Plan, Ppt08(PS4) (Many ideas from Tro, 2.9 & 3.7)

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

1Ppt08(PS4)

NOTE: The handout you have that looks like this will be explained / used throughout this PowerPoint

grams

of AxBy

(in the sample)

Molar mass of AxBy Avogadro's # (# of things/mol of things)

x y

x y

g of A B

mole of A B

236.022 x 10 things

mole of things

formula units

of AxBy

(in the sample)

moles

of AxBy

(in the sample)

Molar mass of A,

Molar mass of B

(g/mol)

moles of A (atoms / ions),

moles of B (atoms / ions)

(in the sample)

atoms / ions of A,

atoms / ions of B

(in the sample)

grams of A (atoms / ions),

grams of B (atoms / ions)

(in the sample)

Avogadro's # (# of things/mol of things)

236.022 x 10 things

mole of things

(Mole) Ratio Using

Subscripts

from the Chemical Formula

(Atom / Ion) Ratio Using

Subscripts


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I. Before we get to moles…subscripts can be used to form “conversion factors”

• How many O atoms are in 125 FU of CO2?

• How many Cs ions are in 3.4 x 1012 FU of Cs3PO4?

atoms O 250 CO of FU 1

atoms O 2 x CO FU 125

22

123 43.4 x 10 FU Cs PO x Cs ions

3 4

3 Cs ions

1 FU of Cs PO

1210.2 x 10

3Ppt08(PS4)

From handout sheet

formula units

of AxBy

(in the sample)

atoms / ions of A,

atoms / ions of B

(in the sample)


Subscripts


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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?

→ 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)

127.2 x 10 P atoms x O atoms10 O atoms

4 P atoms131.8 x 10

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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 ORSample B: Two million Hg atoms

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• What if I asked the same question for:

Sample A: A mole of P4O10 ORSample B: Two moles of Hg

• What if I asked the same question for:

Sample A: One gram of P4O10 ORSample B: One gram of Hg atoms

Can answer without any additional info

or calculation!

Can (still) answer without any additional info or

calculation! (Same “thing”)!

Can NOT answer without additional info and

calculation!

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 CuSO45 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 _____________

– 1 dozen CO2 means ___________________

– 1 dozen Na2O means_______________________

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• “One mole of:– O” means__________________________________

– CO2” means________________________________

– Na2O” means_______________________________

“12 O atoms”“12 CO2 molecules”

“12 Na2O formula units”

“6.022 x 1023 O atoms” (in a sample)

“6.022 x 1023 CO2 molecules ( “ )

“6.022 x 1023 Na2O FU’s ( “ )

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

# C atoms 2.68 mol C x

• How many atoms are in (a sample of) 2.68 mol C?23 6.022 x 10 C atoms

1 mol C

• How many moles of C is 8.45 x 1020 atoms of C?

20# mol C 8.45 x 10 C atoms x 23

1 mol C

6.022 x 10 C atoms

24 C atoms1.613... x 10

0.001403...mol C

• E.g.: 1 mole of C (atoms) = 6.022 x 1023 C atoms

11Ppt08(PS4)

From Handout Sheet



(in the sample)

atoms / ions of A,

atoms / ions of B

(in the sample)


236.022 x 10 things

mole of things

12Ppt08(PS4)

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,• there must be __________ atoms of H present also

– 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, • there must be ____ moles of H atoms present also

2 x (2.34 x 1019)

2

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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?

(atoms) 4 10moles P 5.43 moles P O x (atoms)

4 10

4 mol P

mol P O21.72

(atoms) 21.7 mol P

14Ppt08(PS4)

From Handout Sheet

moles

of AxBy

(in the sample)



(in the sample)

(Mole) Ratio Using

Subscripts


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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?

