November 13/09

Today we learned about empirical formulas and how to create them. Molecular formulas show the actual bond between an atom and a molecule.

Ex. Cl8O4

An Empirical formula would show the bond between an atom and a molecule in its simplist form.

Ex. Cl2O

To find an empirical formula there are many steps involved i will now demonstrate what i learned in chemistry class today.

A sample of an unknown compound is analyzed and it contains 9g of Mg 15g of O and 5g of C

find the empirical formula.

You would want to start out by geting the molar mass of all the elements

Molar Masses:

Mg - 24

O - 16

C - 12

Then you would want to get there mass which you is stated above in the question.

Masses of elements:

Mg - 9g

O - 15g

C - 5g

After you would want to find the number of moles in each element you would do that by geting the mass and dividing it by the atomic mass.

Number of Moles:

Mg- 9 / 24 = .375mol

O - 15 / 16 = .938mol

C - 5 / 12 = .42mol

Then you would get your moles and divide it by the SMALLEST number of moles you got in this case it would .375mol.

Moles / SMALLEST mol :

Mg - .375 / .375 = 1

O - .938 / .375 = 2.5

C - .42 / .375 = 1.1

Then round your answers

Rounded answers:

Mg - 1

O - 3

C - 1

The final step would be get those numbers for each of the elements and put them in as subscripts. So the empirical formula would be MgO3C.

If you still want more help and example on empirical formulas this website should help.

http://dl.clackamas.cc.or.us/ch104-04/empirica.htm

November 11/09

Today in chemistry 11 we learned how to calculate the mass of an elements in compounds.


Find the percentage of each of the following elements C2H2O

First you would have to get all the molar masses of the compound and add them together

2C - 24
2H - 2
1O - 16
total : 42

Then you choose 1 element lets say we go with carbon first so we would get the molar mass of carbon and divide it by the total mass of all the elements 24 / 42 and multiply it by 100 which gives you 57%. You would do the same for hydrogen 2 / 42 x 100 4% and do the same for oxygen 16 / 42 x 100 which gives you 38%.

Percent
C2 would be 57%
H2 would be 4%
O would be 38%


We also learned how to find a certain amount of an element in a certain amount of grams.

For example find the mass of magnessium contained in 50g sample of MgNa
First find the molar masses of all the elements

Magnessium = 24
Sodium = 22
Then add them both together which gives us 46

Second since were looking for the mass of magnessium we get the molar mass of magnessium 24 and divide it by the total mass of the two elements 24 / 46 which gives us .52

After we would get .52 and multiply it by the given grams in the samepl of MgNa which is 50 x .52 which gives us 26. This would tell us how many grams of magnessium are contained in a 50g sample of MgNa which is 26g.



Heres a video pretty much summing up todays lesson and giving us a head start on the next lesson empirical formulas and percentages.

November 4/09

In today's class we learned about density and there relationship between moles.
Density= Mass/Volume Mass= Volume*Density Volume= Mass/Density

Finding the density of gases at STP is easier then finding it at solids and liquids because we know that the volume of the gas will be 22.4L. We can also find the mass by checking our periodic tables. After you just plug in the numbers and solve for the density of the gas.

Molar Mass g / 22.4L/mol = ?g/L

density of oxygen =
Density = Mass / Volume

32g / 22.4L/mol = 1.43g/L at STP

Finding the density of solids and liquids are much harder you do not have all the information you need such as the volume. Luckily Mr.Doktor gave us a neat chart to help us remember how to get the information we need.
use the formula g / molar mass of ele. 6.02 x 10 to the power of 23 subscripts
Density ---> Mass ----> Moles ---> Molecules ----> Atoms

Example: the density of Boron (solid) is 2.34g/mL how many molecules are in a 60ml piece?
2.34 g/ml * 60.0 mL = 140.4g
140.4 g * 1mol / 10.8 g = 13mol
13mol * 6.02 * 10 to the power of 23 / 1mol = 7.84 * 10 to the power of 24

Neat website that can solve density volume or mass for you:
http://www.1728.com/density.htm

November 2/09

In todays class we did a lab to see if the volume at STP is really 22.4L.

