Chem CLass- December 15, 2009

Today, we missed some class time because of a presentation in a gym for the whole school. We did take notes on chemical equations, however.

CHEMICAL EQUATIONS
- Shows the chemicals used up (reactants) and produced (products) during a chemical reaction
- The numbers in front of the symbols are called coefficients and they refer to the number of moles

BALANCED CHEMICAL EQUATIONS
- There are the same number of each type of atom on each side of the equation
- Mass, atoms and charge are conserved

Ex: Sn + O2 -----> SnO
= 2 Sn + O2 -----> 2SnO

Ex: Na + H20 -----> NaOH + H2
= 2 Na + 2 H20 -----> 2NaOH + H2

Ex: NH3 + O2 -----> NO + H20
= 4 NH3 + 5 O2 -----> 4 NO + 6 H2O

For some extra help, here's a video on balancing chemical equations:

Chem Class- December 11, 2009

Today we got our tests back! We hope everyone did well.

• We got a new outline for our new unit. We are are going to be learning about chemical reactions!
• Mr. D showed uss a small lab with hydrochloric acid (PH of -1)
• He added bromothemal blue and the hydrochloric acid turned a light yellow colour
• Then he added ammonia to the mixture. The top of the mixture turned blue. He added more ammonia and the blue stayed on top of the acid.
• Mr. D turned on the gas valve and we found out that someone hadn't turned it off. It stunk really bad!
• If he lit a match, there would be an explosion
• Next, he showed us another experiment where he mixed in potassium iodie with iron iodide. It made the mixture turn yellow
• He also got out the match and burned magnesium. We saw a spectacular light, but were not supposed to look directly at it. Apparently this was how fireworks were made.
• Finally, he dipped a piece of paper in salt water and alcohol and put it through the bunsen burner. Flames appeared for a short instant but it did not burn the paper. The alcohol was burning.

<---burning magnesium


Signs of a chemical Reaction: colour change, property change, energy change, temperature change, precipitate forming, gas production, heat and light produced...

Chem Class- December 4, 2009

For chem we did our lab on Dilution and Creating Solutions.

 

Basically, the procedure goes like this:

1. Measure desired amount of water
2. Measure needed amount of copper (II) Chloride (__mol/L x ___L = # of mol x ___g/ 1 mol = # of g)
3. Add to the water and stir until it dissolves
4. Compare results to 5 testubes (labelled a, b, c, d, e) for the best match (lighter shades of blue to darker shades of blue)

Copper (II) Chloride

  

We also went over the make up of the upcoming test. It will consist of:

• The mole
• The moles conversions chart
• % composition, % mass, empirical formulas
• Mass of elements in compounds
• Concentration
• Molar volume
• Dilution
• Density
• Molecular Formulas

Chem Class- December 2, 2009

After going over questions #25-28 from our Unit 2 Homework Questions, we took notes on Giving Directions and Dilution of Solutions.

GIVING DIRECTIONS

Outline for experimental procedures

---------> simply find the mass you need.

It isn't hard. You just convert from con'c---->moles---->mass

Remember that C=n/v (Concentration = number of moles over volume)

Then you give the directions like so, filling in the missing information:

a) Measure (Volume in L or mL) of Water

b)Weigh (mass in g) of (element or compound)

c) Add (element or compound) to water and stir until dissolved.

Example: Lizzy is asked to make a 3.00 M solution of NH4Cl.  If she needs 1.00 L, what procedure should she use?

3.00 mol L x 1.00 L = 3.00 mol

3.00 mol mol x 53..5 g/1 mol = 161 g

a) Measure 1.00 L

b) Weigh 161 g of NH4Cl

c) Add NH4Cl to water and stir until dissolved

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 DILUTION OF SOLUTIONS

When you add water, the con'c decreases. If the volume is doubled, the con'c is halved. If the volume is halved, the concentration is doubled.

6.0 L = 2.0 mol/L (Start out with)

12.0 L = 1.0 mol/L (Multiply the volume by 2, divide the volume by 2)

48.0 L = 0.25 mol/L (Multiply the volume by 4, divide the volume by 4)

The number of moles is doesn't change though (In this case 12.0 mol)

Remember that C1V1 = C2V2 (Where C1 is equal to the initial concentration, C2 is equal to the final concentration, V1 is equal to the initial volume and V2 is equal to the final volume)

Ex: If 20.0 mL of  0.75 M HBr is diluted to a total volume of 90.0 mL, what is the molar concentration of the HBr in the resulting solution?

V1 = 20.0 mL

C1 = 0.75 M

V2 = 90.0 mL

C2 = ?

(0.75) (20) = C2 (90)

C2 = 0.17 M

Here's a video on calculating Dilution of Solutions for some more help:

Chem Class- November 30, 2009

Mid-term Exams are over and done with, and now we're continuing from where we left off:

CONCENTRATION:

Solution: A homogeneous mixture

Solute: The one present in smaller amount

Solvent: The one present in greater amount

Concentration = amount of solute/amount of solvent (the larger one is on the bottom)

Some units for concentration are g/ml, g/l, mg/L, mg/mL...

The most common (and useful) unitls are mol/L. This is also known as Molarity and Molar Concentration.

It's a good thing to note that the above only works for aqueous solutions and not for gases.

