A Mole of Molecules Samples of One Mole...
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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 asAvogadro’s number.
One mole of any element contains Avogadro’snumber of atoms.
1 mole Na = 6.02 x 1023 Na atoms1 mole Au = 6.02 x 1023 Au atoms
7.1 A Mole
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One mole of a covalent compound containsAvogadro’s number of molecules.1 mole CO2 = 6.02 x 1023 CO2 molecules1 mole H2O = 6.02 x 1023 H2O molecules
One mole of an ionic compound containsAvogadro’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 conversionfactors.
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 molesof Al.
B. Calculate the number of moles of S in 1.8 x1024 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 ofone mole iscalled molarmass (g/mole).
The molarmass of anelement is theatomic massexpressed ingrams.
7.2 Molar Mass
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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 themolar masses of the elements in the formula. Wecalculate the molar mass of CaCl2 to the nearest 0.1 gas 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/mol
1mole 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 thatinhibits the uptake of serotonin by the brain.What is the molar mass of Prozac?
Learning Check
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17Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Prozac, C17H18F3NO, is a widely used antidepressantthat 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 ingas cook tops and gas heaters.
1 mole CH4 = 16.0 g
The molar mass of methane can be written asconversion 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 aceticacid.
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 acid60.0 g acetic acid 1 mole acetic acid
Solution
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21Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
Mole factors are used to convert between thegrams 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 andsoft drinks. How many moles of aspartame arepresent 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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25Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
7.4 Percent Composition
In a compound, the percent composition is thepercent by mass of each element in the formula.
In one mole of CO2 there are 12.0 g of C and32.0 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 compoundused 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 actualnumber of the atoms in a molecule.
The empirical formula is the simplest wholenumber ratio of the atoms.
The empirical formula is calculated by dividingthe subscripts in the molecular formula by awhole number to give the lowest ratio.
C5H10O5 ∏ 5 = C1H2O1 = CH2Omolecular empirical formulaformula
Types of Formulas
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Some Molecular and EmpiricalFormulas
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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, whatis the molecular formula if SN is the empiricalformula? Explain.
1) SN
2) SN4
3) S4N4
Learning Check
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33Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
If the molecular formula has 4 atoms of N, whatis the molecular formula if SN is the empiricalformula? Explain.3) S4N4
If the molecular formula has 4 atoms of N, andS and N are related 1:1, then there must also be4 atoms of S.
Solution
34Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A molecular formula is equal to or a multipleof the empirical formula.
Thus, the molar mass is equal to or a multipleof the empirical mass. molar mass = a whole numberempirical mass
Multiply the empirical formula by the wholenumber to determine the molecular formula.
Relating Empirical and MolecularFormulas
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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 andan empirical formula of C3H4O3. What is themolecular formula?
Learning Check
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37Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.
A compound has a formula mass of 176.0 and anempirical formula of C3H4O3. What is the molecularformula?
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 Cl71.65%. 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 FormulaContinued
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 smallest2.02 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)
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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 wholenumbers. 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 and60.6 % Cl. If the compound has a molarmass of 351 g, what is the molecularformula?
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 and60.6 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 = 3117.1Molecular formula = (SNCl2)3 = S3N3Cl6
Solution (continued)