 3 mol of H2 molecules = 6.048 g
Need 1 mole of N2
• 1 atom of N = 14.01 amu
• 1 molecule of N2 = 28.16 amu
• 1 mol of N2 molecules = 28.16 g
How many grams of NH3 can you make?
How many molecules of NH3 is this many grams?
7.6 Percent Composition of Molecules
The mass of a mole of CH4 (the molar mass) is:
•
•
•
•
I mole of CH4 contains one mole of C and 4 moles of H
I mole of C atoms = 12.01 g
4 moles of H atoms = 4 x 1.008g/mole of H = 4.032 g.
Molar mass = 12.01g/mole + 4.032g/mole = 16.032 g/mole
The mass % of a given element = the mass of the element in 1 mole / molar
mass of the molecule
Therefore, the mass percent composition of this molecule is:
13
12.01 g. carbon + (1.008 g. / H atom x 4 H atoms) = 16. 042 g. / mole CH4
• % C = (12.01 g / 16.042 g) x 100 = 74.87 %
• % H = (4.032g / 16.042 g) x 100 = 25.13 %
7.7 Empirical Formulas and Molecular Formulas
In determining the molecular formula of compounds, three pieces of
information must be established:
• Which elements are present in a sample of the molecule
• The % composition of each element in the sample
• The mass of the sample analyzed
Problem:
• A sample of an unknown compound is weighed out = 0.2015 g
• Analysis reveals only C, H, O present
• The 0.2015 g sample was determined to contain 0.0806 g C, 0.01353
g H, 0.1074 g O
What is the molecular formula for this compound?
The molecular formula is the number of atoms (expressed as whole
numbers) of each element in a molecule. We begin with the empirical
formula or simplest ratio of the atoms in the molecule before considering
the molecular formula.
Given the grams of the atoms, we must first convert grams to moles, i.e. how
many of each:
Carbon = 0.0806 g x (1mole C / 12.01 g C) = 0.00671 mol C atoms
Hydrogen = 0.01353 g x (1 mole H / 1.008 g H) = 0.01342 mol H atoms
Oxygen = 0.1074 g x (1 mole O / 16.00 g O) = 0.006723 mol O atoms
Now we can calculate a simple whole number ratio of the elements relative
to each other. Taking 0.00671 as the lowest common denominator for these
three numbers, we find the following:
Carbon = 0.00671 mol C atoms / 0.00671 = 1.00
Hydrogen = 0.01342 mol H atoms / 0.00671 = 2.00
14
Oxygen =
0.006723 mol O atoms / 0.0067 = 1.00
Thus, these three elements are combined in a ratio of
• CH2O
• 1:2:1
CH2O represents the empirical formula, or lowest common denominator
ratio.
Empirical Formula Calculated from Percent Composition
Determine the empirical formula of acetic anhydride if its percent
composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen.
If given grams of each rather than % of each, one would proceed as above
by converting grams to moles.
Recall that the formula is a whole number ratio of the elements present, so
one must convert from % composition to numbers of atoms of each
element. This will require passing through grams in order to convert to
numbers of atoms.
The process is as follows:
• Assume you have 100 g of the compound
• Use the % to convert % composition to grams of each element
• Convert the grams of each element to moles
• Divide by the lowest common denominator to establish the whole
number ratio of atoms in the empirical formula
47% C x 100g = 47g
47% O x 100g = 47g
6.0% H x 100g = 6g
47g x (1 mol C / 12.01g) = 3.91 mol
47g x (1 mol O / 16.00g) = 2.94 mol
6g x (1 mol H / 1.008g) = 5.59 mol
Dividing by the lowest common denominator, 2.94, the empirical formula
would be: C1.3H2O.
Note however, that one cannot have fractional atoms. Therefore one must
find a factor to multiply these ratios to give whole numbers. Multiplying
the subscripts by 3 and rounding gives the empirical formula C4H6O3.
15
Problem:
The composition of an unknown compound is:
H
P
O
3.09%
31.60%
65.31%
• What is the empirical formula?
7.8 Calculation of Molecular Formulas
Molecular formula from empirical formula:
Numerous compounds have the same empirical formula, e.g.
CH2O
C2H4O2
C3H6O3
C6H12O6
All of these compounds have a CHO ratio of 1:2:1. Thus, CH2O
is the empirical formula, i.e. the lowest common denominator ratio.
The molecular formula will be a multiple of the empirical formula.
Extrapolating the empirical formula to the molecular formula requires one
more piece of information, namely the molar mass.
Consider: If the molar mass for the molecule in the previous problem was
determined to be 180 g, then we would identify this compound as C6H12O6.
This was determined as follows:
• CH2O = 30g / mol, the molar mass of the empirical formula
However, the mass of the actual compound is 180g.
• 180g /30g = 6
• Six times heavier gives us (CH2O)6 or the formula C6H12O6 .
Another example:
16
A compound (benzopyrene) with a molar mass of 252 g, was found to have
an empirical formula of C5H3.
• The formula mass of C5H3 = 63 g.
• 252 g / 63 g = 4.0
• (C5H3)4 = molar mass of 252 g
• Molecular formula = C20H12
Molecular formula from percent composition:
Problem:
The composition of an unknown compound is:
C
H
85.7%
14.3%
• What is the empirical formula?
• What is the formula mass?
The actual molar mass was determined to be 84g.
• What is the molecular formula?
17