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Chapter 12: Chemical Quantities

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Section 12.2: Using Moles (part 2) Remember Avogadro s Principle In terms of moles, states that equal volumes of gases at the same temperature and pressure ... – PowerPoint PPT presentation

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Title: Chapter 12: Chemical Quantities


1
Chapter 12 Chemical Quantities
  • Section 12.2 Using Moles (part 2)

2
Remember
  • Avogadros Principle
  • In terms of moles, states that equal volumes of
    gases at the same temperature and pressure
    contain equal numbers of moles of gases.
  • What is STP?

3
MOLAR VOLUME
  • For a gas
  • the volume that a mole of a gas occupies at a
    pressure of one atmosphere (equal to 101 kPa) and
    at a temperature of 0.00ºC STP
  •  
  • At STP, the volume of 1 mol of any gas is 22.4
    L

4
Steps
  • Vol A ? Mol A ? Mol B ? grams B
  • 22.4 L Coeff.
    Mass
  • OR
  •  
  • Grams A ? Mol A ? Mol B ? Vol A
  • mass Coeff.
    22.4 L

5
Practice Problems
  • 1) What mass of glucose (C6H12O6) must be broken
    down in your body to produce 2.5 L of CO2 at STP?
  • C6H12O6 6O2 ? 6H2O 6 CO2
  •  
  • 2.5 L CO2 x 1 mol CO2 x 1 mol C6H12O6 x 180 g
    C6H12O6
  • 22.4 L CO2 6 mol CO2
    1 mol C6H12O6
  • 3.35 g C6H12O6

6
Practice Problems (Cont)
  • 2) What volume of oxygen is required to react
    with 100 g of iron at STP?
  • 4 Fe 3O2 ? 2 Fe2O3
  •  
  • 100 g Fe x 1 mol Fe x 3 mol O2 x 22.4
    L O2
  • 55.847 g Fe 4 mol Fe
    1 mol O2
  •  
  • 30.08 L O2

7
IDEAL GAS LAW
  • Pressure P, volume V, temperature T, and the
    number of particles n of gas are related by
  • PV nRT
  •  

8
IDEAL GAS LAW (Cont)
  • R can be determined using the definition of molar
    volume at STP
  • P 101.3 kPa
  • V 22.4 L
  • n 1 mol
  • T 273.15 K
  • (101.3 kPa)(22.4L) (1 mol)(R)(273.15 K)
  •  
  • R (101.3 kPa)(22.4L) 8.31 kPaL
  • (1 mol) (273.15 K) molK

9
  • R is a constant 8.31 kPa . L
  • mol . K
  • If the given pressure is in kPa, use the value
    for R above. If the given pressure is in atm,
    then use the value
  • R .08205 atm . L
  • mol . K
  •  
  • If the given pressure is in mm Hg, then use
    the value
  • R 62.36 mm Hg .
    L
  • mol . K
  • YOU WILL NOT HAVE TO MEMORIZE THESE!

10
  • Do NOT let the algebraic formula get you! These
    are all plug chug problems. ?

11
Practice Problem
  • 1) How many moles of gas are contained in a 10 L
    tank at 300 KPa and 500C?
  •  
  • PV nRT ? n PV/ RT
  •  
  • 50273 323 K
  • (300 kPa)(10L) n (8.31kPaL)(323K)
  • molK
  •  
  • n (300 kPa)(10L)
    1.12 mol
  • (8.31kPaL)(323K)
  • molK

12
Practice Problems (Cont)
  • 2) What volume of gas is contained if 2 moles of
    gas are at 20oC and 2 atm?
  •  
  • 2 atm x 101 KPa 202 KPa
  • 1 atm
  •  
  • PV nRT ? V nRT/ P
  • 20273 293 K
  •  
  • V 2 mol (8.31kPaL)(293K)

  • molK

  • 202 kPa
  • V
    24.1 L

13
THEORETICAL YIELD
  • The amount of product predicted to form is called
    the theoretical yield
  •  
  • The actual yield is usually less than the
    predicted (theoretical yield)
  • - Theoretical yield is determined through
    calculation.
  • - Actual yield may be affected by the collection
    techniques, apparatus used, time, and chemist
    skills

14
  • Efficiency of a reaction can be expressed as
    percent yield
  •  
  • PERCENT YIELD actual yield x 100
  • theoretical yield
  •  
  • Manufacturers want to produce chemicals as
    efficiently and inexpensively as possible

15
Practice Problems
  • 1) What is the percent yield if the theoretical
    yield is 4.5 g and the actual is 3.8 g?
  •  
  • yield 3.8 x 100
  • 4.5
  • 84.4

16
  • 2) 10 g of H2 react with excess O2. When the
    reaction is over 85 g of H2O are recovered, what
    is the percent yield?
  • 2H2 O2 ? 2H2O
  • Actual yield 85g
  • 10 g H2 x 1 mol H2 x 2 mol H2O x 18.01
    g H2O
  • 2.0158 g H2 2 mol H2
    1 mol H2O
  •  
  • Theoretical yield 89.34 g H2O
  •  
  • yield 85 g / 89.34 g x 100 95.14

17
Mass Percent
  • Steps
  • 1) Calculate mass of each element
  • 2) Calculate total mass
  • 3) Divide mass of element/ mass of compound

18
Mass Percent (Cont)
  • Ex 1 Calculate the mass of C and H in C2H6
  • 2 mol C x 12 g C 24 g C
  • 6 mol H x 1 g H 6 g H
  •  
  • Total mass C2H6 30 g
  • C 24 g C x 100 80
  • 30 g C2H6
  •  
  • H 6 g H x 100 20
  • 30 g C2H6
  • Should add up to 100

19
Mass Percent (Cont)
  • Ex 2 Calculate the mass of C, H, Br in C6H5Br
  • 6 mol C x 12.011 72.066
  • 5 mol H x 1.0079 5.0395
  • 1 mol Br x 79.904 79.904
  • 157.0095 g
  •  
  • C 72.066 g C x 100
    45.9
  • 157.0095 g C2H5 Br
  •  
  • H 5.0395 g H x 100
    3.2
  • 157.0095 g C2H5 Br
  •  
  • Br 79.904 g Br x 100
    50.9
  • 157.0095 g C2H5 Br
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