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The Mole and Avogadro's Number, continued from page 1
 
 
 
Sample Calculations
 

Example 1.  How many moles of iron are in 50.0 g of iron?

a)  Plug in to the mole formula,  moles   =       50.0 g         =    0.895 mol
                                                               55.845 g/mol
 
b)  Or, you can do the calculation like a conversion problem:

    50.0 g    X      1 mol       =    0.895 mol
       1              55.845 g
 
 
Example 2.  How many moles of carbon atoms and oxygen atoms are in 0.250 mol of CO2 ?

Here we just have to look at the formula.  Since one molecule of CO2 contains one carbon atom, one mole of CO2 molecules will contain one mole of carbon atoms.  If we have 0.250 mol of CO2, there will be 0.250 mol of carbon atoms present.
 
0.250 mol of CO2    X      1 mol of C     =   0.250 mol of C atoms
            1                     1 mol of CO2
 
Since there are two oxygen atoms in one CO2 molecule, there are 2 X 0.250 mol = 0.500 mol of oxygen atoms present in this amount:
 
0.250 mol of CO2    X      2 mol of O     =   0.500 mol of O atoms
            1                     1 mol of CO2
 

Example 3.  How many moles of copper atoms are in a copper penny weighing 3.10 g ?  How many copper atoms are in the penny?

moles   =   3.10 g / 63.546 g/mol   =   0.0488 mol

number of Cu atoms   =   0.0488 mol   X   6.022 X 1023 atoms    =    2.94 X 1022 atoms
                                            1                        1 mol
 

Example 4.  A chemistry class has 15 men and 17 women in it.  How many moles of students are in the class?

OK, that's not something we need to know every day, but can you do the calculation?  Of course!

32 students    X               1 mol                  =    5.314 X 10�?3 mol
       1                  6.022 X 1023 students
 

Example 5.  How much do 1.00 X 1012 (a trillion) gold atoms weigh in grams?

1.00 X 1012 atoms    X              1 mol                X    196.967 g    =    3.27 X 10�?0 g
            1                       6.022 X 1023 atoms               mol
 

Example 6.  The mass of the earth is 5.98 X 1024 kg.  The mass of a baseball is 145 g.  How many baseballs would have a mass equal to the mass of the earth?  What is this number in moles?

5.98 X 1024 kg    X    1 baseball     =    4.12 X 1025 baseballs
          1                     0.145 kg

4.12 X 1025 baseballs     X                 1 mol                   =     68.5 moles
              1                            6.022 X 1023 baseballs
 

Example 7.  A 6.00 M solution of HCl contains 6.00 mol of HCl per liter of solution.  How many moles of HCl are in 125 mL of this solution?

0.125 L    X    6.00 mol    =    0.750 mol
    1                  1 L
 

Example 8.  What is the mass in grams of 0.100 mol of glucose, C6H12O6 ?

grams  =  moles  X  molecular weight  =  0.100 mol  X  180.156 g/mol  =  18.0 g
 

Example 9.  A sample of a certain hydrocarbon contains 0.090 mol of carbon and 0.36 mol of hydrogen.  What is the empirical formula of the compound?

The subscripts in formulas are the relative numbers of each element in the compound.  For example, one mole of C6H12O6 contains 6 moles of carbon atoms, 12 moles of hydrogen atoms, and 6 moles of oxygen atoms.

The empirical formula is the simplest formula with whole-number subscripts.  The empirical formula of the hydrocarbon is "C0.090H0.36".  To convert to whole-number subscripts, divide each number of moles by the smallest value, which is 0.090.  That gives C1H4 or CH4 as the empirical formula.
 

Example 10.  Analysis of a compound showed it to be 40.0% C, 6.7% H, and 53.3% O by weight.  What is the empirical formula of the compound?

First, for simplicity, assume you have 100 g of compound.  Then, it contains 40.0 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen.  That's convenient!  Now convert each amount to moles in order to get an empirical formula as in Example 9:

moles of C  =  40.0 g / 12.0107 g/mol  =  3.33 mol
moles of H  =   6.7 g / 1.00794 g/mol  =  6.7 mol
moles of O  =  53.3 g / 15.9994 g/mol  =  3.33 mol

This gives "C3.33H6.7O3.33" which we can divide through by 3.33 to obtain the empirical formula CH2O.
 
 
 
Mole Exercises  –�?gt;