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Chapter 7 Chemical Quantities. 7.1 The Mole 7.2 Molar Mass 7.3 Calculations Using Molar Mass 7.4 Percent Composition and Empirical Formulas. 7.1 A Mole. A mole contains 6.02 x 10 23 particles (atoms, ions, molecules, formula unit) - PowerPoint PPT Presentation
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1Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Chapter 7 Chemical Quantities
7.1 The Mole7.2 Molar Mass7.3 Calculations Using Molar Mass7.4 Percent Composition and Empirical
Formulas
2Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A mole contains 6.02 x 1023 particles (atoms, ions, molecules, formula unit)
The number 6.02 x 1023 is known as Avogadro’s number.
One mole of any element contains Avogadro’s number of atoms.
1 mole Na = 6.02 x 1023 Na atoms1 mole Au = 6.02 x 1023 Au atoms
7.1 A Mole
3Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
One mole of a covalent compound contains Avogadro’s number of molecules.1 mole CO2 = 6.02 x 1023 CO2 molecules
1 mole H2O = 6.02 x 1023 H2O molecules
One mole of an ionic compound contains Avogadro’s number of formula units.1 mole NaCl = 6.02 x 1023 NaCl formula units
A Mole of Molecules
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Samples of One Mole Quantities
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Avogadro’s number is written as conversion factors.
6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles The number of molecules in 0.50 mole of CO2
molecules is calculated as
0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules
1 mole CO2 molecules
= 3.0 x 1023 CO2 molecules
Avogadro’s Number
6Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A. Calculate the number of atoms in 2.0 moles of Al.
B. Calculate the number of moles of S in 1.8 x 1024 S.
Learning Check
7Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A. Calculate the number of atoms in 2.0 moles of Al. 2.0 moles Al x 6.02 x 1023 Al atoms
1 mole Al=1.2 x 1024 Al atoms
B. Calculate the number of moles of S in 1.8 x 1024 S. 1.8 x 1024 S atoms x 1 mole S
6.02 x 1023 S atoms
= 3.0 mole S atoms
Solution
8Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
The mass of one mole is called molar mass (g/mole).
The molar mass of an element is the atomic mass expressed in grams.
7.2 Molar Mass
9Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms = ________
B. 1 mole of Sn atoms = ________
Learning Check
10Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms =39.1 g
B. 1 mole of Sn atoms = 118.7 g
Solution
11Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Molar Mass of CaCl2
For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows.
Formula mass of CaCl2 = [40.1 + 2(35.45)] = 111.1amu Formula mass = molar mass, so
111.1amu = 111.1g/mol
12Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Molar Mass of K3PO4
Determine the molar mass of K3PO4 to 0.1 g.
Molar Mass = [3(39.1) + 31.0 + 4(16)] = 212.3g/mol
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One-Mole Quantities
32.1 g 55.9 g 58.5 g 294.2 g 342.3 g
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A. 1 mole of K2O = ______g
B. 1 mole of antacid Al(OH)3 = ______g
Learning Check
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A. 1 mole of K2O
2 (39.1) + 1 (16.0) = 94.2 g/mol
1mole K2O x 94.2g/mol = 94.2g
B. 1 mole of antacid Al(OH)3
1 (27.0) + 3 (16.0) + 3 (1.0) = 78.0 g/mol1mole Al(OH)3 x 78.0 g/mol = 78.0g
Solution
16Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?
Learning Check
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Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?
17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) =
204 + 18 + 57.0 + 14.0 + 16.0
= 309 g/mole
Solution
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Methane CH4 known as natural gas is used in gas cook tops and gas heaters.
1 mole CH4 = 16.0 g
The molar mass of methane can be written as conversion factors.
16.0 g CH4 and 1 mole CH4
1 mole CH4 16.0 g CH4
Molar Mass Factors
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Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.
Learning Check
20Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid.
2(12.0) + 4(1.0) + 2(16) = 60.0g/mol
1 mole of acetic acid = 60.0 g acetic acid
1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid
Solution
21Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Mole factors are used to convert between the grams of a substance and the number of moles.
7.3 Calculations with Molar Mass
Grams Mole factor Moles
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Aluminum is often used for the structure oflightweight bicycle frames. How many gramsof Al are in 3.00 moles of Al?
3.00 moles Al x 27.0 g Al = 81.0 g Al1 mole Al
mole factor for Al
Calculating Grams from Moles
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The artificial sweetener aspartame (Nutri-Sweet) C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?
Learning Check
24Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Calculate the molar mass of C14H18N2O5.
14 (12.0) + 18 (1.0) + 2 (14.0) + 5(16.0) = 294 g/mole
Set up the calculation using a mole factor. 225 g aspartame x 1 mole aspartame
294 g aspartame
mole factor(inverted)
= 0.765 mole aspartame
Solution
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7.4 Percent Composition In a compound, the percent composition is the
percent by mass of each element in the formula. In one mole of CO2 there are 12.0 g of C and
32.0 g of O (molar mass 44.0 g/mol),
12.0 g C x 100 = 27.3 % C
44.0 g CO2 32.0 g O x 100 = 72.7 % O
44.0 g CO2 100.0 %
26Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
What is the percent carbon in C5H8NNaO4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats?
