Stoichiometry
Stoichiometry is, in its most basic form, the math behind the chemistry. Using stoichiometry, you can find out the mass and moles of an element used in a chemical reaction - but you need to keep a few things in mind.
1. The molar mass of the substances (no need to memorize these though - that's what we have the periodic table for!)
2. The balanced chemical equation - without this, all hope is lost for even attempting to do the calculations.
3. And, be prepared to use lots of dimensional analysis - I promise you, it is your best friend.
What you're going to do next, is use the mole box. With the mole box, you start off with the molar mass of one compound - for example, let's use the chemical equation H2O --> H2 + 1/2O2. The example we'll use will be going from one molar mass to another - so, from grams of H2O, to grams of O2. Let's say that we started off with 10 grams of H2O, and we're trying to decide how many grams of O2 needed to be used to offset that.
So, first step: Get from mass of water, to moles of water.
So, as mentioned before, our original mass of water is 10 grams. What we're going to do (using dimensional analysis!) is divide the mass of water used, by the molar mass. Why? Well, the molar mass of water (16 g) is the mass of ONE mole of water. So, if we divide the mass we have of water, by the mass of one mole of water, that would leave us with the moles of water that we have. Ex: 10 g H2O x 1 mol H2O/16 g H2O = 0.625 mol H2O
Second step: Getting from the moles of one substance, to the moles of another.
This is where the balanced chemical equation comes into play. In a balanced chemical equation, the moles are there - you just need to know to look for them. Like, in our chemical equation - for every one mole of H2O, there's one mole of H2 and two moles of O2 used. So, we would take the moles we have and multiply by the molar ratio - 2 moles O2 for one mole of H2O. Keep in mind, these fractions can be flipped if need be - one mole of H2O for 2 moles of O2. So, the next calculation would be: 0.625 mol H2O x 2 mol O2/1 mol H2O = 1.25 mol O2.
Third step: Getting from moles of substance 2, to grams.
Basically the same idea as the first step, just reversed. Instead of diving by the number of moles, you'll divide by the molar mass of the second substance, giving you a calculation like this: 1.25 mol O2 x 1 mol O2/32 g O2 = 0.039 g of O2.
Using these 3 steps, you can basically get anywhere - from mole 1 to mole 2; gram 1 to mole 2; even mole 1 to gram 2!
Worksheets:
http://www.bishops.k12.nf.ca/science/chem/2202/worksheets/mass_mole/Stoichiometry%20Worksheet.htm
Practice Test:
http://www.avon-chemistry.com/stoich_prac_test_1.htm
1. The molar mass of the substances (no need to memorize these though - that's what we have the periodic table for!)
2. The balanced chemical equation - without this, all hope is lost for even attempting to do the calculations.
3. And, be prepared to use lots of dimensional analysis - I promise you, it is your best friend.
What you're going to do next, is use the mole box. With the mole box, you start off with the molar mass of one compound - for example, let's use the chemical equation H2O --> H2 + 1/2O2. The example we'll use will be going from one molar mass to another - so, from grams of H2O, to grams of O2. Let's say that we started off with 10 grams of H2O, and we're trying to decide how many grams of O2 needed to be used to offset that.
So, first step: Get from mass of water, to moles of water.
So, as mentioned before, our original mass of water is 10 grams. What we're going to do (using dimensional analysis!) is divide the mass of water used, by the molar mass. Why? Well, the molar mass of water (16 g) is the mass of ONE mole of water. So, if we divide the mass we have of water, by the mass of one mole of water, that would leave us with the moles of water that we have. Ex: 10 g H2O x 1 mol H2O/16 g H2O = 0.625 mol H2O
Second step: Getting from the moles of one substance, to the moles of another.
This is where the balanced chemical equation comes into play. In a balanced chemical equation, the moles are there - you just need to know to look for them. Like, in our chemical equation - for every one mole of H2O, there's one mole of H2 and two moles of O2 used. So, we would take the moles we have and multiply by the molar ratio - 2 moles O2 for one mole of H2O. Keep in mind, these fractions can be flipped if need be - one mole of H2O for 2 moles of O2. So, the next calculation would be: 0.625 mol H2O x 2 mol O2/1 mol H2O = 1.25 mol O2.
Third step: Getting from moles of substance 2, to grams.
Basically the same idea as the first step, just reversed. Instead of diving by the number of moles, you'll divide by the molar mass of the second substance, giving you a calculation like this: 1.25 mol O2 x 1 mol O2/32 g O2 = 0.039 g of O2.
Using these 3 steps, you can basically get anywhere - from mole 1 to mole 2; gram 1 to mole 2; even mole 1 to gram 2!
Worksheets:
http://www.bishops.k12.nf.ca/science/chem/2202/worksheets/mass_mole/Stoichiometry%20Worksheet.htm
Practice Test:
http://www.avon-chemistry.com/stoich_prac_test_1.htm