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Page 1: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

The Mole: Avogadro’s number

Page 2: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

How much is:

n  A dozen? n  12 n  A century? n  100 n  A mole? n  6.02 x 1023

(602,000,000,000,000,000,000,000)

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Can you count a mole of pennies?

n  If you could count 5 per second, it would take you

n  6.02 x 10 23 ÷ 5 pennies/second ÷ 60 sec/min ÷ 60 min/hr ÷ 24hrs/day ÷ 365 days/yr =

n  3,800,000,000,000,000 years!

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Can you spend a mole of dollars?

n  If you could spend $1,000,000 every second it would take you

n  6.02 x 10 23 ÷ $1,000,000/sec ÷ 60 sec/min ÷ 60 min/hr ÷ 24hrs/day ÷ 365 days/yr =

n  19,000,000,000 years!

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The Mole Map

Mole

Mass

Gas Volume @ STP

# Particles

x

x

x ÷

÷

÷

Mol

ar

Mas

s

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When measuring amounts, you can count or you can mass them.

If I want 2 dozen baseballs, I can count 24 baseballs Or I can mass 16 kg of baseballs.

Page 7: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

How many tennis balls are in 6 kg?

( 2 dozen)

Page 8: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

How many tennis balls are in a mole?

Page 9: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Since we can’t count a mole of atoms, we MUST mass chemicals to measure moles

6.02 x 10 23 atoms of sulfur 32.07 grams of sulfur

6.02 x 10 23 atoms of carbon 12.01 grams of carbon

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How do we measure moles?

n  mole = number of particles equal to the number of atoms in 12 g of C-12 n  1 atom of C-12 weighs exactly 12 amu n  1 mole of C-12 weighs exactly 12 g

n  The number of particles in 1 mole is called Avogadro’s Number = 6.0221421 x 1023 n  1 mole of C atoms weighs 12.01 g and has

6.022 x 1023 atoms n  the average mass of a C atom is 12.01 amu

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How do we measure moles?

n  The atomic mass on your periodic table is the mass of a mole of atoms of that element.

n  What is the mass of a mole of copper atoms?

n  63.55 g n  So, to count 6.02 x 1023 copper atoms,

we mass out 63.55 g on the scale.

Page 12: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Mole and Mass Relationships

1 mole Sulfur 32.06 g

1 mole Carbon 12.01 g

Substance Pieces in 1 mole Weight of 1 mole hydrogen 6.022 x 1023 atoms 1.008 g

carbon 6.022 x 1023 atoms 12.01 g

oxygen 6.022 x 1023 atoms 16.00 g

sulfur 6.022 x 1023 atoms 32.06 g

calcium 6.022 x 1023 atoms 40.08 g

chlorine 6.022 x 1023 atoms 35.45 g

copper 6.022 x 1023 atoms 63.55 g

Page 13: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Find the mass of: n  A mole of silicon atoms n  28.09 g n  6.02 x 1023 atoms of nitrogen n  14.01 g n  6.02 x 1023 atoms of sodium n  22.99g n  2 moles of sodium atoms n  45.98 g

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How many atoms are in: n  A mole of silicon n  6.02 x 1023 n  14.01 g of nitrogen n  6.02 x 1023 n  2 moles of sodium n  12.04 x 1023 n  45.98 g of sodium n  12.04 x 1023

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How many things are in: n  A mole of footballs n  6.02 x 1023 footballs n  A mole of water n  6.02 x 1023 molecules n  2 moles of pencils n  12.04 x 1023 n  ½ mole of lead n  3.01 x 1023 atoms

Page 16: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Molar mass is the mass of one mole of a substance

n  To find the molar mass of an element, look on the periodic table.

n  To find the molar mass of a compound, add all the masses of its elements

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Chemical Formulas as Conversion Factors

n  1 spider ≡ 8 legs n  1 chair ≡ 4 legs n  1 H2O molecule ≡ 2 H atoms ≡ 1 O

atom

Page 18: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Molar Mass of Compounds •  the relative weights of molecules can be

calculated from atomic weights Formula Mass = 1 molecule of H2O

= 2(1.0 amu H) + 16.0 amu O = 18.0 amu •  since 1 mole of H2O contains 2 moles of H

and 1 mole of O Molar Mass = 1 mole H2O

= 2(1.01 g H) + 16.00 g O = 18.02 g

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Find the molar mass of:

n  Ammonium phosphate n  NH4

+ PO43-

n  (NH4)3PO4

n  =42.03 +12.12 +30.97 +64.00 n  =149.12g/mol n  Carbon dioxide n  CO2 n  = 12.01 + 32.00 n  = 44.01 g/mol

