Moles Text reference for moles: PP 83 to 87 and PP 237 to 249.
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Transcript of 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
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
21
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
35
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
40
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
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