In this post we will have a brief introduction to moles in chemistry. This fundamental concepts is a part of Stoichiometry which is another important aspect of chemistry.
In chemistry an atomic mass unit (amu) is the mass of a proton or neutron in an atom. This number has been calculated to be.
1.66 X 10-24 g
Knowing this number we can calculate how much a single atom weighs. For example, if we want to calculate the weight in grams of oxygen, we know know that helium has an atomic weight of 4. This means that
The amu cancel each other. This number becomes important because if we take the amu of 1 atom of helium (or any element) in grams and divide by the mass of one atom in grams we get the following number no mattter which element we use.
This number above is how many atoms in 4 grams of helium. This number is called Avogadro’s constant but it also referred to as a mole. Knowing this value, it is possible to calculate the mass of single mole of a molecule. For example, if we want to know the mass of a single mole of glucose we would calculate the amu as shown below.
The mass is as follows
|Element||# of Atoms||amu||Total|
|Total =||180.18 amu|
This output tell us that one mole of glucose is 180.18 grams. We can use this information in other ways such as determining how many moles are in a certain number of grams of a substance. If we have 15 grams of magnesium chloride MgCl2. We can calculate how many moles are in this substance as shown below
Step 1 Calculate the amu of the molecule
Mass of MgCl2 = 24.31 amu + 2 * (35.45 amu) = 95.21 amu
Step 2 Determine Conversion Reltionship
1 Mole of MgCl2 = 95.21 grams MgCl2
Step 3 Convert from grams to Moles
We now know that there are about 0.158 moles in 15 grams of magnesium chloride. But we could take this a step further by determining how many molecules are in 15 grams of magnesium chloride as shown below
0.158 * 6.02 * 1023 = 9.84 * 1022
The first number is the number of moles in 15 grams of magnesium chloride and and the second number is one mole.
There are many variations on the calculations that were done here but this is enough to serve as an introduction.