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Chapter 3: CHEMICAL REACTIONS AND EARTH’S COMPOSITION
(Topics to Review)
3.2 The Mole
Avogadro’s Number (NA) = 6.022×1023 (to 4 sig figs)
How big is this?
• If 6.022×1023 hydrogen atoms were laid side by side, the total length would encircle the
earth about a million times.
• The mass of 6.022×1023 Olympic shotput balls is about equal to the mass of the Earth.
• The volume of 6.022×1023 softballs is about equal to the volume of the Earth.
1 mole (abbreviated mol) = 6.022×1023 entities
Similar to: 1 dozen = 12 entities:
1 dozen doughnuts = 12 doughnuts
1 mole of doughnuts = 6.022×1023 doughnuts
Ex. 1 How many eggs are in 3 dozen eggs?
3 x 12 eggs = 36 eggs
Ex. 2 How many eggs are in 3 moles of eggs?
3 x 6.022x1023 eggs = 1.807x1024 eggs
Ex. 3 How many moles of carbon contain 7.25×1024 carbon atoms?
7.25x1024 C atoms ×
1 mol C
= 12.0 mol C
6.022 × 10 23 C atoms
Atomic weights and molar masses:
– The mass of 1 C atom (on average) is 12.01 amu
– The mass of 1 mole of C atoms is 12.01 g (or 12.01 g/mol)
→ 1 mole (6.022×1023) is the amount of atoms of any element that has a mass in grams equal
to the mass of ONE atom in amu.
→ The atomic masses reported in the Periodic Table give the atomic weight (or
molecular/formula weight for compounds) in amu and the molar mass in g/mol.
CHEM161: Chapter 3 – Independent Review
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MOLAR MASS (MM): Mass in grams of 1 mole of any element/compound
– To obtain, multiply the molar mass of each element by the number of each present, then add
up all the constituent parts.
Example: Determine the molar mass of each of the following compounds:
a. O2: 2 (molar mass of O) = 2 (16.00 g/mol) = 32.00 g/mol
b. H3PO4:
H: 3(1.008)
P: 30.97
O: 6(16.00)
c. Al2(SO4)3:
97.994 g/mol
Al: 2(26.98)
S: 3(32.07)
O: 12(16.00)
310.18 g/mol
Note: Don’t worry about rounding molar masses since they rarely limit the number of sig
figs for your final calculated answer.
Mole Calculations
Ex. 1 How many moles of Ne are in 0.500 g Ne?
0.500 g Ne ×
1 mol Ne
= 0.0248 mol Ne
20.18 g Ne
Ex. 2 How many Ne atoms are in 0.500 g of Ne?
0.500 g Ne ×
6.022 × 10 23 Ne atoms
1 mol Ne
×
= 1.49x1022 Ne atoms
20.18 g Ne
1 mol Ne
Ex. 3 How many moles of CO2 are in 0.500 g of CO2?
0.500 g CO2 ×
1 mol CO2
= 0.0114 mol CO2
44.01 g CO2
Ex. 4 How many CO2 molecules are in 0.500 g of CO2?
0.500 g CO2 ×
6.022 × 10 23 CO2 molecules
1 mol CO2
×
= 6.84x1021 CO2 molecules
44.01 g CO2
1 mol CO2
Ex. 5 How many oxygen atoms are in 0.500 g of CO2?
0.500 g CO2 ×
1 mol CO2
6.022 × 10 23 CO2 molecules
2 O atoms
×
= 1.37x1022 O atoms
×
44.01 g CO2
1 mol CO2
CO2 molecule
CHEM161: Chapter 3 – Independent Review
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