Chapter 7
Introduction to General, Organic, and Biochemistry 10e
John Wiley & Sons, Inc
Morris Hein, Scott Pattison, and Susan Arena
Quantitative Composition of Compounds
These black pearls are made of layers of calcium carbonate. They can be measured by counting or weighing.
Chapter Outline
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7.1 The Mole
7.2 Molar Mass of Compounds
7.3 Percent Composition of Compounds
7.4 Empirical Formula versus Molecular Formula
7.5 Calculating Empirical Formulas
7.6 Calculating the Molecular Formula from the Empirical Formula
Convenient Ways of Counting Items
• 1 dozen eggs = 12 eggs• 1 ream paper = 500 sheets• 1 gross pencils = 144 pencils
How do chemists count really small things, like atoms?
1 mole atoms =
6.022×1023 atoms
This number is known as Avogadro’s number.
:
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1×10 -10 m
Review Question 1
The Mole
1 mole of anything contains Avogadro’s number (6.022×1023 ) of particles.
• 1 mol atoms = 6.022×1023 atoms• 1 mol molecules = 6.022×1023 molecules• 1 mol ions = 6.022×1023 ions
Avogadro’s number can be used as a conversion factor:
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236.022 10 particles
1 mol
23
1 mol
6.022 10 particles
Review Question 7 & 8
The Mole
Calculate the number of atoms in 2.4 mol Na.
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2.4 mol Na236.022 x 10 Na atoms
1 mol Na
24= 1.4 10 Na atoms
Plan
Calculate
2.5 mol Na Na atoms
236.022 x 10 Na atoms
1 mol Na23
1 mol Na
6.022 x 10 Na atoms
Your Turn!
How many moles of HCl are present in 4.3×1023 molecules?
a. 2.6×1047 mol
b. 0.71 mol
c. 2.6 mol
d. 7.1×1045 mol
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4.3 x10 23 molecules 1 mol = 0.71 mol
6.022 x10 23 molecules
Convenient Ways of Counting
It is often convenient to count using mass.
For example, a canning recipe calls for 150 apples to be peeled and cored.
The average mass of an individual apple is 235 g. How many kg are needed to complete this recipe?
The canner should buy 35 kg of apple.
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235 g 1 kg150 apples × = 35 kg
1 apple 1000 g
Counting Atoms with Mass
• Chemists count atoms by using mass since individual atoms are too small to count.
• Average mass of an atom = atomic mass in amu• Average mass of 1 mole of atoms = atomic mass
expressed in grams (molar mass).
1 mol atoms = atomic mass in grams
1 mol atoms = 6.022×1023 atoms
atomic mass in grams = 6.022×1023 atoms
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Review Question 5
The Mole
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Your Turn!
Which of these is not correct?
a. The mass of 1 atom of C is 12.01 amu.
b. The mass of 1 mole of C atoms is 12.01 g.
c. Avogadro’s number of atoms has a mass of 12.01 amu.
d. Avogadro’s number of atoms has a mass of 12.01 g.
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C6
12.01
The Mole
Calculate the mass of 2.4 mol C.
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Plan
Calculate
12.01g C
1 mol C
1 mol C
12.01g C
2.4 mol C g
2.4 mol C12.01g C
1 mol C
= 29 g C
atomic mass
C 12.01
Molar mass is a conversion factor:
Your Turn!
What is the correct set up to calculate number of moles of atoms contained in 3.52 g Al?
a.
b.
c.
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1 mol Al3.52g Al
26.98 g Al
26.98g Al3.52g Al
1 mol Al
236.022 x 10 Al atoms3.52g Al
1 g Al
atomic mass
Al 26.98
The Mole
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Calculate the number of atoms in 36.0 g C.
Plan
Calculate
236.022 10 atoms C
1 mol C
36.0 g C moles C atoms C
36.0 g C236.022 10 atoms C
1 mol C
24= 1.8 10 atoms C
atomic mass
C 12.01
1 mol C
12.01 g C
1 mol C
12.01 g C
Your Turn!
What is the mass of 3.01 ×1023 atoms of lead?
a. 104 g
b. 414 g
c. 0.500 g
d. 1.04×1048 g
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atomic mass
Pb 207.2
3.01 x10 23 atoms 1 mol 207.2 gram = 104 gram
6.022 x10 23 atoms 1 mol
Your Turn!
1 gram of which of the following elements would contain the largest number of atoms?
a. nitrogen
b. hydrogen
c. phosphorus
d. oxygen
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atomic mass
N 14.01
H 1.01
P 30.97
O 16.00
1 gram H 1 mol 6.022 x10 23 atom = 5.96 x10 23 atoms
1.01 g 1 mol
1 gram P 1 mol 6.022 x10 23 atom = 1.94 x10 22 atoms
30.97 g 1 mol
Review Question 3
Your Turn!
