1. No category

Plan, Ppt08(PS4)
(Many ideas from Tro, 2.9 & 3.7)
I. Subscripts can be used to make “conversion factors”—
between two different kinds of FU (atoms or ions)
II. A “mole” of a chemical substance is a certain (huge!)
number of FUs of it (called “Avogadro’s Number”).
III. Avogadro’s Number can be used as a “conversion
factor”--between FUs and moles [of FUs] of a given
substance
IV. The Ratio of FUs = the Ratio of Moles of FUs
→ Subscripts can be used to make “conversion factors”—
between moles of two different types of atoms or ions
V. The mass of one mole of a substance (molar mass) is
related to the mass of one FU of it.
VI. Molar mass can be used as a “conversion factor”-between grams (of a substance) and moles [of FUs] of it
Ppt08(PS4)
1
NOTE: The handout you have that looks like this will
be explained / used throughout this PowerPoint
grams
Molar mass of
AxBy
of AxBy
(in the sample)
Avogadro's #
moles
of AxBy
g of A x B y
(in the sample)
(atoms / ions),
Molar mass of A,
grams of B
6.022 x 10
23
things
Molar mass of B
(in the sample)
(g/mol)
of AxBy
(in the sample)
(Mole) Ratio Using
(Atom / Ion) Ratio Using
Subscripts
Subscripts
from the Chemical
Formula
from the Chemical Formula
moles of A
Avogadro's #
(atoms / ions),
(# of things/mol of things)
(atoms / ions)
(in the sample)
Ppt08(PS4)
atoms / ions of A,
atoms / ions of B
moles of B
(atoms / ions)
formula units
mole of things
mole of A x B y
grams of A
(# of things/mol of things)
6.022 x 10
23
things
(in the sample)
mole of things
2
I. Before we get to moles…subscripts can
be used to form “conversion factors”
• How many O atoms are in 125 FU of CO2?
125 FU CO 2 x
2 O atoms
 250 O atoms
1FU of CO 2
• How many Cs ions are in 3.4 x 1012 FU of
Cs3PO4?
3.4 x 1012
3 Cs ions
FU Cs3PO4 x
 10.2 x 1012 Cs ions
1 FU of Cs3PO4
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From handout sheet
formula units
of AxBy
(in the sample)
(Atom / Ion) Ratio Using
Subscripts
from the Chemical Formula
atoms / ions of A,
atoms / ions of B
(in the sample)
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Subscripts → “conversion factor” (cont.)
• How many O atoms are in a sample of
P4O10 containing 7.2 x 1012 atoms of P?
12
7.2 x 10
10 O atoms
P atoms x
 1.8 x 1013 O atoms
4 P atoms
→ I. Subscripts in a formula can be used to
create “conversion factors” (this for thats)
because they represent ratios of FUs
• Between two DIFFERENT kinds of FU
(here, O atoms and P atoms)
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II. Introduction to Moles
• Which sample has more “basic units” (formula
units) of the substance described:
Sample A: A million P4O10 molecules
Sample B: Two million Hg atoms
OR
Can answer without
any additional info
or calculation!
• What if I asked the same question for:
Sample A: A mole of P4O10 OR
Sample B: Two moles of Hg
Can (still) answer without
any additional info or
calculation! (Same “thing”)!
• What if I asked the same question for:
Sample A: One gram of P4O10 OR
Sample B: One gram of Hg atoms
Ppt08(PS4)
Can NOT answer without
additional info and
calculation!
6
II. Introduction to Moles;
Ratios of FUs are Important!
• The ratio of atoms in a compound is fixed; that
ratio is given by the subscripts in a formula
– In CO2, the number of atoms of O is always twice the
number of atoms of C
• The ratio of H2O molecules to anhydrous
compound FU’s in a hydrate is fixed.
