Mole Problems Answers to Mole Problems Problems


Solutions to Mole Problems



Mole Problems

Molecular Mass / Formula Mass

1) Find the formula mass of KMnO4.

     1 K    1 x  39.10 g =    39.10 g
     1 Mn  1 x  54.94 g =    54.94 g
     4  O   4 x  15.99 g =    63.96 g
          Total                     158.00 g

2) Find the  formula mass of  Mg3(PO4)2.

      3 Mg    3 x 24.31 g =   72.93 g
      2  P      2 x 30.97 g =   61.94 g
      8  O     8 x 15.99 g = 127.92 g
           Total                     262.79 g

3) Find the molecular mass of  C12H22O11.

    12 C   12 x  12.01 g =  144.12 g
    22 H   22 x    1.01 g  =   22.22 g
    11 O   11 x  15.99 g =  175.89 g
          Total                       342.23 g

Conversion of  Moles, Grams, Molecules, and Atoms

1) How many moles are in 17.7 g of  KMnO4?

     [17.7 g KMnO4][   1 mol KMnO4]  =  0.112 mol KMnO4
                               [158.0 g KMnO4]

    How many formula units are in 17.7 g  KMnO4?

     [17.7 g KMnO4][   1 mol KMnO4][6.023 x 1023 formula unitsKMnO4 ]  =  6.747  x 1022formula units 
                                   [158.0 g KMnO4][                    1 mol KMnO4]                            KMnO4

     
    How many atoms of O are in 17.7 g  KMnO4?

   [  17.7 g KMnO4][   1 mol KMnO4][6.023 x 1023 formula units KMnO4  ][                 4 atoms O] =  
                                  [158.0 g KMnO4][                               1 mol KMnO4][1 formula unit KMnO4]
                                                                                                                         2.70  x 1023 atoms O

2) How many moles are in 41.2 g of  Mg3(PO4)2?

[41.2 g Mg3(PO4)2][       1 mol Mg3(PO4)2]  =  1.57 mol Mg3(PO4)2
                               [  262.79 g Mg3(PO4)2]

    How many formula units are in 41.2 g  Mg3(PO4)2?

  [41.2 g Mg3(PO4)2][   1 mol Mg3(PO4)2  ][6.023 x 1023 formula unitsMg3(PO4)2 ]  =  9.44 x 1023 units
                                     [262.79 g Mg3(PO4)2][                                1 mol Mg3(PO4)2]            Mg3(PO4)2


    How many atoms of P are in 41.2 g  Mg3(PO4)2?

   [41.2 g Mg3(PO4)2][   1 mol Mg3(PO4)2][6.023 x 1023 formula units Mg3(PO4)2][       2 atoms P]         =  
                                      [262.79g Mg3(PO4)2][                               1 mol Mg3(PO4)2][1 form unit Mg3(PO4)2]
                                                                                                                         1.89 x 1024 atoms P

3) How many moles are in 26.9 g of  C12H22O11? 

[26.9 g C12H22O11][     1 mol C12H22O11]  =  0.079 mol C12H22O11
                               [342.23g C12H22O11]

    How many molecules are in  26.9 g of  C12H22O11?

  [26.9 g C12H22O11][  1 molC12H22O11 ][6.023 x 1023 molecules C12H22O11  ] =  4.73 x 1022molecules 
                                     [342.23g C12H22O11][                           1 mol  C12H22O11]                   C12H22O11

    How many atoms of C are in 26.9 g of  C12H22O11?

   [26.9 g C12H22O11][   1 mol C12H22O11][6.023 x 1023 molecules C12H22O11][         12 atoms C ] =  
                                      [126.9 g C12H22O11][                             1 molC12H22O11 ][1 molecule C12H22O11]
                                                                                                                         5.68  x 1023 atoms C

4) How many moles are 19.8 x 1024 molecules C12H22O11?

    [19.8 x 1024 molecules C12H22O11][                           1 mol C12H22O11] = 32.9 mol C12H22O11
                                                                  [6.023 x1023 molecules C12H22O11]
     What is the mass of  19.8 x 1024 molecules C12H22O11?

