Tuesday, December 14, 2010

The Magnificent LAB 4C :)

Hey, this is Jessica, and TODAY we did something that will blow your mind. We were all assigned a hydrate of an unknown composition, and we used a bunsen burner to evaporate the water molecules. WOO it was so exciting.

Before I tell you about the lab, ther first thing you need to know is what a hydrate is. A hydrate is a compound that has a definite number of water molecules incorporated into its crystal structure. The crystals appear dry, but when these compouds are heated strongly, water is given off, leaving the anyhydrous form of the compound.

The first mass we found was the empty crucible. We weighed the crucible on a centigram scale. The mass my partner and I found was 25.23g. Then we found the mass of the crucible and the hydrate, which was 30.11g.

To find the mass of the hydrate alone, this is the formula:
Mass of crucible and hydrate - Mass of empty crucible = Mass of hydrate
30.11g - 25.23g = 4.88g

Next, we heated the crucible and anhydrous salt using a bunsen burner and weighed it again on the centigram scale. The mass we found was 27.79g. After that was found, we heated the same objects again for a second heating and came up with a mass of 27.78g

To find the mass of the anhydrous salt alone, this is the formula:
Mass of the first heating - Mass of the empty crucible = Mass of anhydrous salt

Lastly, to find the mass of the water given off, this is the formula:
Mass of the hydrate - Mass of the anhydrous salt = Mass of the water given off

1. Calculate the percentage of water in a hydrate

%H20 = 2.33g
             ----------  x 100% = 47.7%

2. Calculate the number of moles of the anhydrous salt left behind.

2.55g x    1mole
             --------------- = 0.02 mol

3. Calculate the number moles of water removed by heat from your sample of hydrate.

2.33g x  1 mol
             ----------- = 0.13 mol

4. Calculate the moles of water per mole of the anhydrous salt.

------  -> 1

------ -> 7

5. What is the empirical dormula of the hydrate?

AB .  7H2O

Thursday, December 2, 2010

Empirical & Molecular Formula

Hi everyone! This is Melissa, and today I will be guiding YOU- yes you- through the steps of.... determining the empirical and molecular formula of a compound!

Empirical Formula: Gives the lowest whole number ratio of atoms in a the formula.
Molecular Formula: A multiple of the empirical formula that shows the actual number of           atoms that combine to form a molecule. (I will show you how to determine what the multiple is later on.)

Let's being with the empirical formula...

What is the empirical formula of a compound consisting of 60.0% C and 40.0% H?

We are going to assume that we have 100g of the compound, because 80% + 20% works out to 100%. It is the ratio that counts, so it's okay if we don't really have 100g- the ratio will still be the same as long as the percentages are the same.

1) Find the number of moles of each element...

80g of C x 1mole of O
                                                   --------------  = 6.67 moles of oxygen

20g of H x 1 mole of H
                                        --------------- =  20 moles of H

2) Now divide each of the values of moles by the smallest value of mole that you've calculated. The number that you get will be that elements ratio.

 For Oxygen:      -------- = 1

      For Hydrogen:  ------ = 3

Therefore, your empirical formula is CH3.

If you ever get ratios that are like 2.50 or 5.66, just multiply all the ratios by a number that will give you a whole number. For example, if you had:


You would multiply each of those ratios by 3 because that would give you whole numbers:

5.66 x 3 = 17
3 x 3 = 9
1 x 3 = 3


Now moving on to molecular formula...

It requires you to find the empirical formula, then multiply it by a multiple. 
To find that multiple, you use this formula: 

              molar mass of compound
multiple = -------------------------------------------
                molar mass of the empirical formula

A molecule has an empirical formula C2H5, and a molar mass of 58g/mol.

Molar mass of empirical: (2x12)+(5x1) = 29g/mol
Molar mass of compound: 58g/mol (given to us in the question)

      ------------- = 2

Now we know that our multiple is 2, so we need to multiply our empirical formula by 2.

2(C2H5) = C4H10

C4H10 is our molecular formula.

With practice, finding molecular and empirical formulas will become less confusing!

Wednesday, December 1, 2010

How to Calculate Percentage Composition (The Fun Way)

Hey It's Jessica Here from the Smarticle Particles! I hope you like this video I made. This is how you calculate Percentage Composition in a non-boring way : ) enjoy.