AP Chemistry Chapter 3

Chemical Chemical QuantitiesQuantities(The Mole) (The Mole)

Amedeo AvogadroHypothesis: Equal volumes of different gases at the same temperature and pressure contain equal numbers of particles.

Amadeo Avogadro

H2

hydrogenCH4

methane

The Mole

1 dozen =1 gross =

1 ream =

1 mole =

12

144

500

6.02 x 1023

There are exactly 12 grams of carbon-12 in one mole of carbon-12.

Avogadro’s Number6.02 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amedeo Avogadro (1776-1855).

Amadeo Avogadro

I didn’t discover it. Its just named after

me!

The Mole (Quantities)

1 mole =

1 mole =

1 mole =

atoms molecules formula units

Molar mass (periodic table)

22.4 L for a gas @STP

6.02 x 1023

(elements) (nonmetals) (cation-anion)

Mass:

Volume:

Representative Particles:

1. Convert 3.00 moles of ammonia, NH3, to grams of ammonia.

1. Convert 3.00 moles of ammonia, NH3, to grams of ammonia.

1 mole NH3 = ?? g NH3

Conversion factor:

3.00 mol NH3x = g NH3

You have to find molar mass.(Use Periodic

Table)

mol NH3

g NH3

17.04 g NH17.04 g NH33

NHNH33

14.01 g14.01 g 1.01 g1.01 gx 3x 3

14.01 g14.01 g + 3.03 g =+ 3.03 g =x 1x 1

Molar mass of ammonia.

7

N14.01

1

H1.01

1 mole NH3 = 17.04 g NH3

1. Convert 3.00 moles of ammonia, NH3, to grams of ammonia.

1 mole NH3 = 17.04 g NH3

Conversion factor:

3.00 mol NH3x = g NH3

mol NH3

g NH3

NH3

14.01 1.01x 1 x 314.01 g+ 3.03 g = 17.04 g/mol

1 17.04

On calculator: 3.00 x 17.04 = 51.12

Round to 3 sig. figs.

51.1 g NH3

Ch. 7 (Calculating Chemical Quantities)

3.

2.

3NHmol00.3 17.04 g NH3

1 mol NH314.01 1.01x 1 x 314.01 g + 3.03 g = 17.04 g/mol

1. = 51.1 g NH3

2. How many molecules of carbon dioxide are in 2.00 moles of CO2?

2. How many molecules of carbon dioxide are in 2.00 moles of CO2?

1 mole CO2 = 6.02 x 1023 molecules CO2

Conversion factor:

2.00 mol CO2x = molec. CO2

1 mole = 6.02 x 1023 representative

particles

CO2 is made of all nonmetals sorep. particles are “molecules”

mol CO2

molec. CO2

2. How many molecules of carbon dioxide are in 2.00 moles of CO2?

1 mole CO2 = 6.02 x 1023 molecules CO2

Conversion factor:

2.00 mol CO2x = molec. CO2mol

CO2

molec. CO2

1 6.02 x 1023

On calculator: 2.00 x (6.02 x 1023) = 1.204 x 1024

Round to 3 sig. figs.

1.20 x 1024 molec. CO2

Ch. 7 (Calculating Chemical Quantities)

3.

2

2 23

2 COmol 1

CO molec.10x 6.02COmol002. 1.20 x 1024

molec. CO2

2.

3NHmol00.3 17.04 g NH3

1 mol NH314.01 1.01x 1 x 314.01 g + 3.03 g = 17.04 g/mol

1. = 51.1 g NH3

3. How many moles of oxygen are in 44.8 L of oxygen, O2?

1 mole O2 = 22.4 L O2

Conversion factor:

44.8 L O2 x = moles O2

1 mole = 22.4 L (for any gas @ STP)

L O2

mol O2

1 22.4

On calculator: 44.8 ÷ 22.4 = 2

Show to 3 sig. figs.

2.00 moles O2

Ch. 7 (Calculating Chemical Quantities)

2

22 OL 22.4

O mole 1O L 44.8 2.00 mol O23.

2

2 23

2 COmol 1

CO molec.10x 6.02COmol002. 1.20 x 1024

molec. CO2

2.

3NHmol00.3 17.04 g NH3

1 mol NH314.01 1.01x 1 x 314.01 g + 3.03 g = 17.04 g/mol

1. = 51.1 g NH3

4. How many grams of lithium are in 3.50 moles of lithium?

1 mole Li = g Li

Conversion factor:

3.50 mol Li x = g Limol Li

g Li

1 6.94

On calculator: 3.50 x 6.94 = 24.29

Round to 3 sig. figs.

