The Mole
The mole or Avagadro's number is the number of atoms that allows the atomic mass to be equal to the molar mass.

 
One mole equals 6.02 x 1023
The mole is actually only a number and has NO UNITS.

There are only four conversions that you need to know

molar mass (g) 
mole
        mole        
molar mass (g)
6 x 1023 items 
mole
        mole         
6 x 1023 items

Examples

Molar Mass Mole to Atoms or Molecules Molecules or Atoms to Molecules Grams to Moles Moles to Grams Grams to Number of Atoms or Molecules Number of Molecules or Atoms to Grams Balancing a Chemical Equation Stoichiometry

Find the molar mass of CaCl2.
Determine each type of atom in the molecule
Calcium (Ca)
Chloride (Cl)
Determine the number of each atom
1
2
Determine the molar mass of each atom (g)
40.08 g
35.45 g
Multiply the number of each atom by its mass
1 x 40.08 = 40.08g
2 x 35.45g = 70.90g
Add the masses together 
40.08g + 70.90g
= 110.98g
The molar mass of CaCl2 is 110.98-g.

Find the molar mass of  H2SO4.
Determine each type of atom in the molecule
Hydrogen (H)
Sulfur (S)
Oxygen (O)
Determine the number of each atom
2
1
4
Determine the molar mass of each atom (g)
1.01 g
32.06 g
16.00 g
Multiply the number of each atom by its mass
2 x 1.01g = 2.02g
1 x 32.06g = 32.06g
4 x 16.00g = 64.00g
Add the masses together  2.02g + 32.06g + 64.00g = 98.08 g
The molar mass of H2SO4 is 98.08 g.

Return

Moles to Atoms or Molecules
A.  Find the number of atoms in 7 moles of Sodium.
7 moles of Sodium (6.02 x 1023 atoms of Sodium) = 4.214 x 1024 atoms of Sodium
                                        mole of Sodium

B.  Find the number of molecules in 3 moles of NaCl.
3 moles of NaCl (6.02 x 1023 molecules of NaCl) = 1.806 x 1024 molecules of NaCl
                                      mole of NaCl

C.  Find the number of molecules in 5 moles of carbon dioxide.
5 moles of CO2 (6.02 x 1023 molecules of CO2) = 3.10 x 1024 molecules of CO2
                                      mole of CO2

NOTE: Oxygen is one of seven atoms that are always diatomic as elements.
D.  Find the number of atoms in 4 moles of Oxygen gas.
4 moles of O2 (6.02 x 1023molecules of O2)(   2 atoms of O    ) = 4.816 x 1024 atoms of Oxygen
                                 mole of O2                  molecules of O2
 Return

Atoms or Molecules to Moles
A.  Find the number of moles if you have 5.4 x 1023 atoms of carbon.
5.4 x 1024 atoms of carbon (        mole of carbon       ) = 9 moles of carbon
                                           6 x 1023 atoms of carbon

B.  Find the number of moles if you have 9.6 x 1024 molecules of KOH.
9.6 x 1024 molecules of KOH (          mole of KOH          ) = 16 moles of KOH
                                                6 x 1023 molecules of KOH
Return

Grams to Moles
A.  Find the number of moles if you have 302 g of CaCl2.  Note: you must first find the molar mass.
302 g of  CaCl2 ( mole of CaCl2 ) =  2.72 moles of CaCl2
                           111 g of  CaCl2
B.  Find the number of moles if you have 157 g of H2SO4.  Note: you must first find the molar mass.
157 g of H2SO4( mole of H2SO4 ) =  1.60 moles of H2SO4
                           98 g of H2SO4
Return

Moles to Grams
A.  Find the mass if you have 14.00 moles of CaCl2.  Note: you must first find the molar mass.
14.00 moles of  CaCl2 ( 111-g of CaCl2  ) =  1554-g of CaCl2
                                     mole of  CaCl2

B.  Find the mass if you have 8.30 moles of H2SO4.  Note: you must first find the molar mass.
8.30 moles of H2SO4( 98 g of H2SO4 ) =  813-g of H2SO4
                                      mole of H2SO4
Return

