Chapters 10

1
Chapters 10

• Chemical Quantities

2
Conversion Factors

• Conversion factor A fraction equal to 1 that is
  used to change one unit into another.
• (When the numerator denominator, a
  fraction equals 1.)

3
Dimensional Analysis

• Dimensional Analysis A problem solving method
  where conversion factors are used to cancel
  unwanted units.

4
Conversion Examples

• a) Convert 25 to nickels.

25
4 quarters 1
5 nickels 1 quarter
25 4 5 / 1 / 1
500 nickels
5
Conversion Example 2

• Convert 180 days to seconds.

180 days
24 hours 1 day
60 minutes 1 hour
60 seconds 1 minute
180 24 60 60 / 1 / 1 / 1
15,552,000 seconds or 1.6 x 107 seconds
6
Common Conversions to Know

• 1 base (m, l, g) 100 centi .
• 1 base (m, l, g) 1000 milli .
• 1 kilo 1000 base units (m, l, g)
• Convert 125 cm to km.

1 m 100 cm
1 km 1000 m
125 cm
125 / 100 / 1000
0.00125 km or 1.25 x 10-3 km
7
Conversion Example 4

• Convert 15 m/s to km/hr.

15 m 1 s
1 km 1000 m
60 seconds 1 minute
60 minutes 1 hour
15 60 60 / 1000
54 km/hr
8
Measuring Matter

• How do we describe how much of something we have?
• By count, by mass, by volume.
• We use words like dozen to talk about an
  amount.
• In chemistry, we use the MOLE.

9
Mole

• Mole SI unit for measuring an amount of a
  substance.
• A particle will either be
• An atom, a molecule or a formula unit
• Avogadros Number 6.02 x 1023
• Representative particles smallest unit that
  still has all the characteristics of that
  substance.

1 mole 6.02 x 1023 representative particles
10
Representative Particles

• What is the representative particle of
• Element (ex. Cu) ___________
• Exception The representative particle of the 7
  diatomic elements is a molecule. (ex. H2)
• Covalent compound (ex. H2O) _________
• Ionic Compound (ex. NaCl) ___________

atom
molecule
formula unit
11
Conversions

• 4 moles Ca atoms Ca.

4 moles Ca 6.02 x 1023 atoms Ca
1 mole Ca
2.41 x 1024 atoms Ca
12
Conversions

• 5 x 1018 atoms Cu moles Cu.

5 x 1018 atoms Cu 1 mole Cu
6.02 x 1023 atoms Cu
8.3 x 10-6 moles Cu
13
Conversions

• 9.2 moles F2 molecules F2?

9.2 moles 6.02 x 1023 molecules F2
1 mole
5.5 x 1024 molecules F2
14
Conversions

• 9.2 moles F2 atoms F?

9.2 moles F2 6.02 x 1023 mlcls F2 2
atoms F 1 mole
1 molecule F2
1.1 x 1025 atoms F
15
Conversions

• 3.4 moles C2H4 total atoms?

3.4 moles C2H4 6.02 x 1023 mlcls C2H4 6
atoms 1 mole
C2H4 1 mlcl C2H4
1.22 x 1025 atoms
16
Molar Mass

• Molar Mass The mass of one mole of an element
  or compound.
• Molar mass of a compound the sum of the masses
  of the atoms in the formula.
• Use the atomic masses in grams/mol on the
  periodic table.

17
Molar Mass

• Find the molar mass of each
• Sr
• MgBr2
• Ba3(PO4)2

87.6 g/mol
(24.3) (2 x 79.9)
184.1 g/mol
(3 x 137.3) (2 x 31) (8 x 16)
601.9 g/mol
18
MoleGram Conversions
1 mole molar mass (in grams)
5.3 moles LiOH ___________ grams LiOH
(Molar mass LiOH 7 16 1 24 g/mol)

5.3 moles LiOH 24 g LiOH
1 mole LiOH
127.2 grams LiOH
19
Gram-Mole Conversions

• 68 grams F2 moles F2?

68 grams F2 1 mole F2
38 grams
68 / 38 1.8 moles F2
20
STP

• STP Standard Temperature Pressure
• Standard Temp ? 0oC
• Standard Press ? 1 atm
• (See Reference Tables)

21
Molar Volume of a Gas

• Avogadros Hypothesis equal volumes of gases at
  the same temperature and pressure contain equal
  numbers of particles.
• At STP, 1 mole of any gas occupies a volume of
  22.4 L.

1 mole 22.4 L (of a gas at STP)
22
Mole-Volume Conversions

• 5.4 moles He L He at STP?

5.4 moles He 22.4 L He
1 mole He
5.4 x 22.4 120.96 L He
23
Mole-Volume Conversions

• 5.4 moles CH4 L CH4 gas at STP?

5.4 moles CH4 22.4 L CH4
1 mole CH4
5.4 x 22.4 120.96 L CH4
24
Volume-Mole Conversion

• 560 L SO3 mol SO3 at STP

560 L SO3 1 mole SO3
22.4 L SO3
560 / 22.4 25 mole SO3
25
Molar Mass-Density Conversions
grams liters
grams mole

• Density Molar Mass
• A gaseous compound composed of sulfur and oxygen
  has a density of 3.58 g/L at STP. What is the
  molar mass of this gas?

3.58 g 22.4 L L 1 mole
3.58 x 22.4 80.2 g/mole
26
Molar Mass-Density Conversion

• What is the density of Krypton gas at STP?

