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

Percent Composition %

WELL! Today we learned something new. Something Fresh. Something that sounds harder than it really is!
Percent Composition is a relative measure of the mass of each different element present in the compound.

A formula that may help is : % Composition = Mass of element / mass of compound x 100

Here is an example:
1) Calculate the % composition of NaCl
    1. Find its molar mass: 22.99 + 35.45 = 58.44
    2. Calculate the amount of Na: 1 Na = 22.99
    3. Calculate the weight compared to NaCl: 22.99/58.44 = 39.34%
    4.  Calculate the total Cl present : 1 Cl = 35.45%
    5. Fine the weight compared to NaCl: 35.45/58.44 = 60.66%
* Note: Weight MUST add up to 100%, if not then there is an error in the calculations

Practice Questions:

1.Calculate the percent by weight of each element present in ammonium phosphate [(NH₄)₃PO₄]

2. Bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. Its formula is NaHCO₃. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. 

Well! It is pretty straight forward. This is pretty much what we learned all class!

Here is a very instructive video on how to calculate percent composition if it still confuses you!

http://www.youtube.com/watch?v=xbEeyT8nK84

Mole Conversions

Today we practiced our mole conversions. The first step to this is calculating molar mass.
Add up all the molecular masses of the compound:
Ex. H2O ---->
mass of 1 oxygen atom = 16.0g

mass of 2 hydrogen atoms = 2.0g

16.0+2.0 = 18.0 therefore the molar mass for H2O is 18g/mol

Remember to use Avogadro's number when converting Particles/atoms/molecules

We also did some examples in class from our Mole Conversions Exercise A-C, this involved converting:
Moles to particles/atoms/molecules
Particles/atoms/molecules to moles
Moles to grams

Ex.. Moles to atoms

How many atoms are present in 3 moles of sulphur

3 moles S x 6.022x1023/1 mole = 1.81 x 1025 Atoms


Ex. Atoms to moles

How many moles of Al are present in 2.6 x 1012 atoms of Aluminum

2.6 x 1012 atoms Al x 1 mole/6.022 x 1023 atoms = 4.32 x 10-12 Moles


Ex. Moles to grams

What is the mass in grams of 12 moles of Fluorine gas

12 moles Fl  x 19g/1mole = 228g of Fluorine

Harder Mole Conversions

We learned how to do harder mole conversions.

This means converting from moles to number of atoms for example, where there are multiple steps involved.
One helpful way to do this is to use a mole map.

Here is an example:

Grams----------> Moles----------------> Number of Particles------> #  of Atoms in a Particle



To do this you need to know the operations involved in conversion.

Grams to moles = 1mole/ MMG
Moles to Number of Particles= 1 mole/ 6.022 x 10^23 particles
Number of Particles to Number of atoms = Number of atoms in a particle / 1 molecule of that substance

Going the other way, everything is flipped.

Number of atoms to number of particles= 1 molecule of that substance/ number of atoms in one particle
Number of particles to moles= 1 mole/6.022 x 10^23 particles
Moles to grams = MMG/ 1 mole


Here are 2 examples

Convert 20 g of CO2 into number of atoms of O

First find the molar mass which equals

1 C = 12.0 g/mol
2 O = 32.0 g/mol

so...

20g CO2 x 1 mole/ 44.0g x 6.022 x 10^23/ 1 mole x 2 atoms O/1 molecule CO2=

= 5.47 x 10^ 23 atoms of O



2) Convert 2.34 x 10^ 23 molecules of CaCO3 into grams

= figure out Molar mass of  CaCO3
= 1 Ca = 40.1 g/mol
=1 C =12.0g/mol
=3 O = 16.0 g/mol

= 100.1 g/mol

so ..

2.34 x 10^23 molecules of CaCO3 x 1 mole/6.022 x 10^23 particles x 100.1 g/ 1 mole =

= 38.9 g CaCO3

Thursday was rough.

There were numerous reasons why Thursday was a rough day for students. Especially for those in grade 11 Chemistry. First of all, we had school. There was a mighty blizzard outside that was life-threatening to many, yet we still had to bust our behinds to school. Not only that, we had a Moles quiz. If you think that is bad, which I know it is bad, without even thinking, we also had a sub. I mean, what is Chemistry 11 without Ms. Chen? Don't even answer. To top it all off, we got 2 worksheets. And that was why Thursday was rough.