Grams to Number of Atoms or Molecules
A.  Find the number of molecules if you have 4.37-g of CaCl2.  Note: you must first find the molar mass.
4.37 g of  CaCl2 (mole of CaCl2)(6.02 x 1023 molecules of CaCl2)  =  2.37 x 1022 molecules of CaCl2
                          (111 g of  CaCl2)              (mole of CaCl2)

B.  Find the number of chloride ions if you have 4.37-g of CaCl2.  Note: you must first find the molar mass.
4.37 g of  CaCl2 (mole of CaCl2)(6.02 x 1023 molecules of CaCl2)(     2 ions of Cl    )4.74 x 1022 chloride ions
                          (111 g of  CaCl2)              (mole of CaCl2)            (molecules of CaCl2)

C.  Find the number of oxygen ions if you have 32.1-g of H2SO4.  Note: you must first find the molar mass.
32.1 g of  H2SO4 (mole of H2SO4)(6.02 x 1023 molecules of H2SO4)(     4 ions of O    )7.89 x 1023 oxygen ions
                            (98 g of  H2SO4)             (mole of H2SO4)            (molecules of H2SO4)
Return

Number of Molecules or Atoms to Grams
A.  What is the mass of 8.26 x 1021 atoms of carbon?
8.26 x 1021 atoms of C (        moles of C         )(12.01-g of C) = 0.165 g of Carbon
                                    (6.02 x 1023 atoms of C)   (mole of C)

B.  What is the mass of 10,000 molecules of CaCl2?
10,000 molecules of CaCl2 (            moles of CaCl2           )(111-g of CaCl2 ) = 1.84 x 10-18 g of CaCl2
                                           (6.02 x 1023 molecules of CaCl2)(moles of CaCl2)
Return

Balancing a Chemical Equation
A chemical reaction has the reactants (or substrates) on the left side and the products on the right side.  In chemical reactons:
a) The mass of the reactants must be equal to the mass of the products.
b) The same number of each kind of atom appears on each side of the arrow
c) The arrangement of atoms in reactants is not the same as it is in the products of a chemical reaction.

To balance the chemical equation:
a) write the correct chemical formula for each reactant and for each product.
b) adjust the numbers before each molecule's formula until the atoms are conserved.
    1) always start with the atom that is limiting on one side
    2) continue adjusting until atoms are conserved

A. Find the balanced equation for the decomposition of water into hydrogen and oxygen.
Correct chemical formulas
Note: <> means not equal to
H2O -> H2 + O2
has 
2H + 1O <> 2H + 2O
Adjust number limiting of reactant atoms of O 
by increasing the number of H2O
2H2O -> H2 + O2
4H + 2O <> 2H + 2O
Adjust number limiting of product atoms of H 
by increasing the number of H2
2H2O -> 2H2 + O2
4H + 2O  =  2H + 2O
Balanced Chemical Equation
2H2O -> 2H2 + O2
B.  Balance equation the following chemical reaction:  N2O5 + H2O -> HNO3
Correct chemical formulas and determine the 
number of atoms 
Note: <> means not equal to
N2O5 + H2O -> HNO3
has 
2N + 2H + 6O <> N + H + 3O
Adjust number limiting of product atoms of O 
by increasing the number of HNO3
N2O5 + H2O -> 2HNO3
has 
2N + 2H + 6O = 2N + 2H + 6O
Balanced Chemical Equation
N2O5 + H2O -> 2HNO3
C.  STILL UNDER CONSTRUCTION

Return

Stoichiometry
A.  A variety of phosphorus called red phosphorous is used in match heads. When 0.062 grams of red phosphorus burns with oxygen, 0.142 grams of phosphorus oxide is formed.  What is the empirical formula for this oxide?
 
Mass of Phosphorus used
0.062 g
Molar mass of Phosphorus
31.0 g
Moles of Phosphorus used 0.062 g of P (  mole of P ) = 0.002 moles of P 
                     (31.0 g of P)
Mass of Oxygen used
0.142 g - 0.062 g = 0.080 g of Oxygen
Molar mass of Oxygen
16.00 g
Moles of Oxygen used 0.08 g of O (  mole of O ) = 0.005 moles of O 
                   (16.0 g of O)
Molar Ratio -  P:O  P = (0.002 moles of P) = 2P   --> P2O5 
O    (0.005 moles of O)    5O

 STILL UNDER CONSTRUCTION

Return