Which has a higher mass, mol of K or mol of Au?
a. potassium
b. gold
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atomic mass
K 39.0983
Au 196.96654
1 Mol of K 39.0983 gram = 39.0983 gram
1 mol
1 Mol of Au 196.9665 gram = 196.9665 gram
1 mol
Review Question 2
Your Turn!
Which has more electrons, mol of K or mol of Au?
a. potassium
b. gold
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1 Mol K 6.022 x10 23 atom 19 e- = 1.144 x10 25 E-
1 mol 1 atom
1 Mole Au 6.022 x10 23 atom 79 e- = 4.757 x10 25 E-
1 mol atom
atomic number
K 19
Au 79
Review Question 4
Review Question 9
6.022 x10 23
6.022 x10 23
1.204 x10 24
15.99 g
31.98 g
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1 Mole of O 15.99 gram = 15.99 gram
1 mole
1 Mole O 2 2 Mole of O 15.99 gram = 31.98 gram
1 Mole O 2 1 mole
Review Question 9
a. Mole of O atoms = ? Atoms
b. Mole of O2 molecules = ? Molecules
c. Mole of O2 molecules = ? atoms
d. Mole of O atoms= ? grams
e. Mole of O2 molecules = ? grams
Molar Mass of Compounds
1 mol compound = 6.022×1023 formula units compound
Molar mass of compound = mass of 1 mol compound
The molar mass of a compound is the sum of the atomic masses of each atom in the compound.
What is the molar mass of CO2?
1C 1(12.01g)
2O 2(16.00g)
CO2 44.01g/mol
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atomic mass
C 12.01
O 16.00
Molar Mass of Compounds
What is the molar mass of Al2(CO3)3?
2Al 2(26.98g)
3C 3(12.01g)
9O 9(16.00g)
Al2(CO3)3 233.99g/mol
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atomic mass
Al 26.98
C 12.01
O 16.00
Your Turn!
What is the molar mass of (NH4)3PO4?
a. 141.04g/mol
b. 144.07g/mol
c. 146.09g/mol
d. 149.12g/mol
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atomic mass
N 14.01
H 1.01
P 30.97
O 16.00
3*1 N 3(14.01g)3*4 H 12(1.01g)1 P 1(30.97g)4 O 4(16.00g)
Your Turn!
What is the molar mass of Mg(ClO4)2?
a. 301.01g/mol
b. 191.21g/mol
c. 223.21g/mol
d. 123.76g/mol
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atomic mass
Mg 24.31
Cl 35.45
O 16.00
__ Mg __(____g)__ Cl __(____g)__ O __(____g)
Using Molar Masses of Compounds
Molar mass = 1 mol = 6.022×1023 formula units
Calculate the mass of 0.150 mol Mg(ClO4)2
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Plan
Calculate
4 2
4 2
1 mol Mg(ClO )
223.21 g Mg(ClO )4 2
4 2
223.21 g Mg(ClO )
1 mol Mg(ClO )
0.150 mol Mg(ClO4)2 g Mg(ClO4)2
4 20.150 mol Mg(ClO ) 4 2= 33.5 g Mg(ClO )4 2
4 2
223.21 g Mg(ClO )
1 mol Mg(ClO )
Using Molar Masses of Compounds
Molar mass = 1 mol = 6.022×1023 formula units
Calculate the number of moles in 35 g H2O.
Copyright 2012 John Wiley & Sons, Inc
Plan
Calculate
2
2
1 mol H O
18.02 g H O2
2
18.02 g H O
1 mol H O
35 g H2O moles H2O
235 g H O 2= 1.9 g H O 2
2
1 mol H O
18.02 g H O
atomic mass
H 1.01
O 16.00
Using Molar Masses of Compounds
Molar mass = 1 mol = 6.022×1023 formula units
Calculate the number of molecules in 35 g H2O.
Copyright 2012 John Wiley & Sons, Inc
Plan
Calculate
232
2
6.022 10 molecules H O
1 mol H O
2
2
18.02 g H O
1 mol H O
35 g H2O mol H2O molecules H2O
235.0 g H O
242= 1.2 10 molecules H O
2
2
1 mol H O
18.02 g H O
232
2
6.022 10 molecules H O
1 mol H O
Preview Question 10
How many molecules are present in 1 molar mass of sulfuric acid (H2SO4)?
