the mole concept
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THE MOLE CONCEPT
1/7/2011 THE MOLE CONCEPT | REDG 2
The Mole
It is a quantity referring to 602, 000, 000, 000, 000, 000, 000, 000 units
How big is it?The richest man in the world (Carlos Slim Helú and family of Mexico), has a net worth of $53.5 B.
In peso, this amount is
PhP 2,650,000,000,000
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The Mole
By an incredible chance, you own 1 mole of peso. Who is richer, you or Carlos Slim Helú ?
By how many folds richer is the richer man compared to the less rich man?
Your wealth: 602, 000, 000, 000, 000, 000, 000, 000Carlos’ wealth: 2, 650, 000, 000, 000
The questions means the same as : “how many times is 10 bigger than 2?”
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The Mole
2,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,0002,650,000,000,000 2,650,000,000,000 2,650,000,000,000 2,650,000,000,000
602, 000, 000, 000, 000, 000, 000, 000
602, 000, 000, 000, 000, 000, 000, 000
÷ 2, 650, 000, 000, 000
227, 170, 000, 000
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The Mole 602, 000, 000, 000, 000, 000, 000, 000
In scientific notation, it is
6.02 x 1023
This number is called
Avogadro’s Number
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The Mole
1 mol marbles = 6.02 x 1023 pieces marbles
1 mol sand = 6.02 x 1023 particles sand
1 mol water = 6.02 x 1023 molecules water
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The Mole
How many molecules are there in 2 mol O2?
23
2
2 2
2
23
2
24
2
6.02x10 molecules O?molecule O = 2 mol O
1 mol O
12.04 x 10 molecules O
= 1.204 x 10 molecules O
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The Mole
How many molecules H2O are there in 0.15 mol H2O?
23
2
2 2
2
22
2
6.02x10 molecules H O?molecules H O= 0.15 mol H O
1 mol H O
9.03 x 10 molecules H O
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The Mole
How many moles contain 3.01 x 1023
molecules O2?
23 2
2 2 23
2
2
1 mol O?mol O = 3.01x10 molecules O
6.02x10 molecules O
0.5 mol O
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The Mole
If there is 1 “molecule” of Fe2(SO3)3, how many atoms of Fe are there?
2 3 3
2 3 3
2 atoms Fe?atoms Fe= 1 molecule Fe SO
1 molecule Fe SO
2 atoms Fe
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The Mole
If there is 1 “molecule” of Fe4[Fe(CN)6]3, how many atoms of Fe
are there?
4 6 34 6 3
7 atoms Fe?atoms Fe= 1 molecule Fe Fe CN
1 molecule Fe Fe CN
7 atoms Fe
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The Mole
If there are 6.02 x1023 molecules of Fe4[Fe(CN)6]3, how many atoms of Fe
are there?
234 6 3
4 6 3
23
24
7 atoms Fe?atoms Fe= 6.02x10 molecule Fe Fe CN
1 molecule Fe Fe CN
42.14x10 atoms Fe
= 4.214x10 atoms Fe
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The Mole
If there is 0.75 mol of Fe4[Fe(CN)6]3, how many moles of Fe are there?
4 6 34 6 3
7 mol Fe?mol Fe= 0.75 mol Fe Fe CN
1 mol Fe Fe CN
5.25 mol Fe
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The Mole (Exercise)
1. How many molecules of water are there in 0.10 mol H2O?
2. If there is 0.25 mol of Fe4[Fe(CN)6]3, how many moles of C are there?
3. What is the total number of atoms present in 1 mole of oxygen gas?
4. Which has the greatest number of oxygen atoms 1mol O, 1mol O2, 1mol O3 ? (PROVE)
5. Which has greatest number of oxygen atoms, 1g O, 1g O2, 1g O3 ? (PROVE)
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The Mole (Exercise)23
2222 2 2
2
6.02 x 10 molecules H O?molecules H O = 0.10mol H O 6.02 x 10 molecules H O
1 mol H O
4 46 63 34 6 3
18 mol C?mol Fe Fe CN = 0.25 mol Fe Fe CN 4.5 mol C
1 mol Fe Fe CN
23242
22 2
6.02x10 molecules O 2 atoms O?atom O = 1 mol O 1.204x10 atoms O
1 mol O 1 molecule O
2323
23242
22 2
3
6.02x10 atoms O?atom O = 1 mol O 6.02x10 atoms O
1 mol O
6.02x10 molecules O 2 atoms O?atom O = 1 mol O 1.204x10 atoms O
1 mol O 1 molecule O
6.?atom O = 1 mol O
23242
2 2
02x10 molecules O 3 atoms O 1.806x10 atoms O
1 mol O 1 molecule O
1.
