Wednesday, May 1, 2013

Stoichiometry! Part 2

I left off yesterday with the beginnings of stoichiometry and how to solve one of the easy problems.  Today, I will show you how to do the second question.

To jog your memory, the problem left was:
b) How many grams of sodium chloride are needed to make a 200 mL of a 0.5 M solution?
Time to solve it.

First, let's get rid of all of the unnecessary words in the problem.
b) 200 mL of sodium chloride  0.5 M solution  g of sodium chloride = ?

Pretty easy.  Now, what is sodium chloride?  Sodium chloride is NaCl.  Now that we know this, it's time to find the molecular mass.  To do this, find sodium (Na) and chlorine (Cl) on the periodic table.  If you have a problem finding it, here's a picture of where it is located on the periodic table (the one I am using is from chemistry.about.com):
Ok, now look for the atomic mass for each.  Some periodic tables differ with this.  For some, the atomic mass is above the elemental symbol and for others, it is below.  In the one that I am using, the atomic mass is located below the elemental symbol.  Here's a picture to help you out.
We now know that sodium's atomic mass is 22.989768 (we'll say it's 22.990) and chlorine is 35.4527 (We'll say 35.453 to give us a nice rounded number).  To calculate the molecular mass, just think of NaCl as an equation.  Sodium is Na and chlorine is Cl.
(Na) * 1 + (Cl) * 1 = Molecular Mass
Plug in the atomic masses.
(22.990) * 1 + (35.453) * 1 = Molecular Mass
Now do the math!
22.990 + 35.453 = 58.443
Do you remember the units for molecular mass?  (Go on ahead.  Take a peek back if you forgot.)  The units are g/mol.  So now we know that the molecular mass of sodium chloride (NaCl) is 58.443 g/mol!

We now come the the question of "What on earth is 'M'?"  "M" is the molarity of a solution (the concentration) and is moles per liter.  (Did you catch that?)  Let's now look at what we have:
200 mL of NaCl, 0.5 M of NaCl, and 58.443 g/mol of NaCl
We actually have all the information that we need to solve this problem!  "But how do we finish it!  There's mL and a random M that have nothing to do with the rest!  We can't possibly solve it!"  Au contraire!  Remember what the units of molarity are?  I already told you so go back and look (I even gave you a hint).  The units for molarity are mol/L!  All we have to do is convert mL to L!  To do this, just divide by 1000.  Why 1000?  That's because there are 1000 mL per L.  It's just that simple.  So now we have:
0.200 L of NaCl, 0.5 mol/L of NaCl and 58.443 g/mol of NaCl
Time to turn it into an equation.
0.200 L NaCl * 0.5 mol NaCl * 58.443 g NaCl
                        1 L NaCl          1 mol NaCl
Time to solve!  The L and mol will cancel out leaving us with g.
0.200 L NaCl0.5 mol NaCl * 58.443 g NaCl
                        1 L NaCl          1 mol NaCl
Now multiply!  (Since anything divided by one is the same, we don't need to worry about dividing).
0.200 * 0.5 * 58.443 g NaCl = 5.8443 g NaCl
Now we know that 5.8443 g of sodium chloride are needed to make a 200 mL of a 0.5 M solution!


Time for the "homework" question from yesterday.  As you recall, the question was:
There are 13.2 g NaCl in a bowl.  How many moles are there of NaCl?
I'm just going to do the solving part.  To find out how or why I did the steps I did, go back to yesterday's post.

Solution:
13.2 g NaCl *    1 mol NaCl    = 0.226 mol NaCl
58.443 g NaCl     .

Time for today's problem for you to try on your own:
How many grams of H2SO4 are needed to make 350 mL of a 0.179 M solution?
Just work on that problem and I'll give you the answer soon!  Hint:  Depending on how much you already know about Chemistry, there are actually two possible answers but the difference between the two is just a mere 0.001 (depending on if you use the entire atomic mass or just three places after the decimal like I'm using.)  In case you don't have a periodic table of your own, here's a link to the one that I'm using.

Have a great day!

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