– In CuSO45 H2O, the number of H2O molecules is
always 5 times the number of FU’s of CuSO4
• We will see that whenever chemical reaction
occurs, the ratio of FU’s of substances that
react is fixed as well (coefficients)!
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Moles (of things)
• 1 mole of “things” = 6.022 x 1023 things
(6.022 x 1023 is called “Avogadro’s #”)
– The “things” must be specified! They can be
•
•
•
•
•
Atoms
Molecules
Ions
Formula Units
Other things later (photons, electrons, bonds, etc.)
– The kind of thing must be specified!
•
•
•
•
O atoms
Au atoms
CO2 molecules
Na2S formula units
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Moles (of things) continued
• Dozen analogy
– 1 dozen O means _____________
“12 O atoms”
CO2 molecules”
– 1 dozen CO2 means “12
___________________
“12 Na2O formula units”
– 1 dozen Na2O means_______________________
• “One mole of:
“6.022 x 1023 O atoms” (in a sample)
– O” means__________________________________
“6.022 x 1023 CO2 molecules ( “ )
– CO2” means________________________________
“6.022 x 1023 Na2O FU’s ( “ )
– Na2O” means_______________________________
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Think of a “mole” as a
“moleion”! (analogous to “million”)
• One mole is not one “thing” it’s a moleion
(MOLE-yin) things!
– A “mole of C” is “a moleion C’s”
• i.e., ~6 x 1023 C atoms
– A “mole of CH4” is “a moleion CH4’s”
• i.e., ~6 x 1023 CH4 molecules
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Avogadro’s Number can be used as a
conversion factor (between FUs and moles [of FUs])
• 1 mole of XXX’s = 6.022 x 1023 XXX’s
• E.g.: 1 mole of C (atoms) = 6.022 x 1023 C atoms
• How many atoms are in (a sample of) 2.68 mol C?
# C atoms  2.68 mol C x
6.022 x 1023 C atoms
 1.613... x 1024 C atoms
1 mol C
• How many moles of C is 8.45 x 1020 atoms of C?
# mol C  8.45 x 1020 C atoms x
1 mol C
 0.001403...mol C
6.022 x 1023 C atoms
Ppt08(PS4)
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From Handout Sheet
moles of A
Avogadro's #
(atoms / ions),
(# of things/mol of things)
atoms / ions of B
moles of B
(atoms / ions)
(in the sample)
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atoms / ions of A,
6.022 x 10
23
things
(in the sample)
mole of things
12
Importance of Moles (Part I):
Any Ratio of Nanoscopic “Units” is equal to
the Ratio of the Moles of those Units
• In a sample of H2O:
– The ratio of H atoms : O atoms is always 2 : 1
– If there are 2.34 x 1019 atoms of O,
2 x (2.34 x 1019) atoms of H present also
• there must be __________
– If there are 6 x 1023 atoms of O,
• there must be 2 x (6 x 1023) atoms of H present also
– If there is one mole of O atoms,
2 moles of H atoms present also
• there must be ____
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Subscripts can be used to create
conversion factors between moles (of FUs)
• How many moles of P (atoms) are in (a sample
of) 5.43 moles of P4O10?
moles P (atoms)  5.43 moles P4O10
4 mol P (atoms)
 21.72
x
mol P4O10
 21.7 mol P (atoms)
Ppt08(PS4)
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From Handout Sheet
moles
of AxBy
(in the sample)
(Mole) Ratio Using
Subscripts
from the Chemical Formula
moles of A
(atoms / ions),
moles of B
(atoms / ions)
(in the sample)
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Using two conversion factors to get
from “moles of compound” to “# of
atoms”
• How many P (atoms) are in (a sample of) 5.43
moles of P4O10?