2.
3.
4.
Therefore, 1 mol O3 has the greatest number of O atoms
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The Molar Mass
• is the mass of 1 mol of an element or a compound.
• it is numerically the same as formula mass but the unit of the molar mass is grams per mole (g/mol).
Formula mass Molar mass
H2O 18.02u 18.02 g/mol
NaOH 40.00u 40.00 g/mol
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The Molar Mass
What is the mass of 0.5mol H2O?
22 2
2
2
18.02 g H O?g H O = 0.5 mol H O
1 mol H O
9.01g H O
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The Molar Mass
What is the mass of 2.1 mol KMnO4?
44 4
4
4
158.04 g KMnO?g KMnO = 2.1 mol KMnO
1 mol KMnO
331.9g KMnO
What is the mass of 1.25 x 10-4 mol oxygen gas?
-4 22 2
2
-32
32.00 g O?g O = 1.25x10 mol O
1 mol O
4.00x10 g O
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The Molar Mass
How many moles is 10.0g CO2?
22 2
2
2
1 mol CO?mol CO = 10.0g CO
44.01g CO
0.227 mol CO
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The Molar Mass
How many moles is 25.0g CCl4?
44 4
4
4
1mol CCl?mol CCl = 25.0g CCl
153.81 g CCl
0.163 mol CCl
How many mole is 1.0g Na2C2O4?
2 2 42 2 4 2 2 4
2 2 4
-32 2 4
1mol Na C O?mol Na C O = 1.0g Na C O
134.0 g Na C O
7.46x10 mol Na C O
Recall that the percentage composition tells the fraction of the mass of each of the composing elements of the total mass of the compound. Total mass may mean formula mass or molar mass.
Percentage Composition
Na = 39.34%
Cl = 60.66%NaCl1/7/2011 THE MOLE CONCEPT | REDG 21
Percentage Composition
Na = 39.34%
Cl = 60.66%NaCl1/7/2011 THE MOLE CONCEPT | REDG 22
Percentage Composition
Na = 39.34%
Cl = 60.66%NaCl
mass A% A = x 100
mass A+B+C....
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Percentage Composition
Na = 39.34%
Cl = 60.66%NaCl
mass Na% Na = x 100
mass Na+Cl
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What is the percentage composition of S2Cl2?
In 5.00g S2Cl2 how many grams of sulfur is present?
Percentage Composition
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In 5.00g S2Cl2 how many grams of chlorine is present?
How many grams of sulfur and chlorine are present in 7.000g S2Cl2?
Percentage Composition
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• Empirical Formula shows only the ratio of elements in a compound.
• For example, hydrogen peroxide is H2O2, its empirical formula is HO.
What is the empirical formula of sodium oxalate?
Na2C2O4 = NaCO2
Empirical Formula
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from the given percentage composition, the empirical formula can be determined by:
Empirical Formula
Assume that the percentage given is its massADerive the number of moles from the given
massBGet the molar relationship based on the
smallest number of moleC
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Empirical Formula
What is the empirical formula of the compound that is 66.0% Ca and 34.0%P?
The compound has 66.0g Ca and 34.0g P
66.0g Ca = 1.65 mol34.0g P = 1.10 mol
Ca = 1.65 mol ÷ 1.10 mol = 1.5P = 1.10 mol ÷ 1.10 mol = 1
Since the ratio must be in whole number, doubling the 1.5 makes it 3. and by doubling one component, do the same to the rest
Ca3P21/7/2011 THE MOLE CONCEPT | REDG 29
Determine the empirical formula of the compound that is 34.31% sodium, 17.93% carbon and 47.76% oxygen.
