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### The Mole

- Thou shalt not use 6.02 x 1023 in vain.
- Thou shalt not use the term mole if thou has no true knowledge of the term mole.
- Thou shalt not kill a mole.
- Thou shalt not covet your neighbor's mole.
- Thou shall always remember to celebrate Mole Day.
- Thou shalt not disparage Mole Day.
- Thou shalt not use a mole out of season.
- Thou shalt always honor the one who introduced thou to Mole Day.
- You shalt always keep sacred 10/23.
- Thou shalt always remember these commolments or thou will never properly celebrate Mole Day.

- 1 mole of pennies would cover the Earth 1/4 mile deep!

- 1 mole of hockey pucks would equal the mass of the moon!

- 1 mole of basketballs would fill a bag the size of the earth!

The Mole

1 dozen =

12

1 gross =

144

1 ream =

500

1 mole =

6.02 x 1023

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

Avogadro’s Number

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

I didn’t discover it. Its just named after me!

Amadeo Avogadro

Calculating Formula Mass

Calculate the formula mass of carbon dioxide, CO2.

12.01 g + 2(16.00 g) =

44.01 g

One mole of CO2 (6.02 x 1023 molecules) has a mass of 44.01 grams

Calculations with Moles:Converting moles to grams

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

3.50 mol Li

6.94 g Li

= g Li

45.1

1 mol Li

Calculations with Moles:Converting grams to moles

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

18.2 g Li

1 mol Li

2.62

= mol Li

6.94 g Li

Calculations with Moles:Using Avogadro’s Number

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

6.02 x 1023 atoms

3.50 mol

= atoms

2.07 x 1024

1 mol

Calculations with Moles:Using Avogadro’s Number

How many atoms of lithium are in 18.2 g of

lithium?

18.2 g Li

6.022 x 1023 atoms Li

1 mol Li

6.94 g Li

1 mol Li

(18.2)(6.022 x 1023)/6.94

1.58 x 1024

= atoms Li

Standard Molar Volume

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

-Amedeo Avogadro

At STP (Standard Temperature and Pressure):

1 mole of a gas occupies 22.4 liters of volume

Calculations with Moles:Using Standard Molar Volume

How many moles of hydrogen are in 100 L of hydrogen at STP?

1 mol

100 L

= mol

4.64

22.4 L

Calculations with Moles:Using Standard Molar Volume

How many liters are occupied by 3 moles of oxygen gas at STP?

22.4 L

3 mol

= L

67.2

1 mol

In-depth Mole Calculations

- How many atoms of H are in 2.4 moles of water?

6.02 x 1023 mlcl.

2.4 mol

1 mol

1.44 x 1024 mlcl H2O

1.44 x 1024 mlcl H2O x 2 = 2.89 x 1024

In-depth Mole Calculations

- Given 36.04 g of H2O :
a. How many moles of H2O is that?

b. How many grams of H are in the sample?

c. How many grams of O are in the sample?

d. How many molecules of H2O is that?

e. How many moles of H are in the sample?

f. How many moles of O are in the sample?

g. What is the mass of two molecule of H2O?

2 moles

4.04 g H

32.0 g O

1.204 X 1024 mlcl

4 moles

2 moles

36.04 a.m.u.

Percentage Composition

- the percentage by mass of each element in a compound

159.17 g Cu2S

32.07 g S

159.17 g Cu2S

Percentage Composition- Find the % composition of Cu2S.

100 =

%Cu =

79.852% Cu

%S =

100 =

20.15% S

36 g

8.0 g

36 g

Percentage Composition- Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.

100 =

78% Fe

%Fe =

100 =

22% O

%O =

Percentage Composition

- How many grams of copper are in a 38.0-gram sample of Cu2S?

Cu2S is 79.852% Cu

(38.0 g Cu2S)(0.79852) = 30.3 g Cu

147.02 g

Percentage Composition- Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?

24.51%

H2O

%H2O =

100 =

Empirical/Molecular Formulas

CH3

Empirical Formula

- Smallest whole number ratio of atoms in a compound

C2H6

Molecular formula

Empirical formula

- Find mass (or %) of each element. (may be given)
2. Find moles of each element. (g to moles)

3. Divide moles by the smallest # to find subscripts.

4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

1.85 mol

Empirical Formula

- Find the empirical formula for a sample of 25.9% N and 74.1% O.

25.9 g

1 mol

14.01 g

= 1.85 mol N

= 1 N

74.1 g

1 mol

16.00 g

= 4.63 mol O

= 2.5 O

Examples

- An organic compound is found to contain 92.25% carbon and 7.75% hydrogen. Determine the empirical formula.
- Empirical Formula: CH

- What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?
- Empirical Formula: Fe2S3

C2H6

Molecular Formula

- “True Formula” - the actual number of atoms in a compound

CH3

empirical

formula

?

molecular

formula

1. Find the empirical formula.

2. Find the empirical formula mass.

3. Divide the molecular mass by the empirical mass.

4. Multiply each subscript by the answer from step 3.

14.03 g/mol

Molecular Formula

- The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol?

empirical mass = 14.03 g/mol

= 2.00

(CH2)2 C2H4

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