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

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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

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• 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

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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