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Introductory Chemistry:
A Foundation
FOURTH EDITIONby Steven S. Zumdahl
University of Illinois Copyright©2000 by HoughtonMifflin Company. All rights reserved. 2
Chemical Composition
Chapter 8
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Atomic Masses
• Balanced equation tells us the relative numbers of molecules of reactants and products
C + O2→ CO2
1 atom of C reacts with 1 molecule of O2 to make 1 molecule of CO2
• If I want to know how many O2molecules I will need or how many CO2 molecules I can make, I will need to know how many C atoms are in the sample of carbon I am starting with
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Atomic Masses
• Dalton used the percentages of elements in compounds and the chemical formulas to deduce the relative masses of atoms
• Unit is the amu.
– atomic mass unit – 1 amu = 1.66 x 10-24g
• We define the masses of atoms in terms of atomic mass units
– 1 Carbon atom = 12.01 amu, – 1 Oxygen atom = 16.00 amu
– 1 O2molecule = 2(16.00 amu) = 32.00 amu
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Atomic Masses
• Atomic masses allow us to convert weights into numbers of atoms
If our sample of carbon weighs 3.00 x 1020amu we
will have 2.50 x 1019atoms of carbon
atoms C 10 x 2.50 amu 12.01 atom C 1 x amu 10 x 3.00 20 = 19
Since our equation tells us that 1 C atom reacts with 1 O2molecule, if I have 2.50 x 1019C
atoms, I will need 2.50 x 1019molecules of O 2
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Example #1
• Determine the mass of 1 Al atom 1 atom of Al = 26.98 amu • Use the relationship as a conversion factor
amu 2024 atom Al 1 amu 26.98 x atoms Al 75 =
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Chemical Packages - Moles
• We use a package for atoms and molecules called a mole
• A mole is the number of particles equal to the number of Carbon atoms in 12 g of C-12 • One mole = 6.022 x 1023units
• The number of particles in 1 mole is called
Avogadro’s Number
• 1 mole of C atoms weighs 12.01 g and has 6.02 x 1023atoms
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The Mole
• When two chemicals react with each other, we want to know how many atoms or molecules of each are used do that a formula of the product can be established.
• This means we need a way to count atoms, but HOW?
• The solution to this problem is to define a convenient unit of matter that contains a know number of particles.
• 1 moles = 6.022136736×1023particles
– This value is commonly known has Avogadro’s number.
• Remember that one mole always contains the same number of particles, no matter what the substance.
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Molar Mass
• The mass of one mole of atoms of any element is the molar mass, which is numerically equal to the atomic mass, but in grams.
• Example:
– 1 mole of sodium weighs 22.99 g
• This not only true for atoms, but also for molecules, but we use the molecular mass instead of the atomic mass.
– To determine the molecular mass of a molecule, we add up the atomic masses of the elements that make up the molecule.
• Example:
– CCl4
» We have 1 carbon and 4 chlorines » 12.001 g/mol + 4(35.4527 g/mol) = 153.81 g/mol » Therefore, if we have 1 mole of CCl4it would weigh 153.81 g.
Molecules, Compounds, and The
Mole
• Thinking about moles with respect to molecules and compounds is not all that different than with atoms.
• If I have 1 mole of CH4:
1. How many CH4molecules do I have? • I have 6.022 ×1023molecules of CH
4
2. How much does it weigh?
• 16.032 g
3. How many carbon atoms do I have?
• 6.022 ×1023
4. How many hydrogen atoms do I have?
• 4 × 6.022 ×1023= 2.409×1024
MOLES are the CENTER
# of moles of X # of particles Vol. of an “ideal” gas @ STP Mass a b c d e f a) Multiply by Avogadro’s Number b) Divide by Avogadro’s Number c) Multiply by the molecular weight of X d) Divide by the molecular
weight of X
e) Multiply by 22.4 L / mol f) Divide by 22.4 L / mol
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EXAMPLES
• What is the mass, in grams, of 1.5 mols of silicon?
42.1283 g • How many moles are
represented by 454 g of sulfur? How many atoms?
14.2 mols 8.53×1024atoms
• Calculate the molar mass of CaCO3(limestone).
100.08 g/mol • If you have 454 g of CaCO3,
how many moles do you have? How many molecules? How many atoms of oxygen?
4.54 mols 2.73×1023molecules
8.20×1024atom of O
At first these concepts are difficult to understand, but working problems is the best way to become comfortable with these concepts.
