January 11, 2011

Why Hello, Hello, Hello!! I hope everybody is in good spirit and is enjoying the weather?!?! Maybe, maybe not? I know that I am definetely NOT!! Anyway enough about the cold weather and lets talk some Chemistry! Woooo!!


Today in class we took notes of the....... Molar Volume of a Gas at STP

- Description of a STP: The definition is that STP is a standard condition that compares volume of gases. STP stands for Standard Temperature & Pressure. A STP is one atmosphere of pressure and a temperature of 0 degrees celsius and 273.1 K. At STP one mole of gas = 22.4 L.

- The two CONVERSION FACTORS:
             1. 22.4L of gas                        2. 1 mole of gas
                 1 mole of gas                          22.4L of gas

Here is an example for you to practice:
1) calculate the volume occupied by 3.4g of ammonia at STP
= molar mass of ammonia (NH3) = 17 g/mole
= 3.4g x 1 mole = 0.2 moles
               17g
= molar volume =  22.4 L  =  0.2 moles x 22.4L
                            1 mole                       1 mole
= Volume occupied by 3.4g of Ammonia at STP = 4.5 Litres.


Now here are some very helpful, useful links that can help you with this topic:
1) http://www.docbrown.info/page04/4_73calcs/MVGsaTEST.htm
2) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson94.htm
3) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson95.htm
4) http://jc-schools.net/dynamic/science/worksheets/MolarVolumeSTPPractice.pdf
5) http://www.savitapall.com/gases/assignments/Molar%20Volume.pdf

Now here are some images that will help you visually:






During class, we got a worksheet called "Chemistry: Molar Volume Worksheet" that should have been completed in class. This sheet is very good for review and to study from as those links above are also too.

For homework we have a sheet called "Chapter 4 Review WS" that needs to completed for next class (which is tomorrow). This is review for our test next week on Chapter 4 on Tuesday! That is all for now, so have a great day!

Diluting Solutions

"Hi, fellow students... unfortunately, the blogger today had an accident with her experiment involving diluting solutions! In the mean time, I will be teaching you guys all about this lesson because obviously I know more than her to start with. She made many dumb mistakes in preparing for workable solutions indeed!"

Diluting solutions is actually a simple math involving the formula we have learned....Molarity M = mole/ litre

The KEY CONCEPT of diluting solutions is:

[MOLES of Solutes is constant before, after and forever...]

Moles of solute b4 = moles of solute after

How to get moles?

M x L !

thus, M1L1 = M2L2

(careful, 2 meaning the second solute, not times 2)

I just don't understand how she could not do this simple calculation to get the solution right...keke...I am sure all of you can do the following examples as fast as I  can do.

Eg. I have 2 Litre of 10 M of Hcl, and I need 0.5 Litre of 1 M Hcl.

Consider if I need all of 10 M of Hcl? No..

I need to solve for how much Volume to start with ...L1

M1L1 = M2L2

10 M x L1 = 1M x 0.5L

L1 = 0.05 L

So if I need 0.05 L of the 10 M Hcl solution to make 0.5 L of 1M Hcl... I would take 0.05 L of concentrated Hcl and add how much water?

0.5 - 0.05 = 0.45 L

These are useful practices to get familiar with the calculation:

Calculate the new concentration (molarity) if enough water is added to 100mL of 0.25M sodium chloride to make up 1.5L.
 
note: here V = the volume in Litre!

• M₂=(M₁V₁) ÷ V₂
• M₁ = 0.25M
• V₁ = 100mL = 100 ÷ 1000 = 0.100L (volume must be in litres)
• V₂ = 1.5L
• [NaCl(aq)]_new = M₂ = (0.25 x 0.100) ÷ 1.5 = 0.017M
      (or 0.0.017 mol/L or 0.0.17mol L^-1)

Calculate the volume to which 500mL of 0.02M coppper sulfate solution must be diluted to make a new concentration of 0.001M.