1) 7.10 %C
2) 35.5 %C
3) 60.0 %C
Learning Check
27Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
2) 35.5 %C
Molar mass = 169.1 g
% = total g C x 100 total g MSG
= 60.0 g C x 100 = 35.5 % C 169.1 g MSG
Solution
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The molecular formula is the true or actual number of the atoms in a molecule.
The empirical formula is the simplest whole number ratio of the atoms.
The empirical formula is calculated by dividing the subscripts in the molecular formula by a whole number to give the lowest ratio.
C5H10O5 5 = C1H2O1 = CH2Omolecular empirical formulaformula
Types of Formulas
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Some Molecular and Empirical Formulas
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A. What is the empirical formula for C4H8?
1) C2H4 2) CH2 3) CH
B. What is the empirical formula for C8H14?
1) C4H7 2) C6H12 3) C8H14
C. Which is a possible molecular formula for CH2O?
1) C4H4O4 2) C2H4O2 3) C3H6O3
Learning Check
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A. What is the empirical formula for C4H8?
2) CH2 C4H8 4
B. What is the empirical formula for C8H14?
1) C4H7 C8H14 2
C. Which is a possible molecular formula for CH2O?
2) C2H4O2 3) C3H6O3
Solution
32Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.
1) SN
2) SN4
3) S4N4
Learning Check
33Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.
3) S4N4
If the molecular formula has 4 atoms of N, and S and N are related 1:1, then there must also be 4 atoms of S.
Solution
34Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A molecular formula is equal to or a multiple of the empirical formula.
Thus, the molar mass is equal to or a multiple of the empirical mass. molar mass = a whole number empirical mass
Multiply the empirical formula by the whole number to determine the molecular formula.
Relating Empirical and Molecular Formulas
35Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Determine the molecular formula of a compoundthat has a molar mass of 78.0 and an empiricalformula of CH. 1. Empirical mass of CH = 13.0 g/mol2. Divide the molar mass by the empirical mass.3. 78.0 g/mol = 6.00 13.0 g/mol4. Multiply the subscripts in CH by 6.5. Molecular formula = (CH)6 = C6H6
Finding the Molecular Formula
36Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?
Learning Check
37Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?
C3H4O3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol
176.0 g/mol (molar mass) = 2.00 88.0 g/mol (empirical mass)
Molecular formula = 2 (empirical formula) (C3H4O3 )2 = C6H8O6
Solution
38Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A compound is C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is known to be 99.0 g. What are the empirical and molecular formulas? 1. Write the mass percents as the grams in a 100.00-g sample of the compound.
C 24.27 g H 4.07 g Cl 71.65 g
Finding the Molecular Formula
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Finding the Molecular Formula Continued
2. Calculate the number of moles of each element.
24.27 g C x 1 mole C = 2.02 moles C 12.0 g C
4.07 g H x 1 mole H = 4.03 moles H 1.01 g H
71.65 g Cl x 1 mole Cl = 2.02 moles Cl 35.5 g Cl
40Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Finding the Molecular Formula (continued)
3. Divide each by the smallest 2.02 moles C = 1 mole C 2.024.03 moles H = 2 moles H 2.022.02 moles Cl = 1 mole Cl 2.02
Empirical formula = C1H2Cl1 = CH2Cl
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Finding the Molecular Formula (continued)
4. Calculate empirical mass (EM)empirical mass CH2Cl = 49.5 g/mol
5. Divide molar mass by empirical mass Molar mass = 99.0 g/mol = 2Empirical mass 49.5 g/mol
6. Determine Molecular formula(CH2Cl)2 = C2H4Cl2
42Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.
Learning Check
43Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.
60.0 g C x 1 mole C = 5.00 moles C 12.0 g C
4.5 g H x 1 mole H = 4.5 moles H 1.01 g H
35.5 g O x 1mole O = 2.22 moles O 16.0 g O
Solution (continued)
44Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Solution (continued)
Divide by the smallest number of moles.5.00 moles C = 2.25 moles C
2.22 4.5 moles H = 2.0 moles H 2.222.22 moles O = 1.00 mole O 2.22Note that the results are not all whole numbers. To obtain whole numbers, multiply by a factor to give whole numbers.
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Solution (continued)
Multiply each number of moles by 4C: 2.25 moles C x 4 = 9 moles CH: 2.0 moles H x 4 = 8 moles HO: 1.00 mole O x 4 = 4 moles O
Use the whole numbers as subscripts toobtain the simplest formula
C9H8O4
46Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?
Learning Check
47Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
In 100.0 g, there are 27.4 g S, 12.0 g N and 60.6 g Cl. 27.4 g S x 1 mole S = 0.854 mole S
32.1 g S
12.0 g N x 1 mole N = 0.857 moles N 14.0 g N
60.6 g Cl x 1mole Cl = 1.71 moles Cl 35.5 g Cl
Solution
48Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Dividing by the smallest number of moles
0.854 mole S /0.854 = 1.00 mole S
0.857 mole N/0.854 = 1.00 mole N
1.71 moles Cl/0.854 = 2.00 moles Cl
Empirical formula = SNCl2 = 117.1 g/mol
Molar Mass/ Empirical mass
351 = 3117.1 Molecular formula = (SNCl2)3 = S3N3Cl6
Solution (continued)