Solution Stoicheiometry

heyo subscriber!

Today we just finished basic stoichiometry! Now we are going deeper in to stoichiometry, we are now dealing with solutions.

What is a solution? Well it is a mixture of two different types of substances. One is the solvent which dissolves something called the solute. One type of common solvent is water which makes compounds or elements aqueous when dissolved in water.

Take salt for example:
NaCl + H2O => NaCl(aq)

So yeah, we did basic stoichiometry solutions like this.
Want something harder?

check it out, part 2 and part 3 is in the related videos :)

Thank you! 

Mole to Mole Conversions

- Coefficients in balanced equations tell us the number of mole reacted or produced
- They can be used as conversion factors
- What you need over what you have

ex. N2 + 3H  -----> 2NH3

3.5 x  2  =7 mols
          1 

Percent Yield

  One way to think of percent yield is as a measure of success. To find this number you will need two things: the theoretical yield and the experimental yield. The theoretical yield is the amount of products that should be formed and the experimental yield refers to the amount you're given based on the experiment. Because you are usually given the experimental yield, you will need to find the theoretical yield first...

EX.1:

If you burn 12 g of carbon (C) to produce 48.0 g of carbon dioxide (CO2) what is the percent yield?

1. The first and one of the most important steps is to make sure you have a correct and balanced equation.

C + O₂ --- > CO₂

2. You will then have to find the theoretical yield of carbon dioxide because you are given the experimental value. Take what you have and convert it into moles as it is the most common conversion factor.

12 g of C    x    1 mol of C

                      12.0 g of C

3. Next, use your mole to mole ratio: what you need over what you have. Cancel out what you can.

12 g of C    x    1 mol of C    x    1 mol of CO2

                      12.0 g of C            1 mol of C

4. After having your moles of Carbon Dioxide, use molar mass as your conversion factor and cancel out what you can.

12 g of C    x    1 mol of C    x    1 mol of CO2    x    44.0 g of CO2

                      12.0 g of C            1 mol of C            1 mol of CO2

5. Multiply and divide your equation and you will now have your theoretical yield. Plug it into the percent yield formula.

% YIELD    =    Experimental     x 100

                           Theoretical

% YIELD    =    48.0 g    x    100

                          44.0 g

% YIELD    =    109%

For more examples check out this link:

 http://www.youtube.com/watch?v=TKNxdL7DN1I

 

Mass to Mass Conversions

  In order to complete mass to mass conversions, there is one additional conversion you have to make that involves moles.

EX.1: How many grams of O2 are there when 10 g of Al2O3 decomposes?

1) To start off, create a balanced equation.

Al2O3  ->  Al + O2

2Al2O3  ->  4Al + 3O2

2) Since you are given the mass of Al2O3, use the molar mass of the compound in order to cancel out the grams and left with moles

10 g of Al2O3   x   1 mol Al2O3

                           102.0 g Al2O3

3) Next, you will have to cancel out the moles of Al2O3. One way to look at the co-efficients of a compound is to think of them as moles, which you can use to cancel out Al2O3. Put what you know on the bottom and put what you need on the top.

10 g of Al2O3   x   1 mol Al2O3   x   3 mol O2

                           102.0 g Al2O3      2 mol Al2O3

4) Now that you have the moles of O2, you can now use molar mass as a conversion factor in order to find out what you're looking for which the mass of O2.

10 g of Al2O3   x   1 mol Al2O3   x   3 mol O2    X   32.0 g O2

                           102.0 g Al2O3      2 mol Al2O3     1 mol O2

5) Finally, calculate your answer and leave it in its significant digits.

10 g of Al2O3   x   1 mol Al2O3   x   3 mol O2    X   32.0 g O2   =   4.7 g of O2

                           102.0 g Al2O3      2 mol Al2O3     1 mol O2

For more examples about mass to mass or moles to mass conversions, check out this video:

http://www.youtube.com/watch?v=sTsshxcHhaI