1 mole = 6.022 x 1023 particles

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 (N2), which

is how they appear most frequently in nature = 28.02

Moles x 6.022 x 2023 = particles Particles / 6.022 x 1023 = moles

Moles x molar mass = grams Grams / molar mass = moles

Must first convert to moles Then may convert moles to either mass

or particles This will require 2 steps.

Balance the equation Coefficients represent relative number of

moles

For example: Sn + 2Cl2 SnCl4 means that it takes twice as many moles of chlorine molecules as moles of tin atoms to react to form the product.

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.

Calculate moles of known substance For mass mass stoichioimetry, you will

divide the grams of known by its molar mass.

Adjust molar ratio Multiply moles of known by the unknown

coefficient/known coefficient. This gives you the moles of unknown.

In a mass mass problem, solve for mass of unknown Moles of unknown x molar mass = grams

of unknown

How many grams of tin metal are required to react completely with 100 g of Cl2?

Sn + Cl2 SnCl4

This is a mass mass problem because I am given the mass of one substance (Cl2) and asked the mass of another substance (Sn)

Sn + 2Cl2 SnCl4 100 g Cl2 / 71 g/mol Cl2 = 1.41 moles Cl2 1.41 moles Cl2 x (1 /2) = 0.71 moles Sn 0.71 moles Sn x 119 g/mol Sn = 84.5 g

Sn

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.

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

Sn + 2Cl2 SnCl4 45 g Sn/119 g/mol Sn = 0.378 moles of Sn 0.378 moles Sn x (1/1) = 0.378 moles of

SnCl4 0.378 moles SnCl4 x 261 g/mol SnCl4 = 98.7 g

You should have produced 98.7 g SnCl4 You only produced 90.0 g. ( 90.0 / 98.7) x 100 = 91.2%

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.

Identified by 2 known substances. Work the problem twice, once with

each known. The correct answer is the smaller mole

value for the unknown.

If I have 50 g of tin and 87 g of chlorine, how many grams of tin IV chloride will I produce?

Sn + Cl2 SnCl4

Sn + 2Cl2 SnCl4 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 SnCl4

With chlorine as the known: 87 g Cl2 / 71 g/mol Cl2 = 1.23 moles Cl2 1.23 moles Cl2 x (1 /2 ) = 0.62 moles SnCl4

0.42 < 0.62, so tin is my limiting reactant 0.42 moles SnCl4 x 261 g/mol = 109.62 g

SnCl4