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Moles • Text reference for moles: • PP 83 to 87 and • PP 237 to 249

Moles Text reference for moles: PP 83 to 87 and PP 237 to 249

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Moles

• Text reference for moles:• PP 83 to 87 and• PP 237 to 249

2

The Mole

• A number of atoms, ions, or molecules that is large enough to see and handle.

• A mole = number of things– Just like a dozen = 12 things– One mole = 6.022 x 1023 things

• Avogadro’s constant = 6.022 x 1023 – Symbol for Avogadro’s constant is NA.

3

The Mole

• How do we know when we have a mole?– count it out– weigh it out

• Molar mass - mass in grams numerically equal to the atomic weight of the element in grams.

• H has an atomic weight of 1.00794 g – 1.00794 g of H atoms = 6.022 x 1023 H atoms

• Mg has an atomic weight of 24.3050 g – 24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms

4

The Mole

• Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.

Mgg ?

5

The Mole

• Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.

atom Mg1 Mgg ?

6

The Mole

• Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.

atoms Mg 106.022

atoms Mg mol 1atom Mg 1Mg g ?

23

7

The Mole

• Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures.

Mg g 104.04atoms Mg mol 1

Mgg24.30

atoms Mg 106.022

atoms Mg mol 1atom Mg 1Mg g ?

23

23

8

The Mole

• Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

atoms Mg?

9

The Mole

• Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg? 6

10

The Mole

• Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

atoms Mgmol 1

atoms Mg106.022

Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg?

23

6

11

The Mole

• Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

atoms Mg102.48atoms Mgmol 1

atoms Mg106.022

Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg?

1623

6

12

The Mole

• Example 2-3. How many atoms are contained in 1.67 moles of Mg?

atoms Mg ?

13

The Mole

• Example 2-3. How many atoms are contained in 1.67 moles of Mg?

Mg mol 1.67atoms Mg ?

14

The Mole

• Example 2-3. How many atoms are contained in 1.67 moles of Mg?

Mg mol 1

atoms Mg 106.022 Mg mol 1.67atoms Mg ?

23

15

The Mole

• Example 2-3. How many atoms are contained in 1.67 moles of Mg?

atoms Mg 101.00

Mg mol 1

atoms Mg 106.022 Mg mol 1.67atoms Mg ?

24

23

16

The Mole

• Example 2-3. How many atoms are contained in 1.67 moles of Mg?

atoms Mg101.00

Mgmol 1

atoms Mg106.022Mg mol 1.67atoms Mg?

24

23

17

The Mole

• Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

You do it!You do it!

18

The Mole

• Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

Mg g 4.73Mg mol ?

19

The Mole

• Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

Mg g 24.30

atoms Mg mol 1 Mg g 4.73Mg mol ?

20

The Mole

• Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

Mg mol 02.3

Mg g 24.30

atoms Mg mol 1 Mg g 4.73Mg mol ?

IT IS IMPERATIVE THAT YOU KNOWHOW TO DO THESE PROBLEMS

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Formula Weights, Molecular Weights, and Moles

• How do we calculate the molar mass of a compound?– add atomic weights of each atom

• The molar mass of propane, C3H8, is:

• 3C X 12.01g = 36.03g• 8H X 1.01g = 8.08g• 44.11 g/mole

22

Formula Weights, Molecular Weights, and Moles

• The molar mass of calcium nitrate, Ca(NO3)2 , is:

You do it!You do it!

23

Formula Weights, Molecular Weights, and Moles

• One Mole of Contains– Cl2 or 70.90g 6.022 x 1023 Cl2 molecules

2(6.022 x 1023 ) Cl atoms

– C3H8 You do it!You do it!

24

Formula Weights, Molecular Weights, and Moles

• One Mole of Contains– Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms– C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules 3 (6.022 x 1023 ) C atoms 8 (6.022 x 1023 ) H atoms

25

Formula Weights, Molecular Weights, and Moles

• Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.

8383 HC g 74.6molecules HC ?

26

Formula Weights, Molecular Weights, and Moles

• Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.

83

83

8383

HC g 44.11

HC mole 1

HC g 74.6molecules HC ?

27

Formula Weights, Molecular Weights, and Moles

• Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.

83

8323

83

83

8383

HC mole 1

molecules HC 106.022

HC g 44.11

HC mole 1

HC g 74.6molecules HC ?

28

Formula Weights, Molecular Weights, and Moles

• Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.

molecules 10 02.1

HC mole 1

molecules HC 106.022

HC g 44.11

HC mole 1

HC g 74.6molecules HC ?

24

83

8323

83

83

8383

29

Formula Weights, Molecular Weights, and Moles

• Example 2-6. What is the mass of 10.0 billion propane molecules?

You do it!You do it!

30

Formula Weights, Molecular Weights, and Moles

8313

83

83

238310

83

HC of g 1032.7HC mole 1

HC g 11.44

molecules 106.022

HC mole 1molecules 10 1.00 molecules HC g ?

• Example 2-6. What is the mass of 10.0 billion propane molecules?

31

Formula Weights, Molecular Weights, and Moles

• Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O3? The layer of ozone in the stratosphere is very beneficial to life on earth.

You do it!You do it!

32

Formula Weights, Molecular Weights, and Moles

• Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O3? The layer of ozone in the stratosphere is very beneficial to life on earth.

(a)

moles 25.1O g 0.48

mole 1O g 0.60O moles ?

333

33

Formula Weights, Molecular Weights, and Moles

• Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O3? The layer of ozone in the stratosphere is very beneficial to life on earth.

(b)

323

23

3

O molecules 107.53

mole 1

molecules 106.022moles 25.1O molecules ?

