Formula of a Hydrate

Today we did a laboratory experiment to determine the empirical formula of a hydrate.
The lab can be found on page 45 of the Essential Experiments for Chemistry lab book.

Glossary:
Hydrate: A compound containing water (H20) in it's crystal structure. The general formula is AB xH2O
Anhydrous: The form of a compound without water.
Carbohydrate: Organic compound with the general formula Cx(H2O)y

The sutructure of a hydrate

Objectives:
To determine the percentage of water in a hydrate, the moles of water present in each mole of the hydrate and to write the empirical formula of the hydrate.

Procedure:
Summarized for full procedure refer to lab book

• Put on safety equipment and set up bunsen burner.
• Heat a crucible for 3 min, then let it cool
• Determine the mass of the crucible
• Place the hydrate in the crucible, determine the mass, and then heat for 5 min once the bottom of the crucible turns red.
• Let it cool, then determine it's mass
• Repeat the previous two steps to verify your results.
• Add a few drops of water and record the changes that appear

Results:

1. Determine the mass of the water by subtracting the after heating mass from the before heating.

Then divide the mass of the water by the mass of the hydrate. Multiply your answer by 100 to get a percent.

2. Calculate the mass of the salt by subtracting the mass of the water from the hydrate. Then divide the mass of the salt by the molar mass of the hydrate which is 120.4g. The will give you the number of moles of the salt.

3. Calculate the # of moles of water present in the hydrate by dividing the mass of the water by the molar mass of water, which it 18.0g. 

4. Divide both molar amounts by the smaller molar amount. This should equal one for the salt and seven for water

5. Substitute these numbers into the general empirical formula AB xH2O.

Answer:

AB 7H2O

Here is a video summarizing the above lab
http://www.youtube.com/watch?v=wNqzkNW72rM&feature=related

Empirical Formula of Organic Compounds

salutations drowsy readers!

BE AWAKE AS ARSENIC BASED LIFE FORMS EXIST! wait.

Before we begin what the title implies, what must brush up on our limited knowledge of what we call organic compounds!

(because embedding is disabled, i must give a link)

if the point was vague,organic compounds are basically anything that contains carbon

AND TO BEGIN NOW OUR LESSON!

We can find the empirical formula of an organic compound by BURNING it. When we BURN the organic compound (reacting with oxygen), this leaves us with the BURNT(okay i'll stop) product. From the mass of the products, the moles of each element in the original UNBURNT(last one, i swear!) reactants can be calculated. 

we can do this because of the wonderful law of conservation of mass!
-the law states that the mass of the product is the same as the mass of the reactants-
it is also assumed that all carbon and all oxygen are used up during the combustion.

Sounds confusing? PERHAPS! Let us try an example!

A 10g sample is burned, producing 20g of CO2 and 8.0g of H2O. What is the empirical formula?

• convert the grams into moles
• Mol CO2 = 20g x 1mol CO2 / 44.0g CO2 = 0.455 mol CO2
• Mol H2O = 8g x 1mol H2O / 18.0g H2O = 0.444 mol H2O

• find how many moles of C and H, these are the elements that make up the organic compound
• Mol C = (0.455 mol CO2 x 1 mol C) / 1 mol CO2 = 0.455 mol C
• Mol H = (0.444 mol H2O x 2 mol H) / 1 mol H2O = 0.888 mol H

• find the empirical mass
• divide both moles by the smallest molar amount
• C = 0.455 / 0.455= 1
• H = 0.888 / 0.455 = 2

• SUM IT ALL UP!
• the empirical formula is CH2

• don't forget to check your answer!
• convert from moles to grams, it should add up to 10g
• IF your answer does not add up, remember that oxygen could be a component of the compound
• mass of O = mass of compound - mass of c + mass of H

and as this concludes the BURNT lesson, worry not readers with a BURNING passion to learn, as I leave you a feeling of toasty(wait, what?) security!

-JY

Empirical and Molecular Formula

The Empirical Formula gives the lowest term ration of atoms/ moles in the formula
**all ionic compounds are given in the empirical formula


Ex. C8H12-------> C2H3
     Molecular to empirical
General case of converting grams of substances to empirical formula:
As we have already learned to do convert from grams to moles for each measurement, by divide it by it's atomic mass.
Then dived each molar amount by the smallest molar amount to receive a ration. Round these ratios to whole numbers and the make them your subscripts of your equation.
An example is provided in the video below.

General Case of converting percentage ratios into empirical formula:
Assume you have 100g of the substance. Convert each percentage into grams. Divide the amounts by their atomic mass. It will give you the ratio's which you'll need to round off and then use as your subscripts.



Molecular Formula: Is a multiple of the empirical formula which shows the actual number of atoms that form  each specific compound

Molar Mass / Molar Mass of empirical formula = Molecular Formula

Examples are in the video below!

This video does a far better job of explaining than I ever could so please actually watch it! It's worth it trust me!

http://www.youtube.com/watch?v=gfBcM3uvWfs