[19.8 x 1024 molecules C12H22O11][                           1 mol C12H22O11][342.23g C12H22O11]  = 1.13x104g
                                                             [6.023 x1023 molecules C12H22O11][    1 mol C12H22O11]   C12H22O11

 

5) How many moles of  Mg3(PO4)2  are  27.6 x 1037 atoms of O in Mg3(PO4)2?

[27.6 x 1037 atoms O][1form.unit Mg3(PO4)2][                              1mole Mg3(PO4)2]  =  5.73 x 1013 mol
                                     [                  8 atoms O][6.023 x 1023 formula unitsMg3(PO4)2 ]          Mg3(PO4)2

6) How many moles of KMnO4 are 3.64 x 1025 atoms of Mn in KMnO4?

[3.65 x 1025 atoms Mn][1form.unitKMnO4 ][                              1mole KMnO4]  =  60.6  mol
                                        [               1 atom  O][6.023 x 1023 formula unitsKMnO4 ]          KMnO4    


Percent Composition

1) Find the percentage composition of  KMnO4. 

     1 K    1 x  39.10 g =    39.10 g
     1 Mn  1 x  54.94 g =    54.94 g
     4  O   4 x  15.99 g =    63.96 g
          Total                     158.00 g   

      % K = 39.10 g  x 100 = 24.75% K
                 158.00g

     %Mn = 54.94 g  x 100 = 34.77% Mn
                 158.00g

       % O = 63.96 g  x 100 = 40.48% O
                 158.00g

2) Find the percentage composition of  Mg3(PO4)2.

      3 Mg    3 x 24.31 g =   72.93 g
      2  P      2 x 30.97 g =   61.94 g
      8  O     8 x 15.99 g = 127.92 g
           Total                     262.79 g

     % Mg = 72.93 g  x 100 = 27.75% Mg
                 262.79g

        % P = 61.94 g  x 100 = 23.57% P
                 262.79g

     % O = 127.92 g  x 100 = 48.68% Mg
                 262.79g

3) Find the percentage composition of  C12H22O11.

    12 C   12 x  12.01 g =  144.12 g
    22 H   22 x    1.01 g  =   22.22 g
    11 O   11 x  15.99 g =  175.89 g
          Total                       342.23 g

     % C = 144.12 g  x 100 = 42.11% C
                 342.23g

     % H =   22.22 g  x 100 =   6.49% H
                 342.23g

     % O = 175.89 g  x 100 = 51.40% C
                 342.23g


Empirical Formula

1) What is the empirical formula for a compound if a 83.7 g sample contains 44.28g Al and 39.42 g O?

     [44.28 g Al][  1 mole Al]  =  1.64 mol Al
                        [26.98 g Al]

     [39.42 g O][  1 mole O]  =  2.466 mol O
                        [15.99 g O]

      1.64 mol Al  =   1 Al            x      2  = 2 Al    
      1.64 mol Al

       2.466 mol  O   =  1.5 O      x       2 = 3 O
       1.64 mol Al
 
                                                            Al2O3

2) What is the empirical formula for a compound if a 55.0 g sample contains 22.0 g C, 3.67 g H, and
      29.33 g O? 

     [22.0 g C][  1 mole C]  =   1.83  mol C
                     [12.01 g C]

     [3.67 g H][  1 mole H]  =   3.63  mol H
                     [   1.01 g H]  

     [29.33g O][  1 mole O]  =  1.83  mol O
                      [  15.99 g O]  

     1.83  mol C  =  1 C
     1.83  mol C

     3.63  mol H  =  1.98  H
     1.83  mol C

     1.83  mol O  =  1 O
     1.83  mol O

                               CH2O


Molecular Formula

1) A compound has a percentage composition of 40.01% C,   6.69 % H, and  53.30% O.  Its
     molecular mass is 179.9 amu.  What is it's empirical formula?  What is it's molecular formula?

     Let sample size of compound be equal to 100 g. 

     100 g  x 40.01 % C =  40.01 g C
     100 g  x  6.69 %  H =    6.69 g  H
     100 g  x 53.30 % O =   53.30g O

   [40.01 g C][  1 mole C]  =   3.33  mol C
                     [12.01 g C]

     [6.69 g H][  1 mole H]  =   6.62  mol H
                     [   1.01 g H]  

     [53.30g O][  1 mole O]  =  3.33  mol O
                       [15.99 g O]  

     3.33  mol C  =  1 C
     3.33  mol C

     6.62  mol H  =  1.99  H
     3.33  mol C

     3.33  mol O  =  1 O
     3.33  mol C

                               CH2O empirical formula  

    mass of CH2O is 
    1 C = 1 x 12.01 = 12.01 g
    2 H = 2 x  1.01  =   2.02 g
    1 O = 1 x 15.99 = 15.99 g
              Total           30.11 g

    179.9 g molecular mass of compound  = 5.97
      30.11g empirical mass

   1 C x 6 =  6 C
   2 H x 6 = 12 H
   1 O x 6 =  6  O

   C6H12O6  molecular formula

2) A compound has a percentage composition of  46.68% N and   53.32% O.   Its molecular mass
     is  59.98 amu.  What is the empirical formula?  What is it's molecular formula?

 Let sample size of compound be equal to 100 g. 

     100 g  x 46.68 % N =  46.68 g N
     100 g  x  53.32 % O = 53.32 g O
    

   [46.68 g N][  1 mole N]  =   3.33  mol N
                      [ 14.01g N]

   [53.32 g O][  1 mole O]  =   3.33  mol O
                     [ 15.99 g O]  

     3.33  mol N  =  1 N
     3.33  mol N

     3.33  mol O  =  1 O
     3.33  mol N

                               NO empirical formula  

    mass of NO is 
    1 N = 1 x 14.01 = 14.01 g
    1 O = 1 x 15.99 = 15.99 g
              Total           30.00 g

    59.98 g molecular mass of compound  = 2.00
     30.00g empirical mass

   1 N x 2 =  2 N
   1 O x 2 =  2 O

   N2O2  molecular formula


Hydrates

1)  A 16.59 g sample of hydrated sodium thiosulfate.  The sample is heated to drive off the water.  
     The dry sample has a mass of 10.59 g of sodium thiosulfate.  What is the formula for the hydrate?

16.59 g wet Na2S2O3   -  10.59 g dry  Na2S2O3  =  6.00 g H2O

2 Na  2 x 22.99 g =  45.98 g
2 S    2 x 32.07 g =  64.14 g
3O    3 x 15.99 g =  47.97 g
         total              158.09 g 

2H     2 x  1.01 g =   2.02 g
1O     1 x 15.99g =  15.99 g
        total                 18.01 g 

[10.59 g Na2S2O3][     1 mol Na2S2O3] =   .067 mol Na2S2O3
                              [158.09 g Na2S2O3]

[6.00 g H2O][    1 mol H2O]  =  0.333 mol H2O
                     [18.01 g H2O]

0 .067 mol  = 1 Na2S2O3               0.333 mol  =  4.97 H2O
0 .067 mol                                     0 .067 mol

Na2S2O3 * 5H2O

 

2)  A 35.64 g sample of hydrated barium bromide.  The sample is heated to drive off the water.  
     The dry sample has a mass of 30.42 g of barium bromide.  What is the formula for the hydrate?

35.64 g wet BaBr2   -  30.42 g dry  BaBr2  =  5.22 g H2O

1 Ba  1 x  g =  137.33 g
2 Br  2 x  g =     79.90 g
         total        217.23 g 

2H     2 x  1.01 g =   2.02 g
1O     1 x 15.99g =  15.99 g
        total                 18.01 g 

[30.42 g BaBr2][    1 mol BaBr2] =   0.14 mol BaBr2
                         [217.3 g BaBr2]

[5.22 g H2O][    1 mol H2O]  =  0.29 mol H2O
                     [18.01 g H2O]

0 .14 mol  = 1   BaBr2           0.29 mol  =  2.07  H2O
0 .14 mol                              0 .14 mol

BaBr2 * 2H2O

 
Mole Problems