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 contain
equal numbers of moles of gases.
What is STP?
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**
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
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
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
55.847 g Fe
3 mol O2 x 22.4 L O2
4 mol Fe
1 mol O2
= 30.08 L O2
IDEAL GAS LAW
Pressure P, volume V, temperature T, and the number of
particles n of gas are related by:
PV = nRT
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)
(1 mol) (273.15 K)
= 8.31 kPa•L
mol•K
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!**
Do NOT let the algebraic formula get you! These are all
“plug & chug” problems. 
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
50+273 = 323 K
(300 kPa)(10L) = n (8.31kPa•L)(323K)
mol•K
n = (300 kPa)(10L)
(8.31kPa•L)(323K)
mol•K
= 1.12 mol
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
20+273 = 293 K
V = 2 mol (8.31kPa•L)(293K)
mol•K
202 kPa
V= 24.1 L
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
Efficiency of a reaction can be expressed as percent yield:
PERCENT YIELD = actual yield
theoretical yield
x 100%
Manufacturers want to produce chemicals as efficiently
and inexpensively as possible
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%
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%
Mass Percent
Steps:
1) Calculate mass of each element
2) Calculate total mass
3) Divide mass of element/ mass of compound
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= 6gH
x 100% = 20%
30 g C2H6
* Should add up to 100
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 =
%H=
72.066 g C
x
157.0095 g C2H5 Br
100% = 45.9%
5.0395 g H
x
157.0095 g C2H5 Br
100% = 3.2%
% Br = 79.904 g Br
x
157.0095 g C2H5 Br
100% = 50.9%