• V₂=(M₁V₁) ÷ M₂
• M₁ = 0.02M
• V₁ = 500mL = 500 ÷ 1000 = 500 x 10^-3L = 0.500L (since there are 1000mL in 1L)
• M₂ = 0.001M
• V(CuSO₄)_new = V₂ = (0.02 x 0.500) ÷ 0.001 = 10.00L

Thank you for keeping up with me so far... now relax ....


January 5th

Hahaha,,,, today,, we did a experiment,,,
which is lab 4C  FORMULA OF A HYDRATE.

The procedure was:
First,,, Wear your safety GOGGLES!!
 Then set up the pipestem triangle, iron ring, stsand and bunsen burner....
Heat the crucible with busen burner for around 3 min,, to make sure it is dry,,,,,,

-Let crucible cool down and measure the mass of it.

-put the given hydrate inside the crucible and record the mass.

_heat the crucible(with hydrate) until the buttom of crucible is dull red,,, for about five min

-after that,, let it cool down for approximately five min,,,, and record down the mass of it

-to make sure all water is driven off,,,, reheat it for another five min

-cool it down for another five min again and record the mass

-lastly,,,,  drops a few water to the contents of the crucible and see what happen!! XDD


Links:
Another experiment for formula of a hydrate:
http://chem.lapeer.org/Chem1Docs/HydrateLab1.html

A video clip for the expriment:
http://www.youtube.com/watch?v=OuF4hjTFdsg



HAHAHA,, That's all.....

Have A Nice Day! ^^  XDD

DECEMBER 3rd

Do you remember the CH compounds or the compounds containing Carbon?
They are referred to as the ORGANIC COMPOUND--
The Question is...
How do you calculate the Empirical formula of Organic Compounds?







For example, given:

What's the empirical formula of a compound when a 5.00 gram sample is burned producing 15.0 grams of CO2 and 8.18 g of H2O?

Let the empirical formula of the compound be CxHy.

CxHy + zO2 -> xCO2 + y/2 H2O

1. Find the moles of C and H in the compound.

CO2: 15.0 g x 1mol/44.0 g = 0.341 mol

H2O: 8.18 x 1mol/18.02g = 0.454 mol

2.  0.341 moles of C                      0.341/0.341 = 1 x 3 =3

     0.454 x 2 = 0.908moles of H    0.908/0.341 = 2.66 x 3 = 8

3. Ratio of C:H = 3:8

 therefore, the empirical formula is C3H8.

4. Now, check the masses:

0.341 mol of C x 12g/1mol = 4.092 g of C

0.908 mol of C x 1.0g.1mol = 0.908 g of H

Total mass                           = 5.00 g

Therefore, there is no Oxygen atom in the compound.

*Here is a challenging question for an extra thinking~

Adipic acid is an organic compound composed of 49.31% C, 43.79% O, and the rest hydrogen. If the molar mass of adipic acid is 146.1g/mol, what are the empirical and molecular formulas for adipic acid?

if there were 100 g of the sample, it would contain

49.31 g C
43.79 g O
6.90 g H

convert everything to moles

49.31 g C------>4.11 mol C
43.79 g O------>2.74 mol O
6.90 g H -------->6.83 mol H

now, we have the molar ratio

C:O:H = 4.11 mol C : 2.74 mol O: 6.83 mol H

reduce the ratio to form in which the smallest term is 1
do this by dividing each term by the smallest value (2.74)

4.11 mol C : 2.74 mol O: 6.83 mol H = 1.5 mol C : 1 mol O :2.5 mol H

to eliminate fractions, multiply by the LCD (2 in this case)

1.5 mol C : 1 mol O :2.5 mol H = 3 mol C : 2 mol O :5 mol H

thus, the empirical formula is C3H5O2
======================================…
to find the molecular formula, we have to know how many times the empirical weight is the molecular weight
we do this by dividing the molecular weight by th empirical weight

Empirical Weight (EW)
=(12 g/mol C)(3 mol C) + (1.01 g/mol H)(5 mol H) + (16 g/mol O)(2 mol O)
= 73.05 g

Molecular Weight (MW) = 146.1 g

MW / EW = 146.1 g / 73.05 g = 2

multiply this by the subscripts of the elements in the empirical formula
===============

thus, the molecular formula is C6H10O4
thus, the empirical formula is C3H5O2

Did you get it??

