JOKE OF THE DAY! :)

A group of organic molecules were having a party, when a group of robbers broke into the room and stole all of the guest's joules.  A tall, strong man, armed with a machine gun came into the room and killed the robbers one by one. The guests were very grateful to this man, and they wanted to know who he was. He replied: My name is BOND, Covalent Bond.

Lab 4C

We did a lab on Thursday and it was about the composition of the hydrate. 

The purpose of the lab was to determine the percentage of water in an unknown hydrate, to determine the moles of water present in each mole of this unknown hydrate, when given the molar mass of the anhydrous salt and to find out the empirical formula of the hydrate.

So the procedure of this lab was:

1. put on safety equipment (always the most important)
2. set up the equipment (ring stand, ring, bunsen burner, pipestem triangle, crucible, and crucible tongs)
3. turn on the bunsen burner to dry the crucible
4. remove the burner and allow the crucible to cool for about 3 minutes
5. determine the mass of the empty crucible
6. place 1/3 full of hudrate and determine the mass (record it down)
7. place the crucible and contents on the pipestem triangle and heat until the crucible is a dull red (maintain the temperature for 5 minutes)
8. turn off and allow the crucible to cool for 5 minutes 
9. record the mass 
10. reheat the crucible for another 5 minutes, cool as before, and determine the mass (the mass should be the same as the first one) 
11. add water to the content of the crucible and note any changes that occur

Analysis of Results

1) The percentage of water in the hydrate is calculated by :
  mass of water   divided by mass of hydrate x 100%

2) The number of moles the anhydrous salt left behind is calculated by :

Mass after heating – mass of crucible = mass of anhydrous salt
Moles of anhydrous salt = mass of anhydrous salt x 1 mole = moles of anhydrous salt                                                                               120.4g

3) The number of moles of water removed by the heat is calculated by :

Mass of hydrate = mass of crucible & hydrate – mass of crucible

 mass of water = mass of hydrate – mass of salt

Moles of water = mass of water x 1 mole = moles of water

                                                       18g

4) The moles of water per mole of anhydrous salt is calculated by :

Moles of water divided by moles of salt is the  ratio of water to anhydrous salt

5) The empirical formula of the hydrate is MgSO₄ · 7H₂O Magnesium Sulphate Heptahydrate.

Follow-Up Questions

1) SOURCES OF ERROR

• NOT HEATING IT TO BLUE FLAME
• HEATING IT WAY TOO LONG
• WRONG MEASUREMENT
• USING THE WRONG EQUIPMENT

2) Solution :

  Na => 16.1g x 1 mole  = 0.7 moles of Na    x3     2                           23g

C => 0.35 moles of C                              x3         1
O =>  1.05 moles of O                             x3       3 

Water => 3.5 moles of water                  x3     10

The empirical formula is Na₂Co₃ · 10H₂O

3) Solution:

Molar Mass of Na₂CO₃  x  5.82g = 106g/mol x 5.82 = 2.16g
Molar Mass of Na₂CO₃ · 10H₂O   286g/mol

4) Solution:

Moles of Na 18.53g x 1 mole = 0.806 moles of Na        1 x 2          2                                       23g
 

S = 0.806 moles of S                                            1 x 2          2
O = 1.21 moles of O                                             1.5 x 2      3  

H2O = 2                                                               2.5 x 2       5

The empirical formula is Na₂S₂O₃ · 5H₂O.

VIDEOS!

Empirical Formula : Hydrate

The Empirical Formula of Organic Compounds

Finding the empirical formula of an organic compound (any substance with carbon in it) is pretty similar to calculating the empirical formula to other substance. Except it has one additional step at the end.
It can be found by:
- Burning the compound (reacting it with O2)
- Collecting and weighing the products
- The mass of the products, the moles of each element in the original organic compound can be calculated

There are 4 steps you need to follow if you want to find the empirical formula correctly.
Here is an example question and the steps that follow.


What is the empirical formula of a compound that when a 5g sample is burned produces 15g of CO2 and 8.18g of H2O? 


Step 1
Calculate the moles of CO2 + H2O produced.


mol CO2 = 15g CO2 x (1 mol CO2/44.0g CO2) = 0.341 mol
mol H2O = 8.18g H2O x (1 mol H2O/18.0g H2O) = 0.454 mol


Step 2
Find the moles of C + H in CO2 and in H2O.


mol C = 0.341 mol CO2 x (1 mol C/1 mol CO2) = 0.341 mol C
mol H = 0.454 mol H2O x (2 mol H/ 1 mol H2O) = 0.908 mol H


Step 3
The ratio of moles of C to moles of H in the organic compound is 0.341: 0.908.
Therefore the formula could be: C0.4341H0.908 BUT the ratio must be whole numbers.


Step 4
Change the ratio to a whole number ration by multiplying in turn by 2/2, 3/3, 4/4 until you get a whole number ratio.


Ratio of moles = 0.908 mol H/ 0.341 mol C = (2.66/1) x 3/3 = 8/3
The empirical formula = C3H8

JOKE OF THE DAY

> >A psychotic chemist came home from work and had a big
> >fight with his wife. In the heat of the moment, he
> >grabbed a bottle of some lethal chemical substance and
> >forced her to drink it while he screamed: " Die Ethyl,
> >die". The wife dropped dead on the floor and the
> >neighbors who were watching the scene, decided to call
> >the police. The policemen arrived and arrested the
> >chemist. One of them asked: Was there any reason for
> >you to kill your wife? The chemist replied: " There
> >was no chemistry between us. We never bonded well
> >although we tried.In the compound where we lived, our
> >temperaments collided. She always responded negatively
> >to my comments. Our relationship was unstable. There
> >was no possible solution. She had an attitude and I
> >was explosive. Finally, I overreacted. But now I'm
> >glad it's over. I'm in equilibrium again.I will feel
> >free even behind the irons."