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Find the molar mass of: n  Hydrogen gas

n  H2

n  = 2.02 g/mol

n  Elemental hydrogen n  H n  = 1.01 g/mol

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Find the molar mass of: n  Iron

n  Fe n  = 55.85 g/mol

n  Iron (III) hydroxide n  Fe3+ OH-

n  Fe(OH)3 n  = 55.85 + 48.00 + 3.03 n  =106.88 g/mol

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Converting to and from moles.

n  Converting between moles and mass requires the molar mass of the substance from the periodic table.

n  Element: Ag = 107.97g/mol n  Ionic compound: CaCl2 = 110.98 g/mol n  Covalent compound: NO2 = 46.01 g/mol n  Always keep at least two decimal places on

all values taken from the periodic table.

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Converting to and from moles.

n  To convert from moles to grams, multiply by molar mass:

n  0.500 mol H2O x (18.02g/mol) = 9.01g H2O

n  To convert from grams to moles, divide by molar mass:

n  54g H2O x (1mol/18.02g) = 3.0 mol H2O

Page 24: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Converting to and from moles.

n  For gases, use the fact that at STP, 1 mol of any gas has a volume of 22.4 Liters.

n  What is STP? Standard Temperature and Pressure

n  Standard Temperature = 273K or 0.0°C n  Standard Pressure = 1 atmosphere = 760

mm Hg (barometric) = 101.325 kPa.

Page 25: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Converting to and from moles.

n  To go from moles to volume, multiply by 22.4L.

n  3.00 mol x (22.4L/mol) = 67.2L of gas

n  To go from volume to moles, divide by 22.4L

n  44.8L x (1mol/22.4L) = 2.00 moles of gas

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Converting to and from moles.

n  To convert between moles and particles, simply multiply or divide by Avogadro’s number.

n  2.0 mol x (6.02 x 1023 particles/mol) = 1.2 x 1024 particles

n  3.1 x 1024 particles x (1 mol/ 6.02 x 1023 particles) = 5.0 mol

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Remember unit factors?

Page 28: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Converting to and from moles.

n  A convenient tool for making these conversions is called a “mole map.”

n  With the mole at the center, we can put all of the aforementioned calculations together into one simple picture.

Page 29: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

The Mole Map

Mole

Mass

Gas Volume @ STP

# Particles

x

x

x ÷

÷

÷

Mol

ar

Mas

s

Page 30: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Percent Composition

n  Percentage of each element in a compound n  By mass

n  Can be determined from 1.  the formula of the compound 2.  the experimental mass analysis of the compound 3.  the total mass of each element n  The percentages may not always total to 100% due to

rounding

Percentage=partwhole

×100%

Page 31: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

What percentage of water is Oxygen?

1.  Formula of the compound * H2O

2.  Mass of the compound * 2 (1.01 g/mol) + 16.00 g/mol = 18.02 g/mol

3.  Mass of each element * H is 1.01 g/mol, O is 16.00 g/mol

O 88.79% 100% H2O g 18.02

O g 16.00=×

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Mass Percent as a Conversion Factor

n  the mass percent tells you the mass of a constituent element in 100.0 g of the compound n  the fact that NaCl is 39.0% Na by mass

means that 100.0g of NaCl contains 39.0g Na

n  this can be used as a conversion factor n  100.0 g NaCl ≡ 39.0 g Na

Na g NaCl g 100.0Na g 39.0 NaCl g =× NaCl g

Na g 39.0NaCl g 100.0 Na g =×

Page 33: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Empirical Formulas •  The simplest, whole-number ratio of atoms

in a molecule is called the Empirical Formula – can be determined from percent composition or

combining masses •  The Molecular Formula is a multiple of the

Empirical Formula % A mass A (g) moles A

100g MMA

% B mass B (g) moles B 100g MMB

moles A moles B

Page 34: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Empirical Formulas

Hydrogen Peroxide Molecular Formula = H2O2 Empirical Formula = HO

Page 35: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Glucose Molecular Formula = C6H12O6 Empirical Formula = CH2O

Benzene Molecular Formula = C6H6 Empirical Formula = CH

Page 36: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Finding an Empirical Formula 1)  convert the percentages to grams

a)  skip if already grams 2)  convert grams to moles

a)  use molar mass of each element 3)  divide all by smallest number of moles 4)  round or multiply all mole ratios by

number to make all whole numbers a)  if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.

67, multiply all by 3, etc. b)  skip if already whole numbers

Page 37: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

n  Determine the empirical formula of a compound containing 80.0 grams of carbon and 20.0 grams hydrogen.

