Solutions & Stoichiometry

It's finally here... Solutions + stoichometry!

Okay, so you might not be so excited about this. But hopefully this blog

can clear up any confusions you have about blending what we already

know about stoichometry with this unit's topic: solutions.

• Today, we started class by turning in our Molarity Stiumlations packets, which we're on schedule to go over tomorrow. If you need one of these, they're in the Unit 4 Handouts folder!

• Then, we started on the torture part of this unit: Solution stoich. 

Here's a brief overview of the notes we took and the problems we did!

Disclaimer: Ms. Friedmann's notes, which she wrote in class today, are in the Unit 4 Notes folder 

under  'Notes on Solution Stoichiometry'. Go check these out, as well.

First, we redefined stoichometry: looking at one "thing" in a reaction (either a reactant or a product) and figuring out how much of another "thing" in the same reaction will be used up or produced based on the measurement of "thing" 1. 

This definition might seem a little extensive, but to remind everyone...

A stoichiometry set-up looks just like this! 

Now, we all remember doing (a little bit too many of) these problems!

So, where do solutions come in?

Well, to solve stoichometry problems involving Molarity, we need new algorithms.

Our old algorithm looked like this:

We've used this algorithm to solve all of our stoichiometry problems to date. But now,
with solutions, we need a new algorithm involving molarity.
The new algorithm:

Obviously, the new parts of this algorithm are:

• Calculations using the volume to find other measurements. Use the molarity to convert from volume to moles, and vice versa.
• Calculations using the molarity to find other measurements. Use the volume to convert from molarity to moles, and vice versa.

Let's do a problem:

This is the first problem on the worksheet entitled 'solution stoich', located in the Unit 4 Handouts folder.

CaCl2 (aq) + 2 AgNO3 (aq) -----> Ca(NO3)2 (aq) + 2 AgCl (s)

What volume of 0.150 M AgNO3 is needed to react with 45.0 mL of 1.50 M CaCl2?

What is "thing 1"? CaCl2, because we are given its molarity and volume.

What is "thing 2"? AgNO3, because we are asked to calculate the volume of it needed to react.

There are two ways to start this problem, either with the molarity or the volume of CaCl2. We chose to start with the volume of CaCl2.

Since the volume is in mL, we must convert to L just by shifting the decimal 3 places to the left.

Knowing our first conversion factor must be 1.50 moles CaCl2/1L, because molarity is moles of solute over liters of solvent, we can begin to set this problem up.

0.045 L * 1.50 moles CaCl2/1L

Now, we need to convert from moles of CaCl2 to moles of AgNO3. To do that, we look at the balanced equation. There are 2 moles of AgNO3 for every 1 mole of CaCl2. So...

0.045 L * 1.50 moles CaCl2/1L * 2 moles AgNO3/1 mole CaCl2

Next, we need to get from moles to volume of AgNO3. So, knowing the molarity of AgNO3 and using the corresponding conversion factor...

0.045 L * 1.50 moles CaCl2/1L * 2 moles AgNO3/1 mole CaCl2 * 1L/0.150 moles AgNO3 =

0.90 L of AgNO3!

And that's it! The rest of this sheet, along with the two others recieved in class today, are for homework. Good luck, and feel free to ask questions in the comments section!

Here's an awesome Khan Academy video on solution stoichiometry if you're still confused:

Next scribe is Mary L!