34

Formula Weights, Molecular Weights, and Moles

• Example 2-7. How many (a) moles, (b) molecules, and (c) oxygen atoms are contained in 60.0 g of ozone, O3? The layer of ozone in the stratosphere is very beneficial to life on earth.

(c)

O atoms 1026.2

molecule O 1

atoms O 3O molecules 107.53atoms O ?

24

33

23

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Formula Weights, Molecular Weights, and Moles

• Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.

You do it!You do it!

36

Formula Weights, Molecular Weights, and Moles

• Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.

atoms O 106.49

COLi unit formula 1

atoms O 3

COLi mol 1

COLi unitsform.106.022

COLi g 73.8

COLi mol 1COLi g 26.5atoms O ?

23

3232

3223

32

3232

37

Percent Composition and Formulas of Compounds

• % composition = mass of an individual element in a compound divided by the total mass of the compound x 100%

• Determine the percent composition of C in C3H8.

81.68%

100%g 44.11

g 12.013

100%HC mass

C mass C %

83

38

Percent Composition and Formulas of Compounds

• What is the percent composition of H in C3H8?

You do it!You do it!

39

Percent Composition and Formulas of Compounds

• What is the percent composition of H in C3H8?

81.68%100%18.32%

or

%18.32100%g 44.11

g 1.018

100%HC

H8

100%HC mass

H massH %

83

83

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Percent Composition and Formulas of Compounds

• Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures.

You do it!

41

Percent Composition and Formulas of Compounds

• Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 sig. fig.

100% Total

O 48.0% 100%g 399.9

g 16.012 100%

)(SOFe

O12 O %

S 24.1% 100%g 399.9

g 32.13 100%

)(SOFe

S3 S %

Fe 27.9% 100%g 399.9

g 55.82 100%

)(SOFe

Fe2Fe %

342

342

342

42

Derivation of Formulas from Elemental Composition

• Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula?

• Make the simplifying assumption that we have 100.0 g of compound.

• In 100.0 g of compound there are:– 24.74 g of K– 34.76 g of Mn– 40.50 g of O

43

Derivation of Formulas from Elemental Composition

K mol 0.6327K g 39.10

K mol 1K g 24.74 K mol ?

44

Derivation of Formulas from Elemental Composition

Mn mol 0.6327Mn g 54.94

Mn mol 1Mn g 34.76 Mn mol ?

K mol 0.6327K g 39.10

K mol 1K g 24.74 K mol ?

45

Derivation of Formulas from Elemental Composition

rationumber wholesmallest obtain

O mol 2.531O g 16.00

O mol1O g 40.50 O mol ?

Mn mol 0.6327Mn g 54.94

Mn mol 1Mn g 34.76 Mn mol ?

K mol 0.6327K g 39.10

K mol 1K g 24.74 K mol ?

46

Derivation of Formulas from Elemental Composition

Mn 10.6327

0.6327Mnfor K 1

0.6327

0.6327Kfor

rationumber wholesmallest obtain

O mol 2.531O g 16.00

O mol1O g 40.50 O mol ?

Mn mol 0.6327Mn g 54.94

Mn mol 1Mn g 34.76 Mn mol ?

K mol 0.6327K g 39.10

K mol 1K g 24.74 K mol ?

47

Derivation of Formulas from Elemental Composition

4KMnO is formula chemical thethus

O 40.6327

2.531Ofor

Mn 10.6327

0.6327Mnfor K 1

0.6327

0.6327Kfor

rationumber wholesmallest obtain

O mol 2.531O g 16.00

O mol1O g 40.50 O mol ?

Mn mol 0.6327Mn g 54.94

Mn mol 1Mn g 34.76 Mn mol ?

K mol 0.6327K g 39.10

K mol 1K g 24.74 K mol ?

48

Derivation of Formulas from Elemental Composition

• Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound?

You do it!You do it!

49

Derivation of Formulas from Elemental Composition

• Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound?

ratio number wholesmallest find

O mol 0.1480O g 16.00

O mol1O g 2.368O mol ?

Co mol 0.1110Cog 58.93

Co mol 1Co g 6.541Co mol ?

50

Derivation of Formulas from Elemental Composition

• Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is the empirical formula for this compound?

43OCo

:is formula scompound' theThus

O 43O 1.333 Co 33Co 1

number wholetofraction turn to 3 by both multipy

O1.3330.1110

0.1480O for Co 1

0.1110

0.1110Co for

Determination of Molecular Formulas

• The formulas determined by the previous method produce empirical or simplest formulas. In the case of covalent compounds the empirical formula may or may not represent a real molecule. For example:

• CH2 is an empirical formula by no molecule with this formula exists. C2H4 is a real molecule with empirical formula CH2 !

Determination of Molecular Formulas

• In order to determine a molecular formula from the empirical formula you need some additional information, usually the molar mass of the molecule. Since a molecular formula is a multiple of its empirical formula, you can divide the molar mass of the molecular formula by the molar mass of the empirical to find the multiplier. For example:

Determination of Molecular Formulas

• Suppose the empirical formula is CH2 and we were told that the molar mass of a molecule of this substance is 70 g/mole.

• The molar mass of CH2 is 14 g/mole

• So 70/14 = 5 so the molecular formula is 5 X the empirical or C5H10

Determination of Molecular Formulas

• Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula?

»You do it

55

Determination of Molecular Formulas

• Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? – short cut method

H of g 8.100.1437g 56.1

C of g 48.00.8563g 56.1

H is 14.37% and C is 85.63%

g 56.1 contains mol 1

56

Determination of Molecular Formulas

84HC

:is formula theThus

H mol 8H g 1.01

H mol 1H of g 8.10

C mol 4C g 12.0

C mol 1C of g 48.0

moles tomassesconvert