Chapter 6
Section 6.1
• Look at the example of counting jelly jeans on
p. 172
• We can count these jelly beans by weighing.
• But in order to do this we need the average
mass of the jelly beans.
𝑇𝑜𝑡𝑎𝑙 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐵𝑒𝑎𝑛𝑠
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑀𝑎𝑠𝑠 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐵𝑒𝑎𝑛𝑠
• Objects do not need to have identical masses to
be counted by weighing. We simply need to
know the average mass of the objects.
• For the purpose of counting, the objects behave as though they
were all identical, as though they each actually had the
average mass.
• Look at the example on p. 173
• Two samples containing different types of
components, A and B, both contain the same
number of components if the ratio of the
sample masses is the same as the ratio of the
masses of the individual components of A and B
• Suppose you have a small pile of solid carbon and
want to know how many oxygen molecules are
required to convert all of this carbon dioxide. The
balance equation tells us that one oxygen molecule is
required for each carbon atom.
𝐶 𝑠 + 𝑂2 𝑔 → 𝐶𝑂2 (𝑔)
• To determine the number of oxygen molecules
required, we must know how many carbon
atoms are present in the pile of carbon.
• Since individual atoms are far too small to
count, we must count atoms by weighing
samples containing large numbers of them.
• Atomic Mass Unit (amu) – A small unit of mass
equal to 1.66 𝑥 10−24 grams
• Average Atomic Mass – The weighted
average of the masses of all the isotopes of an
element.
• We can find the average atomic mass of an element
on the periodic table of elements.
• Now lets assume that when we weigh the pile of
natural carbon mentioned earlier, the result is
3.00 × 1020 𝑎𝑚𝑢.
• We know the average carbon atom has a mass
of 12.01 amu, so we can calculate the number
of carbon atoms by using the equivalence
statement
• 1 carbon atom = 12.01 amu
• Or
1 𝑐𝑎𝑟𝑏𝑜𝑛 𝑎𝑡𝑜𝑚
12.01 𝑎𝑚𝑢
1
𝑐𝑎𝑟𝑏𝑜𝑛
𝑎𝑡𝑜𝑚
3.00 𝑥 1020 𝑎𝑚𝑢 ×
12.01 𝑎𝑚𝑢
= 2.50 × 1019 𝑐𝑎𝑟𝑏𝑜𝑛 𝑎𝑡𝑜𝑚𝑠
• We have learned how to use atomic mass units
for mass, because atoms are so tiny. When
working in a lab we use a larger unit, gram.
The gram is the convenient unit for mass.
• Now we will move on to using the moles
• The number equal to the number of carbon
atoms in exactly 12 grams of pure carbon.
• The mole is abbreviated – mol
• Is referred to as Avogardo’s Number =
6.022 × 1023
• Look at p. 181
• We can carry out the calculation by using the equivalence
statement
• 1 mol H atoms = 1.008 g H
• To find the number of moles in sample B. We need to know the
following information
• 1 mol H = 1.008 g H
• The original weight of sample B is 0.500g H
• Once we have that information we can get up the equivalence
statements, exactly like we did with converting units between
English and Metric.
1 𝑚𝑜𝑙 𝐻
0.500 𝑔 𝐻 ×
= 0.496 𝑚𝑜𝑙 𝐻 𝑖𝑛 𝑠𝑎𝑚𝑝𝑙𝑒 𝐵
1.008𝑔 𝐻
1. Calculate the number of moles in 5.67g of Al
2. Calculate the number of moles in 23.6g of Cl
3. Calculate the number of moles in 17.6 g of
Ca
Section 6.2
• The molar mass is obtained by summing the
masses of the component atoms.
• Example: What is the molar mass of 𝑆𝑂2 ?