Empirical Formula
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34.31Na = 1.492 1
22.99
17.93C = 1.493 1
12.01
47.76O = 2.985 2
16.00
NaCO2
Calculate the empirical formula of a compound that contains 1.67g cerium and 4.54g iodine.
Empirical Formula
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1.67Ce = 0.0119 1
140.12
4.54I = 0.0358 3
126.90
CeI3
2-Methylpropene is a compound used to synthetic rubber. A sample of 2-methylpropene contains 0.556g C and 0.0933g H. Determine its empirical formula.
Empirical Formula
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0.556C = 0.0463 1
12.01
0.0933H = 0.0924 2
1.01
CH2
The empirical formula of the compound is NaCO2. What is its
molecular formula?
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Comment on the problem below on determining the molecular formula of the compound
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The empirical formula of the compound is C2OH4
and a molar mass of 88 g/mol. What is the molecular formula of the compound ?
Molecular Formula
mass MF 88n = = 2
mass EF 44.06
So, MF = 2 x EF
2 x C2OH4 = C4O2H8
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An unknown compound has the following percentage composition 47.0% potassium, 14.5% carbon and 38.5% oxygen. If the molar mass of the compound is 166.22g/mol, what is the molecular formula of the compound?
47K = 1.20 1
39.10
14.5C = = 1.21 1
12.01
38.5O = = 2.41 2
16.00
Molecular Formula
mass MF 166.22n = 2
mass EF 83.11
So, MF= K2C2O4
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A sample of a compound that contains only carbon, hydrogen and sulfur was burned in oxygen and 9.682g of CO2, 4.956g H2O and 3.523g of SO2 were obtained.(a) What mass of carbon, hydrogen and sulfur were present before burning? (b) What was the mass of the sample that was burned? (c) What is the empirical formula of the compound?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
(a) What mass of carbon, hydrogen and sulfur were present before burning?
2 CO
12.01%C = x 100 27.29%
44.01in
2 CO 2
2
27.29gCgC = 9.682g CO 2.642gC
100g COin
2 2 CO 2 COgC = mass CO x %Cin in
2 CO 2
2
12.01gCgC = 9.682g CO 2.642gC
44.01g COin
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A sample of a compound that contains only carbon, hydrogen and sulfur was burned in oxygen and 9.682g of CO2, 4.956g H2O and 3.523g of SO2 were obtained.(a) What mass of carbon, hydrogen and sulfur were present before burning? (b) What was the mass of the sample that was burned? (c) What is the empirical formula of the compound?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
(a) What mass of carbon, hydrogen and sulfur were present before burning?
2 H O 2
2
2.02gHgH = 4.956g H O 0.5556gH
18.02g H Oin
2 SO 2
2
32.07gSgS = 3.523g SO 1.763gS
64.07g SOin
2 CO 2
2
12.01gCgC = 9.682g CO 2.642gC
44.01g COin
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A sample of a compound that contains only carbon, hydrogen and sulfur was burned in oxygen and 9.682g of CO2, 4.956g H2O and 3.523g of SO2 were obtained.(a) What mass of carbon, hydrogen and sulfur were present before burning? (b) What was the mass of the sample that was burned? (c) What is the empirical formula of the compound?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
(b) What was the mass of the sample that was burned?
g sample = g C+ g H + g S
= 2.642g + 0.5556g + 1.763g
= 4.961g
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A sample of a compound that contains only carbon, hydrogen and sulfur was burned in oxygen and 9.682g of CO2, 4.956g H2O and 3.523g of SO2 were obtained.(a) What mass of carbon, hydrogen and sulfur were present before burning? (b) What was the mass of the sample that was burned? (c) What is the empirical formula of the compound?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
(c) What is the empirical formula of the compound?