83.8 grams Kr 1 mole mole
22.4 Liters
83.8 / 22.4 3.74 g/L Kr
27
1 mole
Grams (use Per.Tble)
6.02 x 1023 particles
28
Multi-step Problem Example 1

• If you had 5.0 L of CO2 how many grams would that
  be?
• Step 1 L ? moles
• Step 2 moles ? grams
• 5.0 L CO2 1 mole CO2 44.0 g CO2
• 22.4 L CO2 1 mole CO2

9.8 g CO2
29
Multi-step Problem Example 2

• How many molecules are in 60.0 grams of water?
• Step 1 grams ? moles
• Step 2 moles ? molecules
• 60.0 g H2O 1 mole H2O 6.02 x 1023 mlcls
• 18.0 g H2O 1 mole H2O

2.0 x 1024 molecules of H2O
30
Percent Composition

• Percent Composition - by mass of each element
  in a compound
• Percent

Part Whole
x 100
31
Percent Composition

• Percent Comp
• Example Find the mass percent composition of
  Al2(SO4)3

Mass of 1 element Mass of compound
x 100
54 342
x 100
15.8
Al
Al 2 x 27 54 S 3 x 32 96 O
12 x 16 192 Total Comp. 342
96 342
S
x 100 28.1
192 342
O
x 100 56.1
32
Percent Example

• Find the percent composition of NiSO3.
• Ni 58.7 g Ni 58.7
• S 32 g 138.7
• O (3 x 16) 48 g
• Total Comp. 138.7

x 100 42.3
32 138.7
S
x 100 23.1
48 138.7
O
x 100 34.6
33
More Percents

• Which of the following shows a compound that is
  92.3C and 7.7H?
• a) C2H4 b) C3H6
• c) CH4 d) C6H6

34
Empirical Formulas

• Empirical Formula The simplest formula.
• Shows the smallest whole number ratio of elements
  in a compound.
• Covalent formulas will not always be empirical.
• Example CH
• Molecular Formula The actual formula.
• For ionic compounds it will be the simplest
  ratio.
• For molecular compounds it will NOT always be
  the simplest ratio.
• Example C6H6

35
To Calculate Empirical Formula

• Calculate the empirical formula of a 2.5 gram
  compound containing 0.90g Ca and 1.60g Cl.
• Step 1 Convert GRAMS to MOLES.
• Ca 0.90g 1 mole 0.0224 mole Ca
• 40.1 g
• Cl 1.60g 1 mole 0.0451 mole Cl
• 35.5 g

36
Calculating Empirical Formula

• Step 2 DIVIDE the of moles of each substance
  by the smallest number to get the simplest mole
  ratio.
• Ca 0.0224 1 Cl 0.0451
  2.01 2
• 0.0224 0.0224

CaCl2
37
Calculating Empirical Formulas

• Step 3 If the numbers are whole numbers, use
  these as the subscripts for the formula. If the
  numbers are not whole numbers, multiply each by a
  factor that will make them whole numbers.
• Look for these fractions
• 0.5 ? x 2
• 0.33 ? x 3
• 0.25 ? x 4

38
Empirical Formula Example

• What is the empirical formula of a compound that
  is 66 Ca and 34 P?
• (Assume you have 100 grams of a compound and
  replace with grams.)

39
Empirical Formula Example

• Step 1 grams ? moles
• Ca 66g 1 mole 1.646 mole Ca
• 40.1 g
• P 34g 1 mole 1.097 mole P
• 31.0 g
• Step 2 Divide by the smallest.
• Ca 1.646 1.5 P 1.097 1
• 1.097
  1.097

40
Empirical Formula Example

• Step 3 Multiply until you get whole numbers.
• (If you multiply one factor by a number, you
  have to multiply ALL the factors by that number!)
• Ca 1.5 x 2 3 P 1 x 2 2

Ca3P2
41
Determining Molecular Formulas

• A compound has an empirical formula of CH2O. Its
  molecular mass is 180g/mol. What is its
  molecular formula?
• Step 1 Find the mass of the empirical formula.
• C 1 x 12 12
• H 2 x 1 2
• O 1 x 16 16
• Total 30

42
Determining Molecular Formula

• Step 2 Divide the molecular mass by the mass of
  the empirical formula to get the multiplying
  factor.
• Step 3 Multiply each of the subscripts in the
  empirical formula by this factor to get the
  molecular formula.
• 6 (CH2O) ? C6H12O6

180 30
6
43
Determining Molecular Formula

• Find the molecular formula of ethylene glycol
  (CH3O) if its molar mass is 62 g/mol.
• Step 2 62 / 31 2
• Step 3 2 (CH3O) ? C2H6O2

Step 1 12 (3 x 1) 16 31 g/mol
44
Empirical/Molecular Example

• The percent composition of methyl butanoate is
  58.8 C, 9.8 H, and 31.4 O and its molar mass
  is 204 g/mol.
• What is its empirical formula?
• What is its molecular formula?

45
Empirical/Molecular Example
58.8 g C 1 mol C 12.0 g C
4.9 2.5 1.96 9.8 5 1.96 1.96
1 1.96
x 2 5 x 2 10 x 2 2
4.9 mol C
9.8 g H 1 mol H 1.0 g H
9.8 mol H
31.4 g O 1 mol O 16.0 g O
1.96 mol O
Empirical Formula C5H10O2
46
Empirical/Molecular Example

• Empirical Formula ? C5H10O2
• Mass 5(12) 10(1) 2(16) 102 g/mole
• Molecular mass ? 204 g/mol 2
• Empirical mass ? 102 g/mol
• So molecular formula is 2 x emp. form
  2(C5H10O2) C10H20O4