Here is a summary of the steps

1. Find # of moles of the compounds containing C and H

2. Find the # of moles of atom of C, H

3. Determine the ratio of C:H based on the # of moles (divide by the smallest number)

4. IF there are decimals, use either rounding or multiply to get a whole number

5. FINALLY, PLEASE check the masses of each atom of C and H to see if it equals the mass (if given) in the question. If it does not, then there must be a O atom in the compound. Find the mass of O atom and convert to moles in order to determine the C:H:O ratio!

December 1, 2010

Wow, I almost can't believe it, it is already the start of December!
Today is officially December 1st so that means it is 24 days until Christmas & 17 more days until Winter Vacation! We are almost there, so keep going everybody!!

Well enough about Christmas (which is my favourite holiday) and lets talk Science.
Today we learned a new topic..... are you ready for this?..........
Empirical + Molecular Formaula!!!!!!
Now this topic is very detailed so stay with me here;

Lets first start with Empirical Formula; (EF)
- EF is the lowest term ratio of atoms in a formula. Otherwise known as reducing.

ex. C6H12 = molecular formula
C3H6 = CH2 = empirical formala

ex. consider that we have 10.87g of Fe and 4.66g of O. What is the EF?
1st Step. Convert grams to moles
Fe 10.87g x 1 mol / 55.8g = 0.195 mols of Fe
O    4.66g x 1 mol / 16.0g = 0.291 mols of O
2nd Step. Divide both by the smaller molar amount
Fe 0.195 / 0.195 = 1  
O   0.291 / 0.195 = 1.5
3rd Step. Scale ratios to whole numbers
Fe 1 x 2 = 2 
O   1.5 x 2 = 3                   Therefore the answer is: Fe2O3

ex. A compund contains 31.9 % K, 28.9 % Cl, 39.2 % O. What is the EF?
* Assume you have 100g
1st Step. Convert grams to moles
K 31.9g x 1 mol / 39.1g = 0.816 mol  
Cl 28.9g x 1 mol / 35.5g = 0.814 mol
O 39.2g x 1 mol / 16g = 2.45 mol
2nd Step. Divide each by the smaller molar amount
K 0.816 / 0.814 = 1
Cl 0.814 / 0.814 = 1
O 2.45 / 0.814 = 3
3rd Step. Scale ratios to whole numbers
Dont need to do because all ratios are whole numbers already (1, 1, 3) 
Therefore the answer is: KClO3

Now I will explain Molecular Formula; (MF)
- MF is the multiple of empirical forms. It shows the number of atoms that combine to form a molecule.
- Formula: n = mm of the compund / mm of the empirical formula. (mm = molar mass)

ex. A molecule has an empirical formula of C2H5 and a molar mass of 58 g/mol.
What is the MF?
MM of C2H5 = 29.0 g/mol
n = 58 g/mol
      -----------     = 2
      29.0 g/mol
MF = 2(C2H5) = C4H10

ex. The empirical formula of a gas is CH2. What is the MF if MM is 42 g/mol?
MM of CH2 = 14 g/mol
n = 42 g/mol
      -----------     = 3
      14 g/mol
MF = 3(CH2) = C3H6

ex. A compound contains 7.44g of C, 1.24g of H, 9.92g of O. The MM is 180 g/mol. What is the MF?
1st Step. Find the empirical formula since it is not shown in the question.
C 7.44 x 1 mol / 12g = 0.62 mols
H 1.24 x 1 mol / 1g = 1.24 mols
O 9.92 x 1 mol / 16g = 0.62 mols
2nd Step. Then divide each by the smallest molar amount which is 0.62
C = 1    H = 2    O = 1      therefore EF = CH2O
3rd Step. Now you can find the MF.
MM of CH2O = 30 g/mol
n = 180 g/mol
      -----------     = 6
      30 g/mol
MF = 6(CH2O) = C6H12O6