2.642C = 0.2200 4
12.010.5556
H = = 0.5501 101.01
1.763S = = 0.0550 1
32.07
So, EF= C4H10S
1/7/2011 THE MOLE CONCEPT | REDG 40
In an analysis of 5.21g sample of a compound that contains tin and chlorine, the chlorine in the sample was converted into AgCl. The process yields 11.47 g of AgCl. What is the empirical formula of the chloride of tin?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
35.45g Cl? g Cl = 11.47g AgCl = 2.837g Cl
143.32 g AgCl
? g Sn = 5.21 - 2.837 = 2.373g Sn-2
-2
1 mol Sn? mol Sn 2.373g Sn 1.999 x 10
118.71 g Sn
1 mol Cl? mol Cl 2.837g Cl 8.000 x 10
35.45g Cl
-2
-2
-2
-2
1.999 x 10Sn 1
1.999 x 10
8.000 x 10Cl 4
1.999 x 10
So, EF= SnCl4
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Caproic acid, the substance responsible for the aroma of dirty gym socks and running shoes, contains carbon, hydrogen, and oxygen. On combustion analysis, a 0.450 g sample of caproic acid gives 0.418 g of H2O and 1.023 g CO2 . What is the empirical formula of caproic acid? If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
2
2
2
2
12.01gCgC = 1.023g CO 0.2792gC
44.01g CO
2.02gCgH = 0.418g H O 0.0469gC
18.02g H O
gO = 0.450g sample - (0.2729gC + 0.0469gH)
= 0.124g O
0.2792C = 0.02325 3
12.010.0469
H = = 0.0464 61.01
0.124O = = 0.00775 1
16.00
So, EF= C3H6O
What is the empirical formula of caproic acid?
1/7/2011 THE MOLE CONCEPT | REDG 42
Caproic acid, the substance responsible for the aroma of dirty gym socks and running shoes, contains carbon, hydrogen, and oxygen. On combustion analysis, a 0.450 g sample of caproic acid gives 0.418 g of H2O and 1.023 g CO2 . What is the empirical formula of caproic acid? If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
EF= C3H6O
If the molar mass of caproic acid is 116.2 g/mol, what is the molecular formula?
mass MF 116.2n = 2
mass EF 58.09
MF= C6H12O2
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Cinnamaldehyde, a compound found in common cinnamon oil, contains carbon, hydrogen and oxygen. The combustion of a 6.50g sample yields 19.49g of CO2 and 3.54g H2O. What is the percentage composition of cinnamaldehyde?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
What is the percentage composition of cinnamaldehyde?
2
2
2
2
12.01 g C?g C 19.49 g CO 5.319gC
44.01 g CO
2.02 g H?g H 3.54 g H O 0.397 g H
18.02 g H O
?g O g sample - (gC gO)
6.50 g - (5.319 0.397)g 0.784 g O
12.1% 100 x 6.50
0.784 O %
6.11% 100 x 6.50
0.397 H %
81.8% 100 x 6.50
5.319 C %
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A plastic derived from methyl methacrylate contains carbon, hydrogen and oxygen. The combustion of a 12.62g sample of a plastic yields 27.73g of carbon dioxide and 9.09g of water. What is the percentage composition of the water?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
Solution and answer on the document Demanding Problems
1/7/2011 THE MOLE CONCEPT | REDG 45
Sulfur-containing compounds are undesirable component in some oils. The amount of sulfur in an oil can be determined by oxidizing the sulfur to SO4
2- and precipitating the sulfate ion as BaSO4, which can be dried and weighed. From a 6.300g sample of an oil, 1.063g of BaSO4 was obtained. What is the percentage of sulfur in an oil?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
Solution and answer on the document Demanding Problems
1/7/2011 THE MOLE CONCEPT | REDG 46
Muscone, the odor-bearing constituent of must, has a molecular formula C16H30O. (a) If 5.000g of muscone is burned in oxygen and if all the
compound was converted into CO2, what mass of CO2 is obtained?
(b) If all the hydrogen is converted into H2O, what mass of H2O is obtained?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
Solution and answer on the document Demanding Problems
1/7/2011 THE MOLE CONCEPT | REDG 47
In the analysis of 8.61g sample of a compound that contains chromium and chlorine, the chlorine in the sample was converted into AgCl.the process yields 20.08g AgCl. What is the empirical formula of the chloride of chromium?
Demanding Problems on Percent Composition, Empirical & Molecular Formulas
Solution and answer on the document Demanding Problems