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Chapter 7 Chemical Quantities

Chemical Quantities

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Chemical Quantities. Yeah, Baby!!. Chapter 7. STOP. 2 lb bag of sugar 5 lb bag of potatoes 2 liters of soda-pop 3 gross of M&M’s 10 gallons of gasoline 1 dozen Krispy Kream doughnuts. We describe quantities by weighing them, how much space they take up, or by counting them. - PowerPoint PPT Presentation

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Page 1: Chemical Quantities

Chapter 7

Chemical Quantities

Page 2: Chemical Quantities

2 lb bag of sugar

5 lb bag of potatoes 2 liters of soda-pop

3 gross of M&M’s 10 gallons of gasoline

1 dozen Krispy Kream doughnuts

•We describe quantities by weighing them, how much space they take up, or by counting them.•We do the same in chemistry

2 g of NaCl (weight)

3 mL of H20 (volume)5 moles of MgI (counting)

STOP

Page 3: Chemical Quantities

• A mole is a quantity of particles, just as…

1 dozen = 12 things

1 mole = 6.02 x 1023 particles

• “Particles” usually measured in moles are atoms, molecules, ions, and formula units

Page 4: Chemical Quantities

Important # to remember…

Avogrado’s Number

1 mole =6.02 x 1023

Page 5: Chemical Quantities

Examples: moles atoms

How many atoms of Al are in 1.50 mol of Al?

Conversion:1 mole = 6.02 x 1023 atoms

1.50 mol of Al

1 mole= 9.03 x 1023 atoms

of Al

6.02 x 1023 atoms

Page 6: Chemical Quantities

Examples: moles molecules atoms How many atoms of H are there in 3.0 moles

of H2O?Conversions:1 mole = 6.02 x 1023 moleculesH2O molecule = 2 atoms of Hydrogen

1 mole

3 moles of H2O

1 H2O molecule

= 3.6 x 1024 atoms H

6.02 x 1023 molec. 2 atoms H

Page 7: Chemical Quantities

Question 1:

• Which contains more molecules:– 1.00 mol H2O2

– 1.00 mol C2H6

– 1.00 mol CO

• H2O2

• C2H6

• CO• All contains 6.02 x 1023 molecules

Page 8: Chemical Quantities

Question 2:

• Which contains more atoms:– 1.00 mol H2O2

– 1.00 mol C2H6

– 1.00 mol CO

• H2O2

• C2H6

• CO• All contains 6.02 x 1023 atoms

Page 9: Chemical Quantities

Question 3:

• Determine the number of atoms in 3.00 mol Sn

A. 1.81 x 1025 atoms SnB. 1.81 x 1024 atoms SnC. 4.98 x 10-24 atoms SnD. None of the above

Page 10: Chemical Quantities

Molar Mass

• Determined simply by looking at the periodic chart

• Molar mass = Atomic Mass

Ca20

40.08 Molar Mass

* Thus, 1 mol Ca = 40 g

Page 11: Chemical Quantities

Question 4

• Calculate the mass of 1.00 mol of copper

A. 29 g CuB. 63 g CuC. 64 g CuD. 1 g Cu

Page 12: Chemical Quantities

Examples: grams molesCalculate the number of moles in 367 g of silver?

Conversion:1 mole Ag = 108 g

367 g Ag

108 g Ag = 3.40 mol Ag

1 mol Ag

Page 13: Chemical Quantities

Molar Mass (compound)• Mass of 1 mole in a compound• Determined by adding the molar mass of

each atom in that compound

What is the gram molecular mass of H2O?2 atoms H = 1 g x 21 atom O = 16 g x 1

= 18 g

Thus… 1 mole of H2O = 18 g

Page 14: Chemical Quantities

Question 5:

• Calculate the molar mass of diatomic nitrogen

A. 7 g N

B. 14 g N2

C. 28 g N2

D. 14 g N

Page 15: Chemical Quantities

Examples: moles gramsWhat is the mass of 3.40 moles of H2O?

Conversion:1 mole H2O = 18 g

3.4 mol H2O

1 mole H2O = 61.2 g H2O

18 g

Page 16: Chemical Quantities

Examples: moles grams What is the mass of 5.60 moles of ammonium

carbonate?

