the mole chapter 7 chemical quantities determine the percent composition of fe(oh) 2 fe – 1 x 55.8...

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The Mole Chapter 7 Chemical Quantities

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Page 1: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

The Mole

Chapter 7

Chemical Quantities

Page 2: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Determine the percent composition of Fe(OH)2

Fe – 1 x 55.8 = 55.8

O – 2 x 16 = 32

H – 2 x 1 = 2

Molar mass = 89.8

Page 3: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Divide element mass by molar mass, then multiply by 100.

Fe –55.8 x 100

89.8

O – 32 x 100

89.8

Page 4: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Divide element mass by molar mass, then multiply by 100.

H –2 x 100

89.8

Page 5: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Percent compositions

% composition Fe = 62.1%

% composition 0 = 35.6%

% composition Fe = 2.2%

Add them up. Do they total 100?

Page 6: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Empirical formulas

Which of the following are empirical formulas?

HCl

K2C2O4

Pb2O4

Page 7: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Empirical formulas

Only one is an empirical formula.

HCl

The subscripts cannot be reduced.

Page 8: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Find the number of molecules in 24 g of CO

24 g (1 mole)

( 28 g )

= 0.86 moles of CO

Page 9: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Find the number of molecules in 24 g of CO

0.86 mol (6.02 x 1023)

(1 mol)

5.2 x 1023

molecules of CO

Page 10: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Find the density of one mole of CO gas at STP.

Density = Mass divided by volume

D = 28 g = 1. 25 g/L

22.4 L

Page 11: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

What is Avogadro’s number?

6.02 x 1023

units of a substance

Page 12: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

What is the difference between molecular

mass and molar mass?

Molecular mass is in atomic mass units; molar mass is in grams.

Page 13: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

What is the difference between a 6 M solution of

H2SO4 and a 1 M solution of H2SO4 ?

The 6 M (molar) sulfuric acid is much stronger. It has 6 moles of the acid dissolved in 1 liter of solution. The 1 M has 1 mole dissolved in 1 liter.

Page 14: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Convert 872 g of ammonium oxide

into moles.First find the correct formula for ammonium oxide.

(NH4)2O

Page 15: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Find the molar mass of ammonium oxide

(NH4)2O

N – 2 x 14 = 28

H – 8 x 1 = 8

O – 1 x 16 = 16

Page 16: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Find the molar mass of ammonium oxide

(NH4)2O = 52 grams

Set up a conversion factor to change from 872 g to moles.

Page 17: The Mole Chapter 7 Chemical Quantities Determine the percent composition of Fe(OH) 2 Fe – 1 x 55.8 = 55.8 O – 2 x 16 = 32 H – 2 x 1 = 2 Molar mass =

Conversion factor

872 g ( 1 mol)

52 g

= 16.8