SOLUTION Stoichiometry

Welcome to our next unit, Solution Stoichiometry!!!!!! We have just finished Quantitative stiochiometry, but according to our resident chemist, there are many different kinds of stoichiometry.

Before we start, you need to know some things................

• Solutions are homogenous mixtures composed of a solute and a solvent
• The solute is the chemical that is present in a lesser amount and dissolves.
• The solvent is the chemical present in larger amounts that has the solute dissolved in it

Chemicals that are dissolved in water( solutes, remember?) are called aqueous and we put a little aq beside their formulas.
ex. NaBr(aq)

Molarity

Molarity is basically a fancy way of saying concentration. There are many ways to express molarity including...g/L, mL/L, % by volume, % by mass, and most importantly for us mol/L. This formulas is represented by a capital m like this         M
Now lets do some examples!!!



Example 1: What is the concentration of .47mol of NaBr in 300mL of water????

okay so first we have to change mL into L so we divide by 1000 which gives us .3 L. Yay
next we use the formula
                                                  M= mol
                                                           L

                                                  M= .47mol
                                                          .3L

                                                  M=1.6 mol/L

Now lets try a slightly trickier one...................


Ex 2:  you accidentally poured 30g of salt(NaCl) into the 800mL of water you are using to cook pasta. What is the concentration(M) of NaCl in your pasta water????

Again we have to change the water in to L so we divide by 1000 to get 0.8L of water. thats done now we have to change the NaCl from grams to moles. You know how to do this if you think back.......................

Thats right we have to use the molar mass!!!

                                                   30g x 1 mol = .513 mol NaCl
                                                             58.5g

Now we can do the real problem just like we did the last one
                                              M= mol
                                                     L
                                              M= .513 mol
                                                      0.8 L
                                              M= .64 mol/L


yay now you know Solution Stoichiometry ( at least some......)!!!!!!!!!!!!!!!!!!!!!


heheheheheheheheh......