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Example #2
Use the Periodic Table to determine the mass of 1 mole of Al
1 mole Al = 26.98 g Use this as a conversion factor for
grams-to-moles Al mol 0.371 g 26.98 Al mol 1 x Al g 10.0 =
Compute the number of moles and number of atoms in 10.0 g of Al
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 15 Use Avogadro’s Number to determine the
number of atoms in 1 mole
1 mole Al = 6.02 x 1023atoms
Use this as a conversion factor for moles-to-atoms atoms Al 10 x 2.23 Al mol 1 atoms 10 x 6.02 x Al mol 0.371 23 = 23
Example #2
Compute the number of moles and number of atoms in 10.0 g of AlCopyright©2000 by Houghton
Mifflin Company. All rights reserved. 16 Use Avogadro’s Number to determine the
number of atoms in 1 mole
1 mole Al = 6.02 x 1023atoms
Use this as a conversion factor for atoms-to-moles Al mol 0.370 atoms 10 x 6.02 Al mol 1 x atoms Al 10 x 2.23 23 23 =
Compute the number of moles and mass of 2.23 x 1023atoms of Al
Example #3
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 17 Use the Periodic Table to determine the
mass of 1 mole of Al
1 mole Al = 26.98 g Use this as a conversion factor for
moles-to-grams Al g 9.99 Al mol 1 g 26.98 x Al mol 0.370 =
Compute the number of moles and mass of 2.23 x 1023atoms of Al
Example #3
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Molar Mass
• The molar mass is the mass in grams of one mole of a compound
• The relative weights of molecules can be calculated from atomic masses
water = H2O = 2(1.008 amu) + 16.00 amu = 18.02 amu
• 1 mole of H2O will weigh 18.02 g, therefore
the molar mass of H2O is 18.02 g
• 1 mole of H2O will contain 16.00 g of oxygen and 2.02 g of hydrogen
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 19 • Percentage of each element in a compound
– By mass
• Can be determined from ° the formula of the compound or ± the experimental mass analysis of the
compound
• The percentages may not always total to 100% due to rounding
100%
whole
part
Percentage
=
×
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Percent Composition
• Percent Composition
– The percentage of the mass of a compound represented by each of its constituent elements.
• According to the law of constant composition, any sample of a pure compound always consists of the same elements combined in the same proportion by mass.
• Example: What is the percent composition of N and H in ammonia (NH3)? 82.27% 100 NH g 17.03 N g 14.01 NH of mol 1 of Mass NH of mol 1 in N of Mass NH in N Percent Mass 3 3 3 3= = × = 17.76% 100 NH g 17.03 H g 3(1.008) NH of mol 1 of Mass NH of mol 1 in H of Mass NH in H Percent Mass 3 3 3 3= = × = Copyright©2000 by Houghton
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Empirical and Molecular Formulas
• We have discussed molecular formulas • Empirical formulas
– A molecular formula showing the simplest possible ratio of atoms in a molecule.
• From percent compositions we can determine empirical and molecular formulas by the following procedure. % A → g A → x mols A % B → g B → x mols B y molB A mol x Find mole
ratio Ratio givesformula
AxBy
Copyright©2000 by Houghton
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• Eugenol is the active component of oil of cloves. It has a molar mass of 164.2 g/mol and is 73.14% C and 7.37% H; the remainder is oxygen. What are empirical and molecular formulas for eugenol?
• Where to start?!
1. What is the % of oxygen? • 100% - (73.14% + 7.37%) = 19.49% 2. How many grams of C, H, and O?
• Assume you have 100 g of the compound therefore your percentages are the number of grams of C, H, and O.
– 73.14 g of C – 7.37 g of H – 19.49 g of O
3. Change the grams into moles.
• 73.14 g of C × 1 mole of C / 12.011g of C = 6.09 mols of C • 7.37 g of H × 1 mole of H / 1.0079 g of H = 7.31 mols of H • 19.49 g of O × 1 mole of O / 15.9994 g of O = 1.22 mols of O
4. Find the mole ratios (divide the large amount(s) of moles by the smallest amount of moles). • 6.09 mols of C / 1.22 mols of O = 5 mols of C to 1 mol of O
• 7.37 mols of H / 1.22 mols of O = 6 mols of H to 1 mol of O 5. So what does this mean?
• C5H6O is the empirical formula
6. How to get the molecular formula? • Molecular weight of the empirical formula.
– 82 g/mol
• Divide the molecular weight of eugenol by the molecular weight empirical formula.
– 164.2 g/mol / 82 g/mol = 2
• So there are 2 units of the empirical formula.
– (C5H6O)2= C10H12O2
Hydrated Compounds
• If ionic compounds are prepared water solution and then isolated as solids, the crystals often have molecules of water trapped in the lattice.
• Compounds in which molecules of water are associated with the ions of the compound are called hydrated compounds.
• Examples:
– NiCl2• 6H2O
– CaSO4• 2H2O
• But what about the molar mass? (Do I add the water?)
– YES!