That concludes of all what we did today, now I see that in the last blog Percent Composition was not explained so I will explain that right now. Percent Compostion is what we did last class.
Percent Composition: (PC)
- PC is percentage by mass of a "species" in a chemical formula.

ex. What is the PC of CO2?
*Assume you have one mole
Total MM = 44.0 g/mole
% of C = 12.0 divided by 44.0 x 100% = 27.3 %
% of O = 32 divided by 44 x 100% = 72.7 %
Notice that they both add up to a hundred percent, in these equations the two or three answers should always add up to a hundred percent or at least very close to.

ex. Compound contains 5.1g of Cl, 22.0g of C, and has a total mass of 44.1g and contains some amount of O. Calculate the PC.
Mass of Oxygen = 44.1 - 5.1 - 22.0 = 17.0 g
% of Cl = 5.1 divided by 44.1 x 100% = 11.6 = 12%
% of C = 22.0 divided by 44.1 x 100% = 49.9%
% of O = 17.0 divided by 44.1 x 100% = 38.5%
All add up to 100.4 percent. This is a way to check your work too by the way.

ex. Calculate the PC of the underlined components of (NH4)2SO3
Total MM = 116.1
MM of all the NH4 = 36
% of NH4 = 36 divided by 116.1 x 100% = 31.0%

Whew, alot of numbers as you can see :). Now its time to treat you with some video's & practice questions for your own practice!
- Worksheets:
Empirical + Molecular Formula
1) http://eaglepoint.lexingtonchristian.org/hs/science/dhumphreys/Shared%20Documents/Honors%20Chemistry/Empirical%20Formula%20Worksheet%205.pdf
2http://www.ccboe.net/Teachers/nutial_laura/files/CBDD907F339B420CB25588150BD481A9.doc
3) http://www.ezclasssites.com/data/captainkirk/EmpiricalFormulaWS.pdf
4) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson59.htm
5) http://lhs2.lps.org/staff/sputnam/practice/UnitV_EmpForm.htm
6) http://westwood.sjsd.net/~dshoesmith/FOV1-0003789A/FOV1-000378A1/Percent%20Composition%20and%20Molecular%20Formula%20Worksheet.doc?FCItemID=S01C8C50E
7) http://www.cbv.ns.ca/rv/campbell/Resources/M%20F%20EF.pdf

Percentage Composition
1) http://misterguch.brinkster.net/PRA023.pdf
2) http://www.fordhamprep.org/gcurran/sho/sho/worksheets/worksht58a.htm
3) http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson58.htm
4) http://misterguch.brinkster.net/001_024.doc
5http://cmsweb1.loudoun.k12.va.us/52820831134912597/lib/52820831134912597/Moles/Homework/masspercomp.pdf
6http://www.ccboe.net/Teachers/epperson_steve/files/36F0A3C2EC0648DBB63EF8981578B7A0.doc
7http://asd1.schoolwires.com/1741206142323657/lib/1741206142323657/Percentage_Composition_Worksheet.pdf

- Videos
1) http://www.youtube.com/watch?v=xbEeyT8nK84
2) http://www.youtube.com/watch?v=_H009sTvYE0
3) http://www.youtube.com/watch?v=gfBcM3uvWfs

Ooops almost forgot...... HW:
1) Empirical and Molecular Formula
2) Reminder: Quiz next week regarding this material and the material will be covered on Frida (next class).

Thank You and Have A Great Weekend!!!!!

25th November, 2010

Well Hi there~~
Today,,,,, we just finished our quiz on mole conversions,,,,,,  sooooo,, actually,,
WE DIDN"T LEARN ANY NEW THING TODAY!!! =[

Anyways,,,,, I think doing a revision about mole conversions here would be better before we start on something new~~

So here,,,,

Converting gram to mole is : x1mole/MMg

Moles to particles: x 6.022x10^23/1 mole

Particles to #atoms in particle: x #of atoms/1 molecule


P.S> don't forget to that we have homework.
the two worksheets: The wonderful world of mole
                                 Harder mole conversions!

Have a fun weekend. =]