# P atoms = 5.43 mol P4O10
6.022 x 1023 P (atoms)
4 mol P (atoms)
x
x
1 mole P
mol P4O10
= 1.307... x 1025  1.31 x 1025 P atoms
Ppt08(PS4)
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From Handout Sheet
moles
of AxBy
(in the sample)
(Mole) Ratio Using
Subscripts
from the Chemical
Formula
moles of A
Avogadro's #
(atoms / ions),
(# of things/mol of things)
atoms / ions of B
moles of B
(atoms / ions)
(in the sample)
Ppt08(PS4)
atoms / ions of A,
6.022 x 10
23
things
(in the sample)
mole of things
17
Why is the unit of “mole” so
important? (Part II)
• A mole of any substance is a macroscopic
amount
– It can be seen, handled, and weighed.
• A formula unit of any substance is a nanoscopic
amount
– It cannot be seen, handled, or weighed
• Using moles is a way to scale up from the
nanoworld to the macroworld. For convenience!
– The scaling factor is Avogadro’s number!
– A mole of C is 6.022 x 1023 times as big a sample as
one atom of C.
• It will have 6.022 x 1023 times as much mass……
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Why is the number of things in a
mole 6.022 x 1023? (…the relation to mass)
• It’s simple: Because 1 g is 6.022 x 1023 times as
much mass as 1 amu!
• When the “mole” was defined, the actual value of
Avogadro’s number was not known!
– Avogadro’s number was defined to be the number of
atoms of 12C in exactly 12 grams of 12C.
– 1 atom of 12C has a mass of exactly 12 amu (definition)
– Determining Avogadro’s number was the same as
determining the mass of 1 amu, in grams!
• Strict analogy:
“a gram is to an amu, as a mole is to a formula unit”
Ppt08(PS4)
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“A gram is to an amu, as a mole is to
a formula unit”
• If one atom of H weighs 1 amu, then one mole of
H atoms will weigh 6.022 x 1023 amus, which
equals 1 gram.
• If one atom of He weighs 4 amus, then one mole
of He atoms will weigh 4 x (6.02 x 1023 amus)
which equals 4 grams.
• If one atom of “Y” weighs y amus, then one mole
of “Y” atoms will weigh y x (6.02 x 1023 amus)
which equals y grams. TOTALLY GENERAL!
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The upshot of this definition?
Molar mass is obtainable from Periodic Table
• The mass of one mole of any chemical species will
equal y grams if one formula unit weighs y amus:
# of amus
# of grams

1 FU
1 mole (of FUs)
• Use the number (for average atomic mass) from the
periodic table to “tell you” the number of grams in one
MOLE of atoms of that element (called “molar mass”).
o Ne: 20.18 amu/atom  20.18
____ g contains1 mol of Ne atoms
183.9 g contains1 mol of W atoms
o W: 183.9 amu/atom  _____
• Convenience
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“Proof” That “g/mol (of FUs)” =
“amu/FU”
12.01 amu 1.6606 x 10 -24 g
6.022 x 10 23 FU of C
12.01 g
x
x

1FU of C
1 amu
1mole C
1mol of C
This is numerically
multiplying by “1”
(if all sig figs were used for both qtys)
6.022 x 10 23 FU of H2 O
18.0 2 amu 1.6606 x 10 -24 g
18.0 2 g
x
x

1FU of H2O
1 amu
1 mole H2 O
1 mol of H2 O
Ppt08(PS4)
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Example
• If you have a sample of 5.6 g of Cu, how
many moles of Cu (“moleion” Cu atoms) will it
contain?
5.6 g Cu
 0.0881...  0.088 mol Cu
# mol Cu 
63.55 g Cu / mol Cu (atoms)
OR
5.6 g Cu x
1 mol Cu
63.55 g Cu (atoms)
• How many atoms will it contain?