Empirical and Molecular Formula

Empirical Formula gives the lowest term ratio of atom in the formula


Finding the empirical formula is somewhat the reverse of finding percentage composition. ( see below)
First you will be given the percentages of each element in a particular compound.

Assuming that you will always be given a 100 gram sample we will convert these percentages to grams. The next step is to find the ratio of moles of each element and then calculate the simplest ratio of subscripts that maintain that ratio.

It is much easier done than said!

Lets start with this data: 36.5% Na, 25.4% S, and 38.1% O
First, convert each percentage to grams : 36.5 g Na, 25.4 g  S, 38.1 g O.
Next, divide each by the grams/mole of that element:

Na: 36.5 g / 23.0 g/mol = 1.58 mol  Na
S: 25.4 g / 32.1 g/mol = 0.791 mol S
O: 38.1 g / 16.0 g/mol = 2.38 mol O

Now we can set up the ration of moles of each element:

Na         S       O
1.58   0.791   2.38

To convert these decimal numbers into whole numbers and maintain the same ration between them, just divide each by the smallest of the subscripts.

1.58/0.791 

0.791/0.791

 2.38/0.791

Our final formula would look like:  Na2SO3   ( sodium sulfite)




MORE EXERCISES!!! :)

Molecular Formula is a multiple of the empirical formula 

Molecular Formula = Molar Mass / Molar Mass of empirical formula

 To find the molecular formula, first find the empirical formula for the data you have been given. Then compare the formula mass for your empirical formula with the formula mass for the molecular formula. You may have to multiply the subscripts in your empirical formula by some factor.

Sample Problem.
Given: 38.7% C, 9.70% H, 51,6% O and a molecular formula mass of 62.0 find the molecular formula:

 First, find the empirical formula ===>  CH3O

 Next find the formula mass: C 1 x 12.0 = 12. 0
                                           H 3 x 1.01 = 3.03
                                           O 1 x 16.0 = 16.0
                                  Formula mass  = 31.0

 Now divide the molecular mass by this formula mass : 62.0 / 31.0 = 2
 Multiply each subscript in your formula by this factor ( 2 here)
Your final molecular formula: C2H6O2

 To check your answer find the formula mass of your final formula. It should be the same as the molecular formula mass.

 Practice:
 Given the following data, find the correct molecular formula:
34.6% C , 9.00% H, 36.4% O and a molecular mass of 176

The correct answer would be:  C8H16O4

The empirical formula would be: C2H4O with a formula mass of 44

The multiplication factor would be 4 ( 176/44.0)

ONE MORE TIME!

VIDEO!!!!!!!!!!!!!! (:

Percentage Composition

The percent composition (percentage composition) of a compound is a relative measure of the mass of each different element present in the compound.
To calculate the percent composition (percentage composition) of a compound

1. Calculate the molecular mass MM, of the compound
2. Calculate the total molar mass of each element present in the formula of the compound
3. Calculate the percent composition (percentage composition)

             total mass of element present ÷ molecular mass) x 100
Example:
NaCl

• Calculate the molecular mass (MM):
  MM = 22.99(Na) + 35.45(Cl) = 58.44g/mol
• Calculate the total molar mass of Na present:
  1 Na is present in the formula, molar mass = 22.99g/mol
• Calculate the percent by weight of Na in NaCl:
  %Na = (molar mass Na ÷ MM) x 100 = (22.99 ÷ 58.44) x 100 = 39.34%
• Calculate the total mass of Cl present:
  1 Cl is present in the formula, molar mass = 35.45g/mol
• Calculate the percent by weight of Cl in NaCl:
  %Cl = (molar mass Cl ÷ MM) x 100 = (35.45 ÷ 58.44) x 100 = 60.66% 

 

The answers above are probably correct if %Na + %Cl = 100, that is,
    39.34 + 60.66 = 100.

Calculate the percent by weight of each element present in ammonium phosphate [(NH₄)₃PO₄]

• Calculate the molecular mass (MM) of (NH₄)₃PO₄:
  MM = 3x[14.01 + (4 x 1.008)] + 30.97 + (4 x 16.00) = 3 x [14.01 + 4.032] + 30.97 + 64.00 = (3 x 18.042) + 30.97 + 64.00 = 54.126 + 30.97 + 64.00 = 149.096
• Calculate the total mass of N present:
  3 N are present, mass = 3 x 14.01 = 42.03
• Calculate the percent by mass of N present in (NH₄)₃PO₄:
  %N = (mass N ÷ MM) x 100 = (42.03 ÷ 149.096) x 100 = 28.19%
• Calculate the total mass of H present:
  12 H are present in the formula, mass = 12 x 1.008 = 12.096
• Calculate the percent by mass of H present in (NH₄)₃PO₄:
  %H = (mass H ÷ MM) x 100 = (12.096 ÷ 149.096) x 100 = 8.11%
• Calculate the total mass of P present:
  1 P is present in the formula, mass = 30.97
• Calculate the percent by mass P in (NH₄)₃PO₄:
  %P = (mass P ÷ MM) x 100 = (30.97 ÷ 149.096) x 100 = 20.77%
• Calculate the total mass of O present:
  4 O are present in the formula, mass = 4 x 16.00 = 64.00
• Calculate the percent by mass of O in (NH₄)₃PO₄:
  %O = (mass O ÷ MM) x 100 = (64.00 ÷ 149.096) x 100 = 42.93%

    The answers above are probably correct if %N + %H + %P + %O =100, that is,
    28.19 + 8.11 + 20.77 + 42.93 = 100


  How to calculate percent composition?
Hmm...
(: