moles/stoichiometry

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Mole calculations

 

1 mole (mol) = # of carbon atoms in 12 grams of C-12

                        = 6.022 x 1023

                       = the Avogadro’s number

Example

In 16 g of O there is 1 mole of O or 6.02 x 1023 O atoms

 

In 32g of O there are 2 moles of O or 2 (6.02 x 1023) O atoms

 

In 4 g of H there are 4 moles of H or 4 (6.02 x 1023) H atoms

 

In 18 g of H2O there are

1 mole of H2O or 6.02 x 1023 H2O molecules

2 moles of H atoms or 2 (6.02 x 1023) H atoms 

1 mole of O atoms or 6.02 x 1023 O atoms

 

In 36 g of H2O there are

2 mole of H2O or 2 (6.02 x 1023) H2O molecules

4 moles of H atoms or 4 (6.02 x 1023) H atoms

2 moles of O atoms or 2 (6.02 x 1023) O atoms

 

Calculations using molar mass

 

The mass of 1 mole of element X = the atomic mass of X in grams

 

By convention

The mass of 1 atom of C-12 = 12 amu

Therefore the mass of 1 mole of C-12 = 12 amu x 6.022 x 1023 = 12 g

 

From the unit factor (1 mole X = atomic mass X) two conversion factors are derived

 

     1mol X                                        atomic mass X

------------------                and          --------------------

atomic mass X                                       1 mol X

 

Example

What is the mass in grams of 6 atoms of Am-243

                                                 

                                                 1 mole           243g

6 atoms of Am-223 -------------------------- ---------- = 2.42 x 10 -21 g

                                    6.022 x 1023 atoms   1 mole

 

Example

How many atoms are present in 10.5 mg of Si?

 

                         1g        6.022 x 1023 atoms

10.5 mg Si ------------ ------------------------- =                   atoms of Si

                    1000mg            28.09g

 

Example

How many moles are present in 10.5 ng of Si?

 

                     1g       1mol

10.5 ng Si -------- ---------- =                   mol of Si

                   109ng   28.09g

 

Example

How many grams are present in 10 moles of Si?

 

                   28.09g

10 mol Si ----------- =                      g of Si

                    1mol

 

Example

How many H atoms are in 1μg of H2O?

 

                        1g        1mol   6.022 x 1023 atom     2 H atoms

1 μg of H2O --------- -------- ------------------------ --------------- =                 H atoms

                     106 μg     18g               1 mol                 1 atom

 

 

Molar Volume of gas

At STP 1 mole of any gas occupies 22.4 L

 

Calculate empirical formula from % composition

 

Example

What is the empirical formula of a substance made up of approximately 75% mercury and 25% chlorine?

 

First, turn the percentages into grams assuming there is 100 g of substance

 

75% Hg = 75 g of Hg

25% Cl = 25 g of Cl

 

Second, change the grams into moles

 

                                                                  75g

Moles of Hg = --------------  = 0.375 mol Hg

                           200 g/mol

 

                             25g

Moles of Cl = ------------ = 0.700 mol Hg

          35 g/mol

 

Third, divide the moles of each element by the smallest number of moles obtained

 

                                                        0.375 mol

For Hg           -------------- = 1               

                        0.375 mol

 

                        0.700 mol

For Cl             -------------- = 2

                        0.375 mol                       

Fourth, assign the values obtained in third step as subscripts and write the empirical formula omitting the ones

 

Hg 1 Cl 2  

 

Empirical formula = HgCl2

 

Determining Empirical formula and molecular formula

 

Molecular formula = (empirical formula) n

 

Example

The % composition by mass of substance X is 71.65% Cl, 24.27%C, and 4.07% H

What is its empirical and molecular formula? (The molar mass of X is 98.96g/mol)

 

First, turn the percentages into grams and calculate the moles for each element

                   

                                     1 mol

71.65g Cl ----------- = 2.021mol Cl

                   35.45g

 

                 1 mol

24.27g C --------- = 2.021mol C

                 12.01g

 

                1 mol

4.07g H ---------- =  4.04 mol H

                1.008g

 

Second, divide the moles of each element by the smallest number of moles obtained and find empirical formula

