Molarity of Solutions

REVIEW
Solution: homogeneous mixture where one substance is dissolved in another substance
      Solute: small quantity
      Solvent: larger quantity

We need to compare the amount of solute dissolved in a certain volume of a solution.
Molarity= number of moles of solute in one liter of a solution

FORMULA
molarity =    mass of solute    ....also known as.... M= mol
                    volume of solution                                              L


This formula can be altered so that any three of these elements can be found as long as you have the other two.

The molar concentration triangle!

Now lets look at a few examples.
Example:  If there is 5 moles of NaCl in 2.5 L of solution, what is the molar
               concentration?
M=  mol        5 mol
         L          2.5 L                        = 2 mol/L NaCl

               

How many moles of Ca(OH)2 is there in 1.30L of a .75 M solution?
         
                 M=mol/L
           = mol=ML
           = mol= (1.30L)(.75M)
                    = 0.975 (SIG. FIGS!)
                    = 0.98 mol Ca(OH)2




                 What volume in mL of 0.1 mol/L NaOH must you take to contain .030 mole of NaOH?
           
                  M= mol/L
                  L= mol/M
                  L= .030mol / 0.1 mol/L = 0.3L
                                                       = 0.3 *     1L     = 300 mL
                                                                   1000 mL

Hopefully now you understand the basics of it.  Its now time to practice!  This game includes some examples of molarity and mole conversions.  Good luck!

Billionaire-molar concentration

Lab 4C: Formula of a Hydrate

Objectives
1. to determine the percentage of water in an unknown hydrate
2. to determine the moles of water present in each mole of this unknown hydrate, when given the molar mass of the anhydrous salt
3. to write the empirical formula of the hydrate

Supplies
Equipment                                                    ring stand and ring
lab burner                                                     centigram or digital balance
crucible and lid                                              lab apron
crucible tongs                                                safety goggles
pipestem triangle
Chemical Reagents
Approximately 5g of a hydrate
water

Procedure
1. Put on safety goggles
2. Set up pipestem triangle, iron ring, stand and bunsen burner like in 4C-1 in your workbork (Page 46)
3. Heat the crucible for 3 min and let it cool for 3 min
4. Determine the mass of empty crucible and record the mass in Table 1
5. Place hydrate into the crucible. Determine and record the mass of the crucible and hydrate
6. Place the crucible and contents on the pipestem triangle. Increase heat till the bottom is red and maintain this temperature for 5 min.
7. Turn off the burner and let the crucible cooldown for 5 min. Then determine and record the mass of the crucible and its contents.
8. Reheat the crucible for 5 min then let it cool for 5 min and then determine and record the mass. Step 6 and 7 should be within 0.03g, if it's not, reheat it again.
9. After the final mass reading, and crucible is cooled, add a drop of water into the contents of the crucible and note any observation/changes.

Table 1 

Analysis of Results
1.      g H₂0       x 100%   =    % of water  
      g hydrate

2. mass of salt = mass after heating - mass of empty crucible
    moles of salt = mass of salt x 1 mol  
                                                159.6g
3. mass of hydrate = mass of crucible/hydrate - mass of crucible
    mass of H₂0 = mass of hydrate - mass of salt
    moles of H₂0 = mass of H₂0 x 1 mol  
                                                     18g
4. moles of H₂0  = _____          
      moes salt
5. CuSO₄  x  __ H₂0  (copper (II) sulphate pentahydrate)

Empirical Formula of Organic Compounds

Organic Compounds are classified as any substance that contains carbon.
They can be found by:

ʘ Burning (as it reacts with oxygen gas [O2])

ʘ Collecting and weighing the products

                From the mass of the product, the mass of each element in the original Organic Compound can be found.

Keep in mind that the Empirical Formula of a Organic compound is CxHy.

The balanced chemical equation for the burning of a organic compound can be written as:

CxHy + 2O2 => XCO2 + Y/2H2O

The aim is to find the x and the y (which are the simplest ratio of atoms of mole in the elements)

From the equation, we can see that all of the C and H atoms in CxHy we used while making CO2 and H2O. It can therefore be said that the moles of carbon and hydrogen in the product are equal to the moles of Carbon in CO2 and the moles of Hydrogen in H2O.

Example:
A 3.27g sample is burned and produces 10.27 g of CO2 and 4.20g of H20. What is it's empirical formula?
Step 1: Find the moles of each
CO2  =>  C = 12                   = 44     10.27g CO2 x 1mole CO2/44g CO2 = 0.233 moles of CO2
                O = 16 x 2 = 32

H2O  =>  H = 1 x 2 = 2         = 18     4.20g H2O x 1mole H2O/18g H2O = 0.233 moles of H2O
                O = 16
Step 2: Find the moles of Carbon and Hydrogen
0.233 mol CO2 x 1 mol C/1 mol CO2 = 0.233 mol H
0.233 mol H2O x 2 mol H/1 mol H2O = 0.466 mol C
Step 3: Divide by smaller number
C 0.233 / 0.233 = 1
H 0.466 / 0.233 = 2
Therefore Empirical Formula = CH2
Step 4: Check Masses
0.466 mol H  x 1g H/1 mol H    = 0.466 g H
0.233 mol C x 12g C/1 mol C  = 2.80 g H
0.466g H + 2.80g C = 3.27g
If there is a difference between the mass of carbon and hydrogen versus the mass of the compound, there is oxygen in the compound. To find the amount of oxygen,
Mass of O = Mass of compound - Mass of C + H

For further explanations, please watch this video:

http://www.youtube.com/watch?v=9qGo9YTdnMg