24.3 g Li

The Mole (Quantities)

1 mole =

1 mole =

1 mole =

atoms molecules formula units

Molar mass (periodic table)

22.4 L for a gas @STP

6.02 x 1023

(elements) (nonmetals) (cation-anion)

Mass:

Volume:

Representative Particles:

5. How many moles of lithium are in 18.2 grams of lithium?

1 mole Li = g Li

Conversion factor:

18.2 g Li x = mol Lig Li

mol Li

1 6.94

On calculator: 18.2 ÷ 6.94 = 2.6225

Round to 3 sig. figs.

2.62 mol Li

Representative Particles:Using Avogadro’s Number

6. How many atoms of lithium are in 3.50 moles of lithium?

3.50 mol Li

= atoms Li1 mol Li

6.02 x 1023 atoms Li 2.11 x 1024

Use 6.02 x 1023 when looking for atoms, molecules, or formula units

x

Representative Particles:Two-part problem

7. How many atoms of lithium are in 18.2 g of lithium?

18.2 g Li

= atoms Li

1 mol Li 6.02 x 1023 atoms Li

1.58 x 1024

6.94 g Li 1 mol Li

(18.2)/6.94 x (6.02 x 1023)

x x

%Na

X 100 =26.00 g

7.86 g= 30.23 % Na

%Cl X 100 =26.00 g

18.14 g= 69.77 % Cl

8. A sample of sodium chloride, (NaCl) commonly called table salt contains 7.86 g of sodium and 18.14 g of chlorine. Find the % composition.Find total mass of NaCl 7.86 g

+ 18.14 gTotal mass = 26.00 g

Take the mass of the element divided by total mass.

%C X 100 =44.11 g

36.03 g=

44.11 g C3H8

CC33HH88

12.01 g 1.01 gx 8

36.03 g + 8.08 g =

x 3

81.68 % C

%H X 100 =44.11 g

8.08 g= 18.32 % H

9. Find the % composition of propane, (C3H8).

6

C12.01

1

H1.01

10. A 12.8 g sample of a gas contains 6.4 grams of sulfur and 6.4 grams of oxygen. What is the empirical formula for this gas?

Formulas

molecular formula = (empirical formula)n where n = integer

molecular formula = C6H6 = (CH)6

empirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.

Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas (continued)

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole be empirical (lowest whole number ratio).number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

All can be divided by 6

Empirical Formula Determination

1. Convert grams values to moles for each element.

2. Divide by lowest moles.

3. If necessary: Multiply each number by an integer to obtain all whole numbers.

6.4 g S =32.07 g S

1 mol Sx 0.1995 mol S = 0.20 mol S

6.4 g O = 0.4 = 0.40 mol O16.00 g O

1 mol Ox

10. A 12.8 g sample of a gas contains 6.4 grams of sulfur and 6.4 grams of oxygen. What is the empirical formula for this gas?

16

S32.07

8

O16.00

0.20 mol

0.20 mol

2. Divide by lowest moles.

1. Convert to moles.

2

1

S OSO2Empirical Formula

3.2 g H =1.01 g H

1 mol Hx 3.16 mol H

19.4 g C = 1.6212.01 g C

1 mol Cx

11. An unknown clear colorless liquid with no odor is analyzed and found to contain the following. Determine the empirical formula.

1

H1.01

8

O16.00

1.6 mol

1.6 mol

2. Divide by lowest moles.

1. Convert to moles.

1

2

H C OEmpirical Formula

3.2 % Hydrogen19.4% Carbon77.4% Oxygen

Assume 100 g sample

= 3.2 g H= 19.4 g C= 77.4 g O

77.4 g O = 4.8416.00 g O

1 mol Ox

1.6 mol 3

6

C12.01

H2CO3

= 3.2 mol H

= 1.6 mol C

= 4.8 mol O

%Na

X 100 =58.44 g

22.99 g=

58.44 g NaCl

NaClNaCl22.99 g 35.45 g

x 1

22.99 g + 35.45 g =

x 1

39.34 % Na

%O X 100 =58.44 g

35.45 g= 60.66 % Cl

8. A sample of sodium chloride, (NaCl) commonly called table salt contains 7.86 g of sodium and 18.14 g of chlorine. Find the % composition. 11

Na22.99

17

Cl35.45

The Mole (Quantities)

1 mole =

1 mole =

1 mole =

atoms molecules formula units

Molar mass (periodic table)

22.4 L for a gas @STP

6.02 x 1023

(elements) (nonmetals) (cation-anion)

Mass:

Volume:

Representative Particles: