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

Percent Composition

Percent Composition: The percent in mass that a certain element in a compound contains.  In other words, the percentage of the molar mass that is taken up by a certain element.

          % Composition =    mass of element      *100
                                      mass of compound


What percentage of glucose (C6H12O6) is oxygen?

Molar mass of Glucose:
        6  C @ 12g/mol =     72g/mol
       12 H @   1g/mol =     12g/mol
        6  O @ 16g/mol =  + 96g/mol
                                        180g/mol

Calculate Percent Composition

  96g/mol    *100 = 53.3 %
180g/mol  

Now add the step.  How about if there is more then just one molecule?  Pull out you mole map because this is where you will need it!

What is the percent composition of 5.00g of Vitamin A (C20H30O)?

Step 1:  Molar mass:
      20  C  @  12g/mol = 240g/mol
      30  H  @    1g/mol =   30g/mol
        1  O  @  16g/mol = +16g/mol 
                                        286g/mol

Step 2:  Percentage of total molar mass
  240g/mol C     *100= 83.9%              30g/mol H     *100= 10.5%             16g/mol O     *100= 5.6%                  
286g/mol total                                    286g/mol total                                   286g/mol total

Step 3: Check your work
(always be sure to check that the percentages that you came up with add up to 100%)

  83.9%
  10.5%
 + 5.6%
  100.0%

Now one question is done.  Yes, I know it is a long process.  Going through all the steps is very helpful though and reminds you what is going on.  Do them even when it does start feeling tedious!


Now that you have seen some examples, try out this quiz!!!  Make sure you have some scrap paper and a periodic table at hand!

Percent Composition Quiz

Calculating the Empirical & Molecular Formula

What is Empirical formula:

• Empirical formula gives the LOWEST TERM RATIOS of atoms or moles
• All formula of ionic compounds are empirical
• e.g. NaCl, MgO, Al2O3

How to determine the Empirical formula given the mass of each atoms?

2 simple steps: 

1. Convert all the mass of elements (in grams) to moles
2. Find the simplest whole number ratio between these amounts

e.g. Determine the empirical formula of a 100 g compound of phosphorus and oxygen that contains a 43.64% of phosphorus by mass.

Solution:

In 100 g, there are 43.64 g of phosphorus and 56.36 g (100 - 43.64 g) of oxygen.

Amount of phosphorus = 43.64 g / 30.97 g mol-1 = 1.409 mol

Amount of oxygen = 56.36 g / 16.00 g mol-1 = 3.523 mol

The whole number ratio of phosphorus to oxygen bay be found by dividing through by the smaller number:

Ratio of P : O = 1.409 : 3.523 = 1 : 2.5 (dividing by 1.409) = 2 : 5 (multiplying by 2)

Therefore, the empirical formula is: P2O5

What is Molecular formula:

• Molecular formula gives all atoms which makes up a molecule
• Ionic or covalent compound formulas can be molecular
• e.g. C3H8, Na2CO3

IMPORTANT EQUATIONS!!!

Molecular formula (MF) = Empirical formula (EF) x Whole Number (N)

MF mass = EF mass x N

Mass of one mole = EF mass (in grams) x N

Tool kit

Here is an awesome online empirical formula calculator.  You can check if your answers are correct by typing in the mass percentage of element and the name of the element.

http://firstyear.chem.usyd.edu.au/calculators/empirical_formula.shtml

Practice questions

If you want to practice on solving empirical formula problems, check out these:

http://kurtniedenzu.cmswiki.wikispaces.net/file/view/Empirical+Formula+Worksheet+5.pdf

http://www.fremont.k12.ca.us/cms/lib/CA01000848/Centricity/ModuleInstance/6112/Empirical_WS.pdf

Two Step Mole Conversions

Two Step Conversions: Grams --> Formula Units/particles/molecules/atoms
The process is much the same except now you have to go two steps.  If you want to go from grams to formula units/particles/molecules/atoms then you multiply the number by 1 mol/grams and by 6.022 x 10^23/1 mol.

Ex.  There are 250g worth of pennies.  How many molecules of atoms of copper are present?
Atomic mass= 58.9
molar mass= 58.9g/mol

250g x   1 mol    x 6.022 x 10^23 -- 2.6 x 10^24 molecules C
           58.9g/mol         1 mol

Ex.  An Olympic sized pool has 2.5 x 10^9 g of water in it.  How many atoms of oxygen are present in an Olympic sized pool?
water=H2O
molar mass= 2 O @ 16.0
                     1 H @   1.0
                                        =33g/mol

2.5 x 10^9g x 1mol x 6.022 x 10^23 x   2 atoms   -- 9.1 x 10^31 atoms O
                      33.0g         1 mol            1 molecule  

Two Step Conversions: Formula Units/particles/molecules/atoms --> Grams
If you want to go from formula units/particles/molecules/atoms to grams you go through a similar process.  This time you multiple the number of molecules by 1 mol/6.022 x 10^23 and then by grams/1 mol.  If it is atoms and you have to account for multiple atoms in each molecule, you have to account for that and divide as necessary.

Ex. There are 300 atoms in a sheet of aluminum.  What is its mass in grams?
atomic mass= 27.0
molar mass  = 27.0g/mol

300 atoms x         1 mol        x 27.0g -- 1.3 x 10 ^-20g
                     6.022 x 10^23    1 mol

Ex. There are 2500 molecules of table salt (NaCl) in a salt shaker.  How many grams of salt is there present?
molar mass= 1 Na @ 23.0
                     1 Cl  @ 35.5
                                        = 58.5g/mol

2500 molecules x        1 mol        x 58.5g -- 2.4 x 10^-19g
                            6.022 x 10^23   1 mol

Mole Conversions

Particle/Atom/Formula Unit ---> Moles

ex. How many moles of Ag are present in 3.0 x 10¹⁶ atoms of Silver?
3.01 x 10²⁴ Ag atoms x         1 mol Ag               
                                     6.022x10²³ Ag atoms
= 5.0 x 10^-8 mol Ag

Moles ---> Particles/Atoms/Formula Units
ex. How many atoms are present in 2 moles of Potassium?
2 mol K x 6.022 x 10²³ atoms K
                        1 mol K
=1x10²⁴ mol K

Moles ---> Grams
ex. What is the mass in grams of 3.5 moles of Sulfur?
3.5 mol S x 32.1 g S
                   1 mol S
= 110 g S

Grams ---> Moles
ex. How many moles are there in 2.49 grams of Tin?
2.49 g Sn x   1 mol Sn   
                   118.7 g Sn
= 0.0210 mol Sn