1. What is the molar mass of H2SO4?
2 H 2(1.01g)
1 S 1(32.06g)
4 O 4(16.00g) = 98.08g/mol
2. How many atoms in one molecule? 2+1+4=7
How many atoms in 1 molar mass? ___
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atomic mass
S 32.06
H 1.01
O 16.00
98.08 g 1 mol 6.022 x10 23 molecules = 6.022 x10 23 molecules
98.08 g 1 mol
Your Turn!
How many moles of molecules are present in 146 g of glucose (C6H12O6)?
a. 180. mol
b. 0.810 mol
c. 26300 mol
d. 4.88×1023 mol
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atomic mass
C 12.01
H 1.01
O 16.00
g = molecules
__ C __(____g)__ H __(____g)__ O __(____g) = _____ g/mol
Your Turn!
What is the mass of 1.20 ×1023 molecules of CH3OH?
a. 161g
b. 38.5g
c. 32.1g
d. 6.39g
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atomic mass
C 12.01
H 1.01
O 16.00
molecules = g
__ C __(____g)__ H __(____g)__ O __(____g) = _____ g/mol
Your Turn!
How many molecules are present in 4.21 moles of HBr?
a. 2.53×1023 molecules
b. 2.53×1024 molecules
c. 6.99×10-24 molecules
d. 3.97×102 molecules
e. 6.99×1024 molecules
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atomic mass
H 1.01
Br 79.90
Percent Composition of Compounds
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Percent composition is a list of the mass percent of each element in a compound.
Na2CO3 is
43.38% Na 11.33% C 45.29% O
How do we calculate the mass percent of Na2CO3?
First determine the molar mass of Na2CO3
2(22.99g Na) + 1(12.01g C) + 3(16.00 g O) = 105.99g/mol Na2CO3
Then find ratio of the mass of each element to the mass of the compound.
2 3
2 22.99 g Nax100 43.38% Na
105.99 g Na CO
2 3
1 12.01 g Cx100 11.33% C
105.99 g Na CO
2 3
3 16.00 g Ox100 45.29% O
105.99 g N a CO
Calculating Percent Composition
Calculating Percent Composition
A compound is found to consist of 2.74g of iron and 5.24g of chlorine. What is the percent composition of the compound?
1. Calculate the mass of the product formed:
2.74g Fe + 5.24g Cl = 7.98g product
2. Calculate the percent for each element.
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2.74 g Fex100 = 34.3% Fe
7.98 g
5.24 g Clx100 = 65.7% Cl
7.98 g
Your Turn!
What is the percent carbon in acetic acid, HC2H3O2?
a. 41.01% C
b. 20.00% C
c. 6.73% C
d. 39.99% C
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atomic mass
C 12.01
H 1.01
O 16.00
__ C __(____g)__ H __(____g)__ O __(____g) = _____ g/mol
First determine the molar mass of Na2CO3
Your Turn!
A 6.00g sample of calcium sulfide is found to contain 3.33g of calcium. What is the percent by mass of sulfur in the compound?
a. 80.2% S
b. 55.5% S
c. 44.5% S
d. 28.6% S
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Empirical Formula versus Molecular Formula
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The empirical formula is the lowest whole number ratio of atoms in a compound.
C6H12O6
CH2O
(CH2O)6
The molecular formula for a substance is the actual number of atoms of each element.
Note that the molecular formula is a whole number multiple of the empirical formula.
Review Question 11
Empirical Formula versus Molecular Formula
It is possible for several different molecules to have the same empirical formula.
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Your Turn!
What is the empirical formula for the compound P4O10?
a. P4O10
b. P2O5
c. PO2.5
d. PO3
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Calculating Empirical Formulas
Steps for calculating an empirical formula:
1. Assume 100g of compound and express the mass of each element in grams.
2. Convert the grams of each element to moles.
3. Find the mole ratio of each element. Round to nearest whole number if it is close to the whole number.
4. If necessary, multiply the ratios by the smallest whole number that will convert them to a whole number.
Copyright 2012 John Wiley & Sons, Inc
Review Question 12
Calculating Empirical Formulas
Calculate the empirical formula of a compound that is 63.19% Mn and 36.81% O.
1. Assume 100 g of material.
63.19 g Mn
36.81 g O
2. Convert grams of each element to moles:
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atomic mass
Mn 54.94
0 16.00
1 mol Mn63.19 g Mn× =1.1502 mol Mn
54.94 g Mn1 mol O
36.81 g O× = 2.3006 mol O16.00 g O
Calculating Empirical Formulas
3. Change the numbers of atoms to whole numbers by dividing by the smallest number.
The simplest ratio of Mn:O is 1:2.
