Lab 4C December 9, 2010

On Thursday December 9, 2010, we did a lab in Chemistry Class. It was 4C: Formula of a Hydrate.

First, we heated the crucible for 5 minutes to make sure it was dry and there was no moisture contained in it. Then we took some hydrate and put it in the crucible and weighed it, and recorded it.

We heated the hydrate for 5 minutes and recorded our observations. The hydrate became bluish white in colour. After 5 minutes we weighed it again and found that it had gotten lighter because all of the water it contained was gone. It was now a anhydrous salt.

A hydrate is a water containing compound.
 Copper (II) Sulfate Pentahydrate
                                           CuSO4 * 5H2O





An anhydrous salt is a salt that has been heated up until all of the water it contains has evaporated.

We repeated the experiment twice and found that both times the mass was the same

After we added a couple of drops of water and a reaction took place.
The salt turned blue and gas was released
We learned that this water caused the salt to turn back into a hydrate.

Calculating Empirical Formula of Organic Compounds December 3rd, 2010

The empirical formula of an organic compound can be found by:

-burning the compound(reacts with O2)
- collecting and weighing products
-from mass of products, moles of each element in the original organic compound can be calculated.
What is the difference between empirical and molecular?

What is the empirical formula of a compound whose composition is 45.28% Carbon, 6.289% Hydrogen, 35.22% O and 13.21% Nitrogen?

mol C = 45.28 g x 1 mol/12.0 g C = 3.777 mol

mol H = 6.289 g H x 1 mol/ 1 g = 6.289 mol H

mole O = 35.22 g O x 1 mole/16.0 g = 2.2 mol

mole N = 13.21 g N x 1 mole/ 14.0 g = 0.944 mol



Divide by smallest molar amount

3.777/0.944= 4 x 3 = 12

6.289/ 0.944 = 6.6 x 3 = 20

2.2/ 0.944= 2.33 x3 = 7

0.944/0.944 = 1 x3 =3

= C12H20O7N3






A compound contains 92.26% Carbon and 7.74 % H

What is the empirical formula?

*Assume there are 100 g

92.26 g C x 1 mole/ 12.0 g = 7.688 mol

7.74 g H x 1 mole/1.0 g = 7.74 mol


Divide by smallest amount

7.74/7.688 = 1

7.688/7.688 = 1

Ratio of C to H = 1:1

= C1H1

= CH

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

Cheers!

Empirical Formula

The Empirical formula gives the lowest term ratio of atoms (or moles) in the formula.
All Ionic compounds are empirical formula.




A compound contains 80% C and 20 % H?
What is its empirical formula?

*Assume you have 100 g

80.0 g of C x  1  mole/ 12.0 g = 6.6666 mol

20.0 g of H x 1 mole/1 = 20.0 mol

Divide by smallest mole amount


20.0/6.6666= 3

6.6666/6.6666=1

= ratio of C to H is 1 to 3 = CH3

Here are some examples to help you!!!1
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm
A great video!!!
http://www.youtube.com/watch?v=FWozjZ20JyA


Enjoy!!!

Chapter test is December 7, 2010