Unit 19
Calculations with the Mole
•Conversion Factor Approach (5.4)
•Mole-to-Mass and Mass-to-Mole Conversions (5.4)
•Mole and Mass Relationships in Chemical
Equations (5.4)
Conversion Factor Approach (5.4)
•A common approach for calculating a wide range of quantities is called
the conversion factor approach. Its premises are:
•Multiplying any quantity by the number “1” returns the same quantity
•A factor which has a numerator equal to its denominator is equal to “1”
•A starting quantity can be multiplied by any number of conversion factors to simply
change the units on the original quantity
•A simple example
•Convert 15 feet to yards (you can do this in your head – this just shows the approach)
1 yd
3 ft
•The conversion factor is 1 yd = 3 feet. The factor that has 3 ft or 1 yd is equal to “1”
since the numerator and denominator are the same in each factor. Thus, our initial
measured quantity may be multiplied by either one without changing the measured
length.
15 ft 1yd
•The calculation sets up as: ? yd 

 5 yd
1
3 ft
The conversion factor is written in the orientation that cancels the original unit – feet –
leaving yd for the final answer.
Mole-to-Mass and Mass-to-Mole Conversions (5.4)
•From the previous unit, we know that we can identify one mole of a material with Avogadro’s
number of particles (6.02 x 1023 particles) and also a mass equal to the formula mass,
molecular mass, or molar mass (all the same number) in grams.
•The mole then provides a convenient set of conversion factors that can be used in working
problems.
•Consider NH3:
1 mol NH3 = 6.02 x 1023 NH3 molecules = 17 g NH3
•Example problems:
•How many grams of NH3 are in 12.5 mol of NH3 ?
•How many mol of NH3 are in 145 g of NH3 ?
? g NH 3 
? mol NH 3 
•How many NH3 molecules are in 35.0 g of NH3 ? ? molecules NH 3 
Desired Quantity
12.5 mol 17 g NH 3

 212 g NH 3
1
1 mol NH 3
145 g NH 3 1 mol NH 3

 8.53 mol NH 3
1
17 g NH 3
35.0 g NH 3 6.02 1023 molecules NH 3

 1.24 1024 molecules NH 3
1
17 g NH 3
Given Quantity
Conversion Factor
Mole and Mass Relationships in Chemical Equations (5.4)
•The conversion factor approach may also be used to address mass and
mole relationships across a chemical equation. Consider the equation:
C3H8 (g) + 5 O2(g)
→
3 CO2 (g)
+
4 H2O (g)
•A question is posed that asks how many grams of CO2 could be produced
from 50.0-g of C3H8 reacted with an excess of oxygen.
•The mass of CO2 produced will be dependent on the number of grams of C3H8 since
there is plenty of oxygen.
•The coefficients give the mole relationships in the chemical reaction. For every 1 mol
of C3H8 burned 3 mol of CO2 are produced
•Since the relationships between moles and mass is known through the molar masses,
there is sufficient information given to solve the problem.
•The next slide illustrates a method of handling this sort of problem using the conversion
factor approach
Application of the Conversion Factor Approach in Chemical
Equations (5.4)
•The question is: How many grams of CO2 can be produced from completely reacting
50.0-g of C3H8? Consider the following table.
Starting/
Desired Info
50.0-g C3H8
C3H8 (g)
Info from
Equation
1 mol C3H8
44 g C3H8
? g CO2
+ 5 O2(g)
5 mol O2
5 x 32 g O2
→
3 CO2 (g) +
3 mol CO2
3 x 44 g CO2
4 H2O (g)
4 mol H2O
4 x 18 g H2O
•The information in the first row is from the specific problem being presented – the mass
of C3H8 given and the unknown mass of CO2.
•The second row simply reproduces the chemical equation for reference.
•The third and fourth rows give information related to the chemical equation given. The
coefficients in the chemical equation give the mole ratios and the conversion to mass is
done by multiplying the number of moles times the molar mass.
•All quantities in the shaded area represent the same amount of chemical reaction –
they can be taken in any pair to form a conversion factor that is equal to “1”.
Application of the Conversion Factor Approach in Chemical
Equations (5.4)
•The question is: How many grams of CO2 can be produced from completely
reacting 50.0-g of C3H8? Consider the following table.
Starting/
Desired Info
50.0-g C3H8
C3H8 (g)
Info from
Equation
? g CO2
+ 5 O2(g)
1 mol C3H8
44 g C3H8
5 mol O2
5 x 32 g O2
→
3 CO2 (g) +
3 mol CO2
3 x 44 g CO2
4 H2O (g)
4 mol H2O
4 x 18 g H2O
•First, set up write down the quantity being sought and the given amount in the
conversion factor setup as:
? gCO2 
50.0 gC3 H 8

1
•To determine the conversion factor to be used, recognize that its denominator
will have to be in g C3H8 to cancel the measured quantity and its numerator
will have to be in g CO2 to come up with the desired quantity. Those numbers
are taken from the shaded area below the equation.
? g CO2 
50.0 g C3 H 8 3  44 g CO2

 150 g CO2
1
1 44 g C3 H 8
Application of the Conversion Factor Approach in Chemical
Equations (5.4)
•This approach is easily used to solve other problems related to the
balanced chemical equation.
Starting/
Desired Info
50.0-g C3H8
C3H8 (g)
Info from
Equation
1 mol C3H8
44 g C3H8
? g CO2
+ 5 O2(g)
5 mol O2
5 x 32 g O2
→
3 CO2 (g) +
3 mol CO2
3 x 44 g CO2
4 H2O (g)
4 mol H2O
4 x 18 g H2O
•Examples:
•How many g of O2 will react with 75 g of C3H8?
75.0 g C3 H 8 5  32 g CO2
? g O2 

 273 g O2
1
1 44 g C3 H 8
•How many moles of water will be formed in carrying out this reaction with 350 g of O2 ?
? mol H 2O 
350.0 g O2 4 mol H 2O

 8.75 mol H 2O
1
5  32 g O2