2.  1 mole = 6.022 x 10 23 particles

3.  The mass of one mole of atoms of a substance.  The average atomic mass from the chart.  For diatomic molecules, don’t forget to multiply by 2.  Molar mass of nitrogen atoms (N) = 14.01  Molar mass of nitrogen molecules (N 2 ), which is how they appear most frequently in nature = 28.02

4.  Moles x 6.022 x 20 23 = particles  Particles / 6.022 x 10 23 = moles

5.  Moles x molar mass = grams  Grams / molar mass = moles

6.  Must first convert to moles  Then may convert moles to either mass or particles  This will require 2 steps.

7.  Balance the equation  Coefficients represent relative number of moles  For example: Sn + 2Cl 2  SnCl 4 means that it takes twice as many moles of chlorine molecules as moles of tin atoms to react to form the product.

8.  Identify your known and unknown substances.  Your unknown is what you are looking for.  Your known is what you know how much of it you have. You can calculate the number of moles of your known substance.

9.  Calculate moles of known substance  For mass mass stoichioimetry, you will divide the grams of known by its molar mass.

10.  Adjust molar ratio  Multiply moles of known by the unknown coefficient/known coefficient.  This gives you the moles of unknown.

11.  In a mass mass problem, solve for mass of unknown  Moles of unknown x molar mass = grams of unknown

12.  How many grams of tin metal are required to react completely with 100 g of Cl 2 ? Sn + Cl 2  SnCl 4

13.  This is a mass mass problem because I am given the mass of one substance (Cl 2 ) and asked the mass of another substance (Sn)  Sn + 2Cl 2  SnCl 4  100 g Cl 2 / 71 g/mol Cl 2 = 1.41 moles Cl 2  1.41 moles Cl2 x (1 /2) = 0.71 moles Sn  0.71 moles Sn x 119 g/mol Sn = 84.5 g Sn

14.  Essentially the same thing.  If you do it lab:  Divide what you actually got by what you calculated you should have gotten.  If it’s a word problem:  Divide what the problem tells you that you actually got by what you calculated you should have gotten.

15.  React 45 g of tin with excess chlorine gas and get 90 grams of tin IV chloride. What is my percent yield?

16.  Sn + 2Cl 2  SnCl 4  45 g Sn/119 g/mol Sn = 0.378 moles of Sn  0.378 moles Sn x (1/1) = 0.378 moles of SnCl 4  0.378 moles SnCl 4 x 261 g/mol SnCl 4 = 98.7 g  You should have produced 98.7 g SnCl 4  You only produced 90.0 g.  ( 90.0 / 98.7) x 100 = 91.2%

17.  The excess reactant is what you will always have enough of.  The limiting reactant is what when its runs out, the reaction stops.  Candle in a room will burn until the candle burns out. Oxygen is excess, candle is limiting.  Candle in a closed jar will burn until oxygen is exhausted. Candle is excess, oxygen is limiting.

18.  Identified by 2 known substances.  Work the problem twice, once with each known.  The correct answer is the smaller mole value for the unknown.

19.  If I have 50 g of tin and 87 g of chlorine, how many grams of tin IV chloride will I produce? Sn + Cl 2  SnCl 4

20.  Sn + 2Cl 2  SnCl 4  With tin as the known;  50 g Sn/119 g/mol Sn = 0..42 moles of Sn  0.42 moles Sn x (1/1) = 0.42 moles of SnCl 4  With chlorine as the known:  87 g Cl 2 / 71 g/mol Cl 2 = 1.23 moles Cl 2  1.23 moles Cl2 x (1 /2 ) = 0.62 moles SnCl 4  0.42 < 0.62, so tin is my limiting reactant  0.42 moles SnCl 4 x 261 g/mol = 109.62 g SnCl 4