4 10# P atoms = 5.43 mol P O x (atoms)

4 10

4 mol P

mol P O

25 25= 1.307... x 10 1.31 x 10 P atoms

23(atoms) 6.022 x 10 P

x 1 mole P

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From Handout Sheet

moles

of AxBy

(in the sample)



(in the sample)

atoms / ions of A,

atoms / ions of B

(in the sample)


236.022 x 10 things

mole of things

(Mole) Ratio Using

Subscripts


17Ppt08(PS4)

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”

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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:

• 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”).

# of amus # of grams

1 FU 1 mole (of FUs)

• Convenience

o Ne: 20.18 amu/atom ____ g contains1 mol of Ne atoms

o W: 183.9 amu/atom _____ g contains1 mol of W atoms

20.18

183.9

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“Proof” That “g/mol (of FUs)” = “amu/FU”

C of mol 1

g 112.0

C mole 1

C of FU 10 x 6.022 x

amu 1

g 10 x 1.6606 x

C of FU 1

amu 112.0 23-24

OH of mol 1

g 218.0

OH mole 1

OH of FU 10 x 6.022 x

amu 1

g 10 x 1.6606 x

OH of FU 1

amu 218.0

22

223-24

2

This is numerically multiplying by “1”

(if all sig figs were used for both qtys)

22Ppt08(PS4)

Example

• If you have a sample of 5.6 g of Cu, how many moles of Cu (“moleion” Cu atoms) will it contain?

• How many atoms will it contain?

(atoms)

5.6 g Cu# mol Cu

63.55 g Cu / mol Cu 0.0881... 0.088 mol Cu

(atoms)

1 mol Cu

63.55 g Cu 5.6 g Cu x OR

# Cu atoms (atoms)

1 mol Cux

63.55 g Cu 5.6 g Cu x

23 6.022 x 10 atoms

1 mol Cu

23Ppt08(PS4)

From Handout Sheet

Molar mass of A,

Molar mass of B

(g/mol)



(in the sample)

atoms / ions of A,

atoms / ions of B

(in the sample)



(in the sample)


236.022 x 10 things

mole of things

24Ppt08(PS4)

Caution

• An amu ≠ a gram !! “Nanoscopic” amounts of

matter• A FU ≠ a mole !!

• But…..# amus # grams

1 FU 1 mole (of FUs)

“Macroscopic” amounts of

matter

(Average) Atomic Mass

Both are “this for that’s”—same ratio

Molar Mass

25Ppt08(PS4)

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

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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!)

28Ppt08(PS4)

Example: Using molar mass as a “conversion factor”

• How many moles of Na2S are in 24.3 g?

SNa g 24.3 2 SNa g 78.05

SNa mol 1x

2

2 SNa mol 310.31 2

• 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)

• 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)

29Ppt08(PS4)

grams

of AxBy

(in the sample)

Molar mass of AxBy

x y

x y

g of A B

mole of A B

moles

of AxBy

(in the sample)

30Ppt08(PS4)

Examples: Using subscripts and Avogadro’s number

31Ppt08(PS4)

moles

of AxBy

(in the sample)



(in the sample)

atoms / ions of A,

atoms / ions of B

(in the sample)


236.022 x 10 things

mole of things

(Mole) Ratio Using

Subscripts


32Ppt08(PS4)

Relationships between “amounts” of a substance AxBy (and component atoms / ions)

grams

of AxBy

(in the sample)

Molar mass of AxBy Avogadro's # (# of things/mol of things)

x y

x y

g of A B

mole of A B

236.022 x 10 things

mole of things

formula units

of AxBy

(in the sample)

moles

of AxBy

(in the sample)

Molar mass of A,

Molar mass of B

(g/mol)



(in the sample)

atoms / ions of A,

atoms / ions of B

(in the sample)



(in the sample)


236.022 x 10 things

mole of things

(Mole) Ratio Using

Subscripts



Subscripts


33Ppt08(PS4)

PS Sign-Posting

• The concepts and skills related toproblems 1-12 on PS4 have been covered in this PowerPoint. Give them a try now!

34Ppt08(PS4)

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”.

35Ppt08(PS4)

Documents

Plan, Ppt08(PS4) (Many ideas from Tro, 2.9 & 3.7)