Empirical formula = MnO2
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2.3006 mol OO = = 2.000
1.1502 mol Mn1.1502 mol Mn
Mn = = 1.0001.1502 mol Mn
Calculating Empirical Formulas
Calculate the empirical formula of a compound that is 72.2% Mg and 27.8% N.
1. Assume 100 g of material.
72.2 g Mg
27.8 g O
2. Convert grams of each element to moles:
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atomic mass
Mg 24.31
N 14.01
1mol Mg72.2g Mg× =2.970 mol Mg
24.31g Mg1mol N
27.8g N× = 1.984 mol N14.01g N
Calculating Empirical Formulas
3. Change the numbers of atoms to whole numbers by dividing by the smallest number.
4. Multiply by a number that will give whole numbers.
Mg: (1.500)2 = 3.00 N: (1.000)2 = 2.00
Empirical formula = Mg3N2
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2.970 mol MgMg = = 1.500
1.984 mol N
1.984 mol NN = = 1.000
1.984 mol N
Your Turn!
What is the empirical formula of an alcohol that is 52.13% C, 13.15% H and 34.72% O.
a. CH2O
b. C4HO3
c. C2H6O
d. C2H3O2
Copyright 2012 John Wiley & Sons, Inc
atomic mass
C 12.01
H 1.01
O 16.00
1. Assume 100 g so52.13g C, 13.15g H and 34.72g O
2. Moles52.13g C/12.01= 4.34 mol C13.15g H/1.01= 13.02 mol H34.72g O/16.00= 2.17mol O
3. Fractional part4.34 mol C /2.17= 213.02 mol H /2.17= 62.17mol O /2.17= 1
4. Whole numubersAlready done
5. Empirical formula: 2 C & 6 H & 1 O, so …
Give it a try
Calculating the Molecular Formulafrom the Empirical Formula
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The molecular formula will be either equal to the empirical formula or some integer multiple of it.
The ratio of the molecular mass to the mass predicted by the empirical formula tells us how many times larger the molecular formula is.
molecular formula mass = = number of empirical formula units
empirical formula massn
Calculating the Molecular Formulafrom the Empirical Formula
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Determine the molecular formula for glyceraldehyde which has a molar mass of 90.08 g/mol and an empirical formula of CH2O.
2
90.09 g n = = 3
30.03 g CH O(CH2O)3 = C3H6O3
molecular formula massn=
empirical formula mass
Calculating the Molecular Formulafrom the Empirical Formula
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Determine the molecular formula of a nitrogen oxide compound (NxOy) with a molar mass of 92.011 g/mol and a empirical formula of NO2.
(NO2)2 = N2O4
2
92.011 g N On 2
46.0055 g NOx y
molecular formula massn=
empirical formula mass
Your Turn!
Copyright 2012 John Wiley & Sons, Inc
atomic mass
C 12.01
H 1.01
Cl 35.45
A. 1(12.01g C) + 2(1.01g H) + 1(35.45 g Cl) = 49.48g/mol CH2Cl
B. 197.92 g/mol / 49.48 g/mol = 4
C. (CH2Cl) *4=
PLAN THE STEPS
A. Empirical formula mass
B. Divide molar by empirical
C. Multiple result by formula
Plan – Set up - Calcualte
What is the molecular formula of a compound with the empirical formula CH2Cl and molar mass of 197.92 g/mol?
a. CH2Cl
b. C2H4Cl2
c. C3H6Cl3
d. C4H8Cl4
Calculating the Molecular Formula
A disinfectant is known to be 76.57% C , 6.43% H, and 17.00% O. It has a molar mass of 188.24 g/mol Determine its molecular formula.
Determine the mass and moles of C, H and O.
C: 76.57%(188.24g) = 144.1 g C/(12.01g/mol) = 12
H: 6.43% (188.24g) = 12.1 g H/(1.01g/mol) = 12
O: 17.00%(188.24g) = 32.00 g O/(16.00 g/mol) = 2
Molecular formula: C12H12O2
Copyright 2012 John Wiley & Sons, Inc
Your Turn!
What is the molecular formula of a substance that consists of 85.60% C and 14.40% H and has a molar mass of 28.08 g/mol?
a. CH2
b. C2H2
c. CH3
d. C2H4
e. C2H6
Copyright 2012 John Wiley & Sons, Inc
atomic mass
C 12.01
H 1.01
Questions
Review Questions – Did in class
Paired Questions (pg 139)– Do 1, 3, 7, 9, 11, 15, 21, 27, 29, 31, 35 , 39, 43– Practice later 2, 6, 12, 16, 20, 24, 28, 32, 36, 40, 44
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