• Sulfur’s mass = 32.07g
• Oxygen’s mass = 16.00 g
• 1 × 32.07 + 2 × 16.00 = 64.07
• The Molar Mass of 𝑆𝑂2 is 64.07 g
1. 𝐶𝑎𝐶𝑂3
2. NaCl
3. 𝐶10 𝐻6 𝑂3
1. 9.95 g 𝐶𝑎𝐶𝑂3
2. 15.5 g NaCl
3. 1.56 g 𝐶10 𝐻6 𝑂3
1. 𝑁𝐻3
2. 𝑁2 𝐻4
3. (𝑁𝐻4 )2 𝐶𝑟2 𝑂7
4. 𝑁𝑎2 𝐶𝑂3
5. 𝐶𝑎3 (𝑃𝑂4 )2
6. 𝑁𝑎2 𝐻𝑃𝑂4
• We use Avogadro's number to convert between moles of one
substance atoms.
1 𝑚𝑜𝑙𝑒
6.022 𝑥1023
or
6.022 𝑥 1023
1 𝑚𝑜𝑙𝑒
1. 5.56g Ba(NO3)2 to
atoms
2. 2.3 atoms Fe to moles
3. 3.7 atoms Cu2SO4 to
grams
4. 0.00473 mol KCl to
atoms
5. 27.2g H3PO4 to atoms
6. 1.343g Pb(CN)2 to
atoms
𝑀𝑎𝑠𝑠 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑖𝑛 1 𝑚𝑜𝑙 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
=
𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑚𝑜𝑙 𝑜𝑓 𝑎 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
Then we multiply this number by 100 to turn it into a percent.
• 𝑀𝑎𝑠𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓 𝐶 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶 𝑖𝑛 1 𝑚𝑜𝑙 𝑜𝑓 𝐶2 𝐻5 𝑂𝐻
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑜𝑛𝑒 𝑚𝑜𝑙𝐶2 𝐻5 𝑂𝐻
• 𝑀𝑎𝑠𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑜𝑓𝐶 =
24.02𝑔
46.07 𝑔
• Mass Percent of C = 52 %
× 100%
× 100%
Using the same compound as the previous example calculate
1. The mass percent of O
2. The mass percent of H
• The empirical formula gives the ratio of the number of atoms of
each element in a compound.
• These ratios must be expressed as whole numbers.
Compound
Formula
Hydrogen peroxide H2O2
Benzene
C6H6
Ethylene
C2H4
Propane
C3H8
28
Empirical Formula
OH
CH
CH2
C3H8
• A compound’s empirical formula can
be determined from its percent
composition.
• A compound’s molecular formula is
determined from the molar mass and
empirical formula.
29
• A compound was analyzed to be 82.67% carbon and 17.33%
hydrogen by mass. What is the empirical formula for the
compound?
• Assume 100 g of sample, then 82.67 g are C and 17.33 g are H
30
• Convert masses to moles:
82.67 g C x mole/12.011 g = 6.88 moles C
17.33 g H x mole/1.008 g = 17.19 mole H
Find relative # of moles (divide by smallest number)
31
Convert moles to ratios:
6.88/6.88 = 1 C
17.19/6.88 = 2.50 H
Or 2 carbons for every 5 hydrogens
C2 H5
32
Empirical Formula is: C2H5
Formula weight is: 29.06 g/mole
If the molecular weight is known to be 58.12 g/mole
Then the molecular formula must be:
C4H10
33
• In order to obtain the molecular formula we must know the
molar mass and the empirical formula.
• We compare the molar mass and the empirical formula using
the following equation
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑛=
𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠
**n is the smallest whole number**
• A molecular formula is always a whole number multiple of the
empirical formula.
• A white powder is analyzed and found to have an empirical
formula of 𝑃2 𝑂5 . The compound has a molar mass of
283.88g/mol. What is the compound’s molecular formula?
1.
Calculate the empirical formula mass
• P (2 x 30.97) = 61.94g
• O (5 x 16.00) = 80.00g
• Total = 141.94g
2. Divide the molar mass by the empirical formula mass, to get the whole
number multiple of the empirical formula.
• 𝑛=
• 𝑛=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠
283.88𝑔
141.94𝑔
=2
• The molecular formula is therefore (𝑃2 𝑂5 )2 or 𝑃4 𝑂10