stoichiometryreviewpart1

work in progress

**__ Stoichiometry __** Terms you should know:


 * Saturated solution: contains the max amount of solute for a given amount of solvent at a constant temperature (i.e. kool-aid with undissolved sugar at the bottom of the glass)


 * Solubility: the amount of solute that can be dissolved in a given quantity of solvent at a given temperature


 * Unsaturated solution: contains less than the max amount of solute for a given amount of solvent at a constant temperature (i.e. packet of sweetener in a cup of coffee)


 * Miscible: when two liquids are able to dissolve in each other (i.e. water and food dye)


 * Immiscible: liquids that are **not** miscible (i.e. oil and water)
 * Henry’s law: S= solubility of gas, P= pressure in STP


 * Supersaturated solution: contains more than the max amount of solute a solvent should theoretically be able to hold at a given temperature. (when crystallization occurs if a seed crystal is added to solution)

**__ Molarity __**
Definition: The number of moles in a solute dissolved per litres of solution.


 * The Holy Trinity of Stoichiometry: **

To find the molarity of a solution, take the number of moles of solute in the solution over the total volume of the solution in litres. So when 2 moles(mol) of glucose are dissolved in 5 litres(L) of water, the solution has a molarity of 0.4M (or mol/L). To find the moles of solute in a solution, multiply the solution molarity by its volume, and to find the total volume of the solution, take moles of solute over the solution's molarity. In brief;  M = mol/L L = mol/M mol = M x L Where L is the volume of the solution in litres. ** __  Dilution  __

The number of moles of solute in the solution remains constant even after dilution. ** And because mol = M (Molarity) x V (volume in litres or millilitres):


 * mol = M 1 x V 1 = M 2 x V 2

** Example: How many millilitres of a stock solution of 2.00M MgSO 4 would one need to prepare a 100.0 mL of 0.400 M MgSO 4 ? M 1 = 2.00M MgSO 4 M 2 = 0.400M MgSO 4 V 2 = 100.0 mL V 1 = ?

M 1 x V 1 = M 2 x V 2 V 1 = (M 2 x V 2 ) / M 1

V 1 = (0.400M x 100.0 mL) / 2.00M V 1 = 20.0 mL

So 20.0 mL of 2.00M MgSO 4 must be diluted by 80. 0 mL of distilled water, in order to get the final volume of 100.0 mL 0.400M MgSO 4.

Percent by volume = (Volume of solute / volume of solution) x 100%
 * __ Percent Solutions  __  **

Percent by mass over volume = (mass of solution (g) / volume of solution (mL)) x 100%