December 4/09

In todays class we learned about Molecular Formula and we did a lab on dilution.

For the lab Mr.Doktor showed us 5 test tubes each that are different colors and we had to compare our test tube to the other 5 and pick one of them.

First we had to measure the amount of water we needed
Then do our calculations to get the amount of Copper (II) Chloride we needed
Then add our solution to the water and stir
Our last step was to pick the test tube that was the most similar to ours.

A Molecular formula is what the equation should actually be.

You will need to know the Empirical Formula before doing the molecular formula lets jut say our empirical formula is CH2O Molecular Formula which is getting the mass in the equation dividing it by the original molar mass and then just multiply the equation and use subscripts.

Example : CH2O has a molecular weight of 180g/mol find the molecular formula.
Original Molar mass = 30g/mol
Mass in equation = 180g/mol
180g/mol / 30g/mol = 6 C = C6 H2 = H12 O = O6

The molecular formula would be C6H12O6

Video showing Molecular formula:

December 2 2009

In todays class we learned about giving directions for dilution of solutions and learn about dilution of solutions itself.

To give directions to a experiment we must first find the information we need.
concentration > moles > mass

concentration = n/v
N represents number of moles
V represents the volume

Once you find all of that you would tell them how much volume to measure how much grams of the solution you would put in and finally add the solution itself.

Example: Son needs to make a 4M solution of CaCl2. If he needs 2 L what procedure will he use?
4 mol x 2 L = 8mol
4 mol x 111g = 444g
Now we have all of the information we need
1) first measure 2L of water in your test tube
2) weigh 444g of CaCl2
3) add the CaCl2 to the water and stir the solution

We also learned about the dilution of solutions which is when you add water to the concentration it decreases. If the volume is doubled then the concentration would be halved. If the volume is halved then the concentration would be doubled.

4 L = 2M <--- what were starting with
8L = 1M <--- multiply 2 volume divide the M by 2
2 L = 4M <------ divide the volume by 2 multiply the M by 2

to solve for dilution of solutions we would use C1V1 = C2V2
C1 = initial concentration
C2 = final concentration
V1 = initial volume
V2 = final volume

An example would be:
If 30mL of .67M of NaBr is diluted to a total volume of 60ml what is the final Molarity of the solution?

C1 = .67M
C2 = ?
V1 = 30ml
V2 = 60ml

C2 = (V1)(C1)/V2 C2 = (30ml)(.67M)/60ml C2 = .34M\

Video helping to solve dilution of solutions:


Heres a link on Dilution of Solutions:
http://www.emsb.qc.ca/laurenhill/science/cv.html

November 30 2009

Today in chemistry we learned about Concentration which is the amount of solute and amount of solvent some units used for this are g/ml g/l mg/l mg/ml

Solution: a homogeneous mixture Solute: the one present in the smaller amount Solvent : the one present in the bigger amount

The unit we use the most is mol/L. Also known as Molarity.
Molarity = Moles/Volume

Molarity = Moles/Volume only works for aqueous solutions and not gases.

Here are some Examples:

Calculate the Molarity of 2.5mol of HSO3 in 3L solution of [HSO3]
[HSO3] 2.5mol/3L = .83M

How many grams of NaCl are contained in 300ml of .0420M of NaCl solution?
.0420M x .0300 L = .0126 mol
.0126 mol x 58.5g = .737g of NaCl

Heres a video on how to calculate Molarity