1 mol Cu
6.022 x 1023 atoms
x

# Cu atoms  5.6 g Cu x
1 mol Cu
63.55 g Cu (atoms)
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From Handout Sheet
grams of A
(atoms / ions),
Molar mass of A,
grams of B
moles of A
Avogadro's #
(atoms / ions),
(# of things/mol of things)
atoms / ions of B
moles of B
(atoms / ions)
Molar mass of B
(in the sample)
(g/mol)
(atoms / ions)
(in the sample)
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atoms / ions of A,
6.022 x 10
23
things
(in the sample)
mole of things
24
Caution
“Macroscopic”
• An amu ≠ a gram !! “Nanoscopic”
amounts of
amounts of
matter
matter
• A FU ≠ a mole !!
• But…..
# amus
# grams

1 FU
1 mole (of FUs)
(Average) Atomic
Mass
Molar Mass
Both are “this for that’s”—same ratio
Ppt08(PS4)
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What is the molar mass of a
compound?
• Molar Mass of a monatomic element is
“directly” from Periodic Table (previous slide)
• Molar Mass of a compound?
– Sum the masses of the moles of atoms it
contains (use chemical formula).
– See next slide
Ppt08(PS4)
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Ppt08(PS4)
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Example: Application of Idea
• What is the formula (unit) mass in amu (i.e.,
mass of one molecule) of cholesterol if 0.5731 mol
weighs 221.6 g?
(no, you do not need the chemical formula of cholesterol here!)
Ppt08(PS4)
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Example: Using molar mass as a
“conversion factor”
• How many moles of Na2S are in 24.3 g?
• 1 mol Na2S has a mass of:
2 x 22.99 + 1 x 32.07 = 78.05 g
(1 FU Na2S has a(n) avg mass of 2 x 22.99 + 1 x 32.07 = 78.05 amu)
1 mol Na 2 S
 0.3113 mol Na 2S
24.3 g Na 2S x
78.05 g Na 2 S
• Makes sense: If one mole of Na2S has a
mass of ~78 g, then a sample with 24.3 g is
less than a mole (about a third of a mole)
Ppt08(PS4)
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grams
Molar mass of
AxBy
of AxBy
(in the sample)
moles
of AxBy
g of A x B y
(in the sample)
mole of A x B y
Ppt08(PS4)
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Examples: Using subscripts and
Avogadro’s number
Ppt08(PS4)
31
moles
of AxBy
(in the sample)
(Mole) Ratio Using
Subscripts
from the Chemical
Formula
moles of A
Avogadro's #
(atoms / ions),
(# of things/mol of things)
atoms / ions of B
moles of B
(atoms / ions)
(in the sample)
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atoms / ions of A,
6.022 x 10
23
things
(in the sample)
mole of things
32
Relationships between “amounts” of a
substance AxBy (and component atoms / ions)
grams
Molar mass of
AxBy
of AxBy
(in the sample)
Avogadro's #
moles
of AxBy
g of A x B y
(in the sample)
(atoms / ions),
Molar mass of A,
grams of B
6.022 x 10
23
things
Molar mass of B
(in the sample)
(g/mol)
of AxBy
(in the sample)
(Mole) Ratio Using
(Atom / Ion) Ratio Using
Subscripts
Subscripts
from the Chemical
Formula
from the Chemical Formula
moles of A
Avogadro's #
(atoms / ions),
(# of things/mol of things)
(atoms / ions)
(in the sample)
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atoms / ions of A,
atoms / ions of B
moles of B
(atoms / ions)
formula units
mole of things
mole of A x B y
grams of A
(# of things/mol of things)
6.022 x 10
23
things
(in the sample)
mole of things
33
PS Sign-Posting
• The concepts and skills related to
problems 1-13 and problem 21 on
PS4 have been covered in this
PowerPoint. Give them a try now!
Ppt08(PS4)
34
If time…
…you can now redo the “#FU’s of H2O in
150. g” question (from Ppt07) the
“traditional” way (g → mol → FU) instead of the
more “obscure” way (g → amu → FU)
Chemists use moles because we often do not need to know the
actual number of FU’s of substances, just the ratio of them. And
“grams” is a much more “convenient” mass unit than “amu”.
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