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CHEMISTRY 161 Chapter 4 The Mole. Macroscopic versus Microscopic Worlds. 2 H 2 + O 2 2 H 2 O. 1 liter water contains about 3.3 X 10 25 molecules. CHEMICAL MASS SCALE. standard / calibration. atomic mass unit (amu, u). one atom of carbon-12 12 u (exactly).

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slide1

CHEMISTRY 161

Chapter 4

The Mole

slide2

Macroscopic versus Microscopic Worlds

2 H2 + O2 2 H2O

1 liter water contains about 3.3 X 1025 molecules

slide3

CHEMICAL MASS SCALE

standard / calibration

atomic mass unit

(amu, u)

one atom of carbon-12 12 u (exactly)

we have to correlate u with kg

slide4

H2O

PSE

O: 15.999 u

H: 1.008 u

H: 1.008 u

H2O: 18.015 u

formula mass: weight of one molecule

slide5

CaO

O: 15.999 u

Ca: 40.08 u

CaO: 56.08 u

formula mass: weight of one molecule

correlation between u and kg

slide7

one mole of a compound contains the same

number of molecules/atoms as the number of atoms

in exactly

12 g of 12C

Avogadro’s number

Na

6.023 x 1023

slide8

6.023 x 1023

1 mole of H2O

molecules

6.023 x 1023

atoms

1 mole of 12C

atoms

6.023 x 1023

1 mole of Na

molecules

6.023 x 1023

1 mole of NaCl

Avogadro’s number

links micro and macroscopic world

slide9

6.023 x 1023

1 mole of H2O

molecules

1 molecule of H2O – 2 H atoms and 1 O atom

6.023 x 1023 molecules of H2O

6.023 x 1023 atoms of O

2 x6.023 x 1023 atoms of H

1 mole of H2O – 2 mole H atoms and 1 mole O atoms

slide10

H2O

Na

O: 15.999 u

H: 1.008 u

H: 1.008 u

H2O: 18.015 u

15.999 g/mol

Na

1.008 g/mol

Na

1.008 g/mol

1 mole of H2O – 18.015 g

slide11

CaO

O: 15.999 u

Ca: 40.08 u

CaO: 56.08 u

1 mole of CaO – 56.08 g

slide12

1 mole of H2O – 18.015 g

6.023 x 1023 molecules of H2O – 18.015 g

one molecule of H2O – 2.99 x 10-23 g

molecular weight of one mole of H2O

18.015 g mol-1

slide13

2 H2 + O2 2 H2O

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

4.03176g 31.9988g 36.03g

x g y g 70.0g

STOICHIOMETRY

slide14

2 H2 + O2 2 H2O

2 molecules 1 molecule 2 molecules

2 moles 1 mole 2 moles

4.03176g 31.9988g 36.03g

x g y g 70.0g

STOICHIOMETRY

slide15
Example I:

How many grams of iron are in a 15.0 g

sample of iron(III) oxide?

1. molecular formula

Fe2O3

2. weight of one molecule

159.7 u

3. weight of one mole Fe2O3

159.7 g

4. 1 molecule Fe2O3 contains 2 atoms of Fe

5. 1 mole Fe2O3 contains 2 moles of Fe

159.7 g

111.69 g

15.0 g

x g

x = 10.5 g

slide16
Example II:

How many grams of oxygen are in a 10.0 g

sample of nickel (II) nitrate?

1. molecular formula

2. weight of one molecule

3. weight of one mole Ni(NO3)2

4. 1 molecule Ni(NO3)2 contains 6 atoms of O

5. 1 mole Ni(NO3)2 contains 6 moles of O

x = 5.25 g

slide17
Example III:

How many atoms are in 10 kg of sodium?

1 mole sodium = 22.98977 g

6.023 x 1023 atoms = 22.98977 g

x atoms = 10,000 g

x = 2.6 x 1026 atoms

slide18
Example IV

How heavy are 1 million gold atoms?

1 mole gold = 196.96654 g

6.023 x 1023 atoms = 196.96654 g

1,000,000 atoms = x g

x = 3.2 x 10-16 g = 0.32 fg

slide19

Mass Percentage Composition

P4O10

=

X 100 %

=

X 100 %

slide20
Example V

A sample was analyzed and contains 0.1417 g of nitrogen and 0.4045 g of oxygen.

Calculate the percentage composition.

=

X 100 %

1. mass of whole sample

2. percentages of elements

slide21

MOLECULAR FORMULA

H2O

EMPIRICAL FORMULA

H2O2

HO

H2O

H2O

P2O5

P4O10

P2*2O2*5

slide22
Example

A sample contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formula.

N2O5

slide23

COMBUSTION

CxHy

CO2

H2O

slide24

COMBUSTION

C3H8

+ O2

+ H2O

CO2

0.013068 g

How many grams of oxygen are consumed?

  • balance equation
  • convert to moles
slide25
The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO2and 4.512 g of H2O.

Calculate the empirical formula of the compound.

How many grams of Al2O3 are produced when 41.5 g Al react?

2Al(s) + Fe2O3(s) Al2O3(s) + 2 Fe(s)

slide26

LIMITING REACTANT

C2H4 + H2O C2H5OH

slide27

limiting reactant

excess reactant

slide28
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to

4 NH3 + 5 O2 4 NO + 6 H2O

2 C2H2 + 5 O2 4 CO2 + 2 H2O

slide29

THE YIELD

2 C2H2 + 5 O2 4 CO2 + 2 H2O

theoretical yield of CO2: 10 g

actual yield of CO2: 8 g

X 100 %

=