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Chapter 6
Section 6.1
• We can count these jelly beans by weighing.
• But in order to do this we need the average
mass of the jelly beans.
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘€π‘Žπ‘ π‘  π‘œπ‘“ π΅π‘’π‘Žπ‘›π‘ 
π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘€π‘Žπ‘ π‘  =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π΅π‘’π‘Žπ‘›π‘ 
• 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.
𝐢 𝑠 + 𝑂2 𝑔 → 𝐢𝑂2 (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 π‘₯ 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 π‘Žπ‘šπ‘’.
• 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 π‘π‘Žπ‘Ÿπ‘π‘œπ‘› π‘Žπ‘‘π‘œπ‘š
12.01 π‘Žπ‘šπ‘’
1
π‘π‘Žπ‘Ÿπ‘π‘œπ‘›
π‘Žπ‘‘π‘œπ‘š
3.00 π‘₯ 1020 π‘Žπ‘šπ‘’ ×
12.01 π‘Žπ‘šπ‘’
= 2.50 × 1019 π‘π‘Žπ‘Ÿπ‘π‘œπ‘› π‘Žπ‘‘π‘œπ‘šπ‘ 
• 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
• How many moles are in 0.500g of H?
• We can carry out the calculation by using the equivalence
statement
• 1 mol H = 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 π‘šπ‘œπ‘™ 𝐻
0.500 𝑔 𝐻 ×
= 0.496 π‘šπ‘œπ‘™ 𝐻 𝑖𝑛 π‘ π‘Žπ‘šπ‘π‘™π‘’ 𝐡
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. πΆπ‘ŽπΆπ‘‚3
2. NaCl
3. 𝐢10 𝐻6 𝑂3
1. 9.95 g πΆπ‘ŽπΆπ‘‚3
2. 15.5 g NaCl
3. 1.56 g 𝐢10 𝐻6 𝑂3
1. 𝑁𝐻3
2. 𝑁2 𝐻4
3. (𝑁𝐻4 )2 πΆπ‘Ÿ2 𝑂7
4. π‘π‘Ž2 𝐢𝑂3
5. πΆπ‘Ž3 (𝑃𝑂4 )2
6. π‘π‘Ž2 𝐻𝑃𝑂4
• We use Avogadro's number to convert between moles of one
substance atoms.
1 π‘šπ‘œπ‘™π‘’
6.022 π‘₯1023
or
6.022 π‘₯ 1023
1 π‘šπ‘œπ‘™π‘’
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
π‘€π‘Žπ‘ π‘  π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› π‘“π‘œπ‘Ÿ π‘Ž 𝑔𝑖𝑣𝑒𝑛 π‘’π‘™π‘’π‘šπ‘’π‘›π‘‘
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘™π‘’π‘šπ‘’π‘›π‘‘ π‘π‘Ÿπ‘’π‘ π‘’π‘›π‘‘ 𝑖𝑛 1 π‘šπ‘œπ‘™ π‘œπ‘“ π‘π‘œπ‘šπ‘π‘œπ‘’π‘›π‘‘
=
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ 1 π‘šπ‘œπ‘™ π‘œπ‘“ π‘Ž π‘π‘œπ‘šπ‘π‘œπ‘’π‘›π‘‘
Then we multiple this number by 100 to turn it into a percent.
• π‘€π‘Žπ‘ π‘  π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘ π‘œπ‘“ 𝐢 =
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ 𝐢 𝑖𝑛 1 π‘šπ‘œπ‘™ π‘œπ‘“ 𝐢2 𝐻5 𝑂𝐻
π‘šπ‘Žπ‘ π‘  π‘œπ‘“ π‘œπ‘›π‘’ π‘šπ‘œπ‘™πΆ2 𝐻5 𝑂𝐻
• π‘€π‘Žπ‘ π‘  π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘ π‘œπ‘“πΆ =
24.02𝑔
46.07 𝑔
• Mass Percent of C = 13.13 %
× 100%
× 100%
Using the same compound as the previous example calculate
1. The mass percent of O
2. The mass percent of H
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