1.The materials were:
2.100ml graduated cylinder
3.lighter with butane gas
3.a sink filled with water
4.scale



The procedure was:
1)fill the sink with water don't over flow it
2)put the graduated cylinder underwater until it is completely filled with water and there no gas bubbles
3)put your lighter underwater so that water gets in the lighter then go weigh it and record the 4)mass make sure you dry your lighter before weighing it
5)then put the lighter underwater at the entrance of the graduated cylinder and release 10ml of butane you should see gas starting to form and water starting to decrease
6)once you released 10ml of butane dry off your lighter and record its mass



Observations:
Mass of lighter : 16.5
Mass of lighter after butane released: 16.3
number of moles in the lighters mass: 0.3g

Percentage error

22.4 - 16.3 / 22.4 * 100 = 27%
some errors we had in this experiment was we had to redo the experiment twice because water got in our lighter and made it heavier even though we released butane gas and we released 20ml of butane gas instead of 10ml.

Chemistry October 29/09

Atoms and Molecules

For monotomic elements: a molecule=an atom
For diatomic elements: a molecule Cl2 an atom CL

Molecules of Compounds

O

H H =H2O

2H atoms
1O atom

= 1 molecule

Ammonium Carbonate: NH3 + CO3= (NH4)2 CO3=14 atoms
Converting moles to molecules and vice versa:

6.02x10^23 molecules/1mole
OR
1mol/6.02x10^23 molecules

Examples:
How many molecules are in a 0.25 mol sample of CO2?
0.25 mol x 6.02x10^23 molecules/1mol = 1.51x10^23 molecules

There are 1.51x10^23 carbon atoms
and there are 3.02x10^23 oxygen atoms because its O(2)

5.1772x10^24 molecules of water = ? moles
5.1772x10^24 molecules x 1 mole/6.02x10^23 = 8.6 mol

Another example:
Find the number of 'H' atoms in 4.0 mol of ammonia.
NH3 convert: mol → molecules → atoms

4.0 mol x 6.02x10^23/1mol = 24.0x10^23 molecules
= 3(24.0x10^23)
= 7.22x10^24 H atoms

If you still don't get this here is a website that can help you:
http://www.chem1.com/acad/webtext/intro/MOL.html

Here's a video showing how to convert moles to atoms:

October 27/09 Chemistry 11

Today in chemistry 11 we did a lab, to determine the ratio of moles of copper produced to moles of iron consumed during the chemical reaction between Iron and Copper (II) Chloride.



The procedure for the lab is :

1)Weigh your empty 100ml beaker and record its mass.
2)Add around 8g of copper(II) chloride to the beaker and record its mass
3)Add 50ml of distilled water to the beaker and stir until the copper(II) chloride dissolves
4)Get two nails that are clean and dry record the two nails masses and record it
5)Put the nails in your beaker that contains the copper(II) chloride for 20 minutes you should start to see copper form at the bottom of your beaker
6)Use the tongs to pick up each nail and then use the scupula to scrape off the remaining copper on the nails into the beaker
7)Dry the nails in the drying oven and record its mass once its dry
8)Decant the liquid from the solid, pour the liquid into another beaker (just in case you lose some of the copper you can still retrieve it from the other beaker)
9)Rinse the solid with 25ml of distilled water then decant again into the beaker you decanted into before
10)Wash the copper with with 25ml of hydrochloric acid then decant the hydrochloric acid and was the copper with 25ml of distilled water decant again after
11)Put your beaker with the copper in it in the drying oven to dry
12)Let the copper with the beaker dry and record the mass of the beaker with the copper in it
13)Wash your hands clean up your lab station put away your materials and lab aprons and safety goggles

After that we recorded our information:

Empty beaker: 92.5g
Beaker with copper (II) chloride: 97.67
Nails: 5.54
Nails after the solution: 5.2
Beaker with copper: 94.24

After we managed to determine the amount of iron used which was .34 and amount of copper produced which was 1.74

We ended up with a ratio of 3:1 the correct answer was 3:2 we had a 20% percentage error.
Some errors where we went wrong in the lab was we did not let our materials completely dry before weighing them and we also did not let our nails sit for the full amount of time in our solutions.