– NiCl2• 6H2O → 237.7 g/mol • 129.6 g/mol from NiCl2 • 108.1 g/mol from 6H2O
– CaSO • 2HO → 172.14 g/mol
Determining the Units of Hydration
• Example: The following reaction takes place. 1.023 g of CuSO4· x H2O + heat → 0.654 g CuSO4+ ? g H2O
• Find the units of hydration. • Step 1: determine the mass of water
1.023 g of CuSO4· x H2O – 0.654 g of CuSO4= 0.369 g of water
• Step 2: determine the number of moles
0.369 g of water × 1 mol of water / 18.02 g of water = 0.0205 mols of water 0.654 g CuSO4× 1 mols CuSO4/ 159.6 g CuSO4= 0.00410 mols CuSO4
• Step 3: determine the molar ratio
• So, we know that we started with 1.023 g of CuSO4· 5 H2O 4 2 4 2 CuSO mol 1.00 O H mol 5.00 CuSO mol 0.00410 O H mol 0.0205 =
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 25 Determine the mass of each element in 1
mole of the compound
2 moles C = 2(12.01 g) = 24.02 g 6 moles H = 6(1.008 g) = 6.048 g
1 mol O = 1(16.00 g) = 16.00 g
Determine the molar mass of the compound by adding the masses of the elements
1 mole C2H5OH = 46.07 g
Determine the Percent Composition from the Formula C2H5OH
Example #4
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 26 Divide the mass of each element by the molar
mass of the compound and multiply by 100% 52.14%C 100% 46.07g 24.02g× = 13.13%H 100% 46.07g 6.048g× = 34.73%O 100% 46.07g 16.00g× =
Determine the Percent Composition from the Formula C2H5OH
Example #4
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Empirical Formulas
• The simplest, whole-number ratio of atoms in a molecule is called the Empirical Formula
– can be determined from percent composition or combining masses
• The Molecular Formula is a multiple of the Empirical Formula % A mass A (g) moles A100g MMA % B mass B (g) moles B100g MMB moles A moles B Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 28 Find the greatest common factor (GCF) of the
subscripts
factors of 20 = (10 x 2), (5 x 4) factors of 12 = (6 x 2), (4 x 3)
GCF = 4
Divide each subscript by the GCF to get the empirical formula
C20H12= (C5H3)4 Empirical Formula = C5H3 Determine the Empirical Formula of
Benzopyrene, C20H12
Example #5
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 29 Convert the percentages to grams by
assuming you have 100 g of the compound
– Step can be skipped if given masses
47gC 100g 47gC 100g× = 47gO 100g 47gO 100g× = 6.0gH 100g 6.0gH 100g× = Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen
Example #6
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 30 Convert the grams to moles
C mol 3.9 12.01g C mol 1 C 47g × = O mol 2.9 16.00g O mol 1 O g 47 × = H mol 6.0 1.008g H mol 1 H g 6.0 × =
Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 31 Divide each by the smallest number of moles
1.3 2.9 C mol 3.9 ÷ = 1 2.9 O mol 2.9 ÷ = 2 2.9 H mol 6.0 ÷ =
Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 32 If any of the ratios is not a whole number, multiply
all the ratios by a factor to make it a whole number
– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then multiply by 3; if ?.25 or ?.75 then multiply by 4
4 3 x 1.3 2.9 C mol 3.9 ÷ = = 3 3 x 1 2.9 O mol 2.9 ÷ = = 6 3 x 2 2.9 H mol 6.0 ÷ = =
Multiply all the Ratios by 3 Because C is 1.3
Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 33 ° Use the ratios as the subscripts in the
empirical formula 4 3 x 1.3 2.9 C mol 3.9 ÷ = = 3 3 x 1 2.9 O mol 2.9 ÷ = = 6 3 x 2 2.9 H mol 6.0 ÷ = = C 4H6O3
Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen
Example #6
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Molecular Formulas
• The molecular formula is a multiple of the empirical formula
• To determine the molecular formula you need to know the empirical formula and the molar mass of the compound
Determine the empirical formula
• May need to calculate it as previous C5H3
Determine the molar mass of the empirical formula
5 C = 60.05 g, 3 H = 3.024 g C5H3= 63.07 g
Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an
empirical formula of C5H3
Example #7
Divide the given molar mass of the compound by the molar mass of the empirical formula
– Round to the nearest whole number
4
07
.
63
252
=
g
g
Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an
empirical formula of C5H3
Example #7
Copyright©2000 by Houghton
Mifflin Company. All rights reserved. 37 Multiply the empirical formula by the
calculated factor to give the molecular formula
(C5H3)4= C20H12
Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an
empirical formula of C5H3
Example #7
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Homework
• Questions and Problems:
• 6, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84