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Grams to moles C mol 6.67

C g 12.01C mol 1

C 80.0g =×

H mol 19.8 H g 1.01H mol 1 H 20.0g =×

Page 39: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Divide by smallest

1.00 mol 6.67

mol 6.67 C ==

3.00 2.96 mol 6.67

mol 19.8 H ≈==

Page 40: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Write Empirical Formula

CH3

Page 41: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

All these molecules have the same Empirical Formula. How are the molecules different?

Name Molecular Formula

Empirical Formula

glyceraldehyde C3H6O3 CH2O

erythrose C4H8O4 CH2O

arabinose C5H10O5 CH2O

glucose C6H12O6 CH2O

Page 42: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

All these molecules have the same Empirical Formula. How are the molecules different?

Name Molecular Formula

Empirical Formula

Molar Mass, g

glyceraldehyde C3H6O3 CH2O 90.09

erythrose C4H8O4 CH2O 120.12

arabinose C5H10O5 CH2O 150.15

glucose C6H12O6 CH2O 180.18

Page 43: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Molecular Formulas

n  The molecular formula is a multiple of the empirical formula

n  To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

Molar Massreal formula Molar Massempirical formula

= factor used to multiply subscripts

Page 44: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

What is the molecular formula for ethane if it has a molar mass of 30.06 g/mol?

CH3= 15.04 g/mol

Molecular formula = 2 x the empirical formula

C2H6

2.0001.99867 g/mol 15.04

g/mol 30.06 ratio ===

Page 45: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Determine the Molecular Formula of Cadinene if it has a molar mass of 204 g and an empirical formula of C5H8

Page 46: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Solutes and Solvents

n  Solution: a homogenous mixture n  Solute: thing that dissolves n  Solvent: thing that does the dissolving

(found in the largest amounts) n  If the solvent is water, then it is called an

aqueous solution

Page 47: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Solubility

n  Why does sugar “disappear” in your iced tea?

n  How do fish breathe underwater? n  Why does soda go flat faster when left

out than when it is refrigerated?

n  It is all based on solubility!

Page 48: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Solubility n  Example: iced tea

n  Solute sugar tea n  Solvent water

Page 49: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

States and Solutions

n  Solutions can be any state of matter n  Solid-solid: alloys (gold jewelry, brass, etc.) n  Solid-liquid: salt water, sugar water n  Liquid-liquid: vinegar, peroxide, rubbing

alcohol n  Liquid-gas: soda, champagne, O2 in H2O n  Gas-gas: air, air tanks (scuba)

Page 50: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

How Things Dissolve

n  Need to find/ create “holes” in water for the dissolving substance to move into n  Need to over come hydrogen bonding

between water (or solvent) molecules

n  Get interactions between water molecules and molecules of the solute n  Ion-dipole interactions n  Dipole-dipole (and H bonding)

Page 51: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Why some coffees “Put hair on your chest.”

n  “Strong” coffee has more coffee dissolved in a given amount (say 1 pot) than “weak” coffee. n  Strong coffee = concentrated n  Weak coffee = dilute

n  Concentration: the amount of solute in a given amount of solvent (or solution).

Page 52: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Molarity (M)

n  Most common way to express concentration n  Molarity is the number of moles of solute

dissolved in each liter of solution n  Formula

n M = moles of solute

liters of solution

n  Dependent on temperature n  The higher the molarity the stronger the

concentration

Page 53: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Practice Problems 1. What is the molarity when 6.0 moles of

glucose is dissolved in water to make 3.0 L of solution.

0.5 L NaCl( )

4.0 moles1 L

! " # $

% & = 2 moles NaCl

2. How many moles of sodium chloride are there in 500 mL of 4.0 M solution?

M =

6.0 moles3.0 L

= 2.0 M

3. What is the volume of 3.0 M solution that contains 15 moles of glucose?

15 moles( )

1 L3.0 moles! " # $

% & = 5.0 L

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How does something so strong become so weak?

n  The answer is dilution. n  The more dilute something is, the lower

the concentration (it’s weaker). n  To accomplish this, add more solvent n  How do we know how much to add?

n M 1V1 = M 2V2 n  Typically start with a highly concentrated

solution and dilute down to what you need

Page 55: The Mole: Avogadro s number · The Mole: Avogadro’s number . How much is: ! ... requires the molar mass of the substance ... • The simplest, whole-number ratio of atoms

Figure 15.8: Process of making 500 mL of a 1.00 M acetic acid solution.

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Mole Day is October 23!