Conversion:ammonium carbonate - (NH4)2CO3

N: 2 x 14 g = 28 gH: 8 x 1 g = 8 gC: 1 x 12 g = 12 gO: 3 x 16 g = 48 g

5.60 mol (NH4)2CO3

1 mol (NH4)2CO3 = 538 g

1 mol (NH4)2CO3 = 96 g

96 g

Page 17: Chemical Quantities

Examples: moles atoms How many atoms is this?

Conversions:1 mole = 6.02 x 1023 molecules1 (NH4)2CO3 molecule = 14 atoms

6.02 x 1023 molecules

1 mole (NH4)2CO3

5.6 moles (NH4)2CO3 14 atoms

1 molecule (NH4)2CO3

= 4.72 x 1025 atoms

Page 18: Chemical Quantities

• When dealing with gasses, volume is determined by using the conversion:

1 mole = 22.4 L

Page 19: Chemical Quantities

Examples: moles volume

How much volume is 5.60 moles of CO2?

Conversion:1 mole = 22.4 L

5.6 mol CO2

1 mol= 125 L CO2

22.4 L

Page 20: Chemical Quantities

Review of conversions for moles

1 mole = 6.02 x 1023 particles

1 mole = _____ grams

1 mole = 22.4 L

STOP

Page 21: Chemical Quantities

Examples:

How many grams of Al are in 2.0 mol Al?

Conversion:1 mole Al = 27 g

2.0 mol Al 27 g Al

1 mol Al= 54 g Al

Page 22: Chemical Quantities

Examples:

How many molecules are in 5 mol of N2O3?

Conversion:1 mole N2O3 = 6.02 x 1023 molecules

5 mol N2O3 6.02 x 1023 molecules

1 mol N2O3

= 3.01 x 1024 molecules N2O3

Page 23: Chemical Quantities

Examples:

What is the mass of 8 L of CO2? (Hint: 2 step conversion)

Conversion:1 mol = 22.4 L1 mol CO2 = 44 g

8 L CO2 1 mol

22.4 L

= 15.7 g CO2

44 g CO2

1 mol CO2

Page 24: Chemical Quantities

Percent Composition:• Describes the relative amounts of each

element in a compound (% by mass)

% mass of =element

Grams of element in compound

Grams of compoundx 100

Page 25: Chemical Quantities

Grams of Compound = 9.41g + 5.99g = 15.40g

What is the % mass of each element when you combine9.41g of Ca with 5.99g of S ?

% Ca =

% S =

9.4115.40

5.9915.40

x 100

x 100

= 61%

= 39%

Page 26: Chemical Quantities

If you know the chemical formula, you can calculate % composition using molar mass

Ex.) Calculate the % comp. of ethane (C2H6)

C2 =

H6 =

C2H6 =

24 g

6 g

30 g

% C = % H =

% C = % H =

x 10024

30x 1006

30

80% 20%

Page 27: Chemical Quantities

% comp. can be used to calculate the # of grams of an element in a specific amount of a compound

Ex.) Calculate the mass of carbon in 30 g of

ethane (C2H6)

From prev. example: %C in C2H6 = 80% of C

30 g C2H6 .80

(80 %) C

= 24 g Cx

Page 28: Chemical Quantities

Empirical Formula• The lowest whole-number ratio of the elements

in a compound

Ie.) The E.F. of hydrogen peroxide (H2O2) is HO

• The E.F. can be determined by % comp.

Page 29: Chemical Quantities

• Assume you have 100 g of this compound 67.6 g Hg 10.8 g S 21.6 g O

• Determine the # of moles of each

Ex.) Calculate the formula for a compound that is 67.6% Hg, 10.8% S, & 21.6% O.

Page 30: Chemical Quantities

21.6 g O 1 mole O

16 g O= 1.35 mol O

10.8 g S 1 mole S

32 g S= .337 mol S

But the subscripts need to be whole #’s

67.6 g Hg 1 mole Hg

200 g Hg= .337 mol Hg

Page 31: Chemical Quantities

• Find the smallest subscript and divide each by it

.337 mol Hg

.337 mol

1.35 mol O

.337 mol

.337 mol S

.337 mol

=

=

=

1 mol Hg

1 mol S

4 mol O

HgSO4

Empirical Formula