 

           2.021

Cl       -------  = 1

           2.021

 

          2.021

C       -------  = 1

          2.021

 

           4.04

H       --------  = 2

           2.021

 

Empirical formula Cl1C1H2 or ClCH2

 

Third, from the formula       Molecular formula = (empirical formula) n

                                                               

         Molecular formula = (empirical formula) n

                                         ? = (ClCH2) n

Molecular formula mass = (empirical formula mass) n

                   98.96g/mol   = (49.48g/mol) n

 

       98.96g/mol  

n = ---------------   = 2

       49.48g/mol

 

Fourth, calculate the molecular formula 

 

(ClCH2)2 = Cl2C2H4

 

Molecular formula = C2H4Cl2

 

Limiting Reactants

The concentration of reactants will dictates how much products can theoretically form. Often when two reactants are present one is in excess and the other limits the amount of products.

 

Example

Given the reaction 4 NH3(g) + 5O2(g) ---> 4 NO(g) + 6 H2O(l)

How many grams of NO are produced when 51g of ammonia reacts with 32g of O2?

 

First, calculate the number of moles for each reactant

 

                 1 mol NH3

51g NH3 --------------- = 3 mol NH3 

                       17g

 

             1 mol O2

32g O2 ------------ = 1 mol O2 

                  32g

 

Second, find the limiting reactant (the one that will limit the amount of products formed)

 

                The ratio of ammonia to oxygen is 4:5 since 1 mole of oxygen is available 0.8 moles of ammonia are required.

 

 4 mol NH3       X mol NH3

-------------- = ----------------                    X = 0.8

  5 mol O2           1 mol O2

 

                Therefore ammonia is in excess and oxygen is the limiting reactant.

 

Third, use the limiting reactant to calculate the number of moles of product.

 

                 4 mol NO      30g NO

1 mol O2 -------------- -------------- = 24 g NO

                  5 mol O2       1 mol NO

 

Example

If a sample of 18.1g of NH3 reacts with 90.4g of CuO, how many grams of N2 will be formed?

2 NH3 + 3 CuO ---> N2 + 3 Cu + 3 H2O

 

First, calculate the number of moles for each reactant

 

                   1 mol NH3

18.1g NH3 --------------- = 1.06 mol NH3 

                         17g

 

                     1 mol O2

90.4g CuO  ------------ = 1.14 mol CuO 

                         80g

 

Second, find the limiting reactant (the one that will limit the amount of products formed)

 

                The ratio of NH3 to CuO is 2:3 since 1.14 mole of CuO is available 0.76 moles of ammonia are required.

 

 2 mol NH3      X mol NH3

-------------- = ------------------                 X = 0.76

 3 mol CuO     1.14 mol CuO

 

                Therefore ammonia is in excess and CuO is the limiting reactant.

 

Third, use the limiting reactant to calculate the number of moles of product.

 

                          1 mol N2       28g N2

1.14 mol CuO -------------- ------------- = 11 g N2

                         3 mol CuO    1 mol N2

 

Percent Yield

                               Actual yield

Percent Yield = ---------------------- x 100

                           Theoretical yield

 

Example

If the actual yield is 6.63g of N2 and the theoretical yield is 10.6g of N2 the % yield of N2 is

 

  6.63g N2

------------- x 100 = 62.5%

  10.6g N2

 

Solubility Rules

1

All compounds containing group 1 cations and the ammonium ion are soluble.

2

All compounds containing NO3-, ClO4-, ClO3-, and C2H3O2- anions are soluble.

3

All chlorides, bromides, and iodides are soluble except those containing Ag+, Pb2+, or Hg22+.

4

All sulfates are soluble except those containing Hg22+, Pb2+, Sr2+, Ca2+, or Ba2+.

5

All hydroxides are insoluble except compounds of the alkali metals, Ca2+, Sr2+, and Ba2+.

6

All compounds containing PO43-, S2-, CO32-, and SO32- ions are insoluble except those that also contain alkali metals or NH4+.

 

Example

2KI(aq) + Pb(NO3)2(aq) ---> 2 KNO3(aq) + PbI(s)

 

 

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