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Exercises:
Caclulate the molar concentration of 2.8 mol of HNO3 in 4.0 L of solution:

[HNO3] = 2.8 mol/4.0 L = 0.70 M

Brackets mean concentration of HNO3...

What volume of 2.40 M HCl can be made from 100.0 g of HCl?

Moles of HCl= 100.0 g x 1 mol/36.5 g = 2.74 mol

2.74 mol/2.40 mol/L = 1.14 L (V= mol/M)

How many grams of CaCl2 are contained in 225 mL of 0.0350 M CaCl2 solution?

0.0350 mol/L x 0.224 L = 7.88 x 10 to the power of -3 mol (Moles of CaCl2)

7.88 x 10 to the power of -3 mol x 111.1g/1 mol = 0.875 g (Mass of CaCl2)

Chem Class- November 18, 2009

Today we went through the homework (empirical formula), reviewed for the mid-term and found out that our teacher is available for extra study help before the exam itself.

We also reviewed a mid-term outline:

____________________________________________

Need to know:             {Nomencalture: a) Binary Ionic
                                                            b) Multivalent
                                                            c)Polyatomic
                                                            d) Acids/Bases
                                                            e) Hydrates
                                                            f) Molecular Compounds
                                                            g) Classical Naming System
        70-75% of the exam {Mole Conversions a) Mole Conversion Table
                                                                       b) Mole--Mass
                                                                      c) Mole--Volume (Gas-STP)
                                                                       d) Density
                                                                       e) Number of Molecules
                                                                       f) Atoms
                                       {Significant Digits
                                       {SI system
                                       {Classification
                                       {Lab Safety
                                       {The labs we've done so far

____________________________________________

Chem Class- November 16

The Empirical Formula:

• Sometimes called the simplest formula
• Is the smallest whole-number ratio of atoms (represented by the subsrcipts)

Molecular formulas are different. They specify how many of each atom there are in a molecule of a compound (represented by the subscripts).

Examples of molecular formulas: P4O10, C6H18O3, N2O4
Examples of empirical formulas: P2O5, C2H6O, NO2

The youtube video we included in the last blog talked about calculating empirical formulas. Basically you:

•  find the mass of each element in the compound
• Find the atomic mass (g/mol)
• Calculate the number of moles
• Divide the number of moles by the smallest amount of moles
• You'll be given a whole number and that is the ratio
• If you are not given a whole number for the ratio here's what you do:

Ratio ends in a decimal of: ~0.5: multipy by 2
Ratio ends in a decimal of: ~0.33 or ~0.6: multiply by 3
Ratio ends in a decimal of: ~0.25 or ~0.75: multipy by 4
Ratio ends in a decimal of ~0.2, ~0.4, ~0.6, ~0.8: multiply of by 5

Also, if you are given a percentage of each element in a compound then assume that you have 100 g of the compound (makes up100%)

That's all there really is to it!:) Just remember that you need the number of moles to solve the problem. We find this helpful:

Chem Class- November 12, 2009

We:

• Went over the answers for the 10-2 Practice Problems
• Did some more review on what we learned so far in the unit (Exams are coming up)
• Learned the percent mass of elements in compounds

Now we can find the % by the mass of each element in a compound. We can also find the mass of an element in a given mass of a compound.

Here's a Video on calculating the Percent Compostition. At about 4:24, however, they start empirical formulas (which we didn't learn yet!)



Here's some sample questions we made up for practice:

1) Find the percent of Iron by mass in FeCl2:
Fe = 19/ Cl = 2(35.5)/ = 19 + 71 = 90 g/mol
% Iron = (19/90) x 100 = 21.1 %

2) Find the mass of Sodium contained in 15.0 g sample of NaOH:
Na = 23/ O = 16/ H = 1/ = 23 + 16 + 1 = 40g/mol
% of Sodium = (23/40) x 100 = 57.5%
15.0 g x 0.575 = 8.6 g

3) Calculate the percentage composition of (NH4)3PO4:
Molar mass =149.0 g/mol (N = 42, H = 12, P = 31, O = 64)
% N = (42.0 g/149.0 g) x 100 = 28.2 %
% H = (12.0g/149.0 g) x 100 = 8.1 %
% P = (31.0 g/149.0g) x 100 = 20.8 %
% O = (64.0 g/ 149g) x 100 % = 43.0 %

Chem Class- November 9, 2009

Today in class, we went over some questions from last time and continued on dealing with the mole with a few practice problems.

Here's one of the examples we did in class: 250 mL of a gas which is known to contain one sulphur atom and an unknown number of fluorides has a mass of 1.63 g at STP.
- We first had to find the molar mass, which was 0.0112 mol (Try it yourself and see if your answer matches. Remember that the units for molar mass are g/mol)
- Here's a newbie. We then had to come up with the chemical formula. There is one sulphur atom, so that leaves us with this so far: SFx. We need to find out the number of Fluride ions

1 x S
? x F
____
146 g/mol -----> 32.1 g/mol
                           x (19.0 g/mol)
                          ___________
                          146 g/mol   -----> 32.1 + 19x = 146
                                                                  19x = 113.9
                                                                      x = 5.99 or 6 atoms

The final chemical formula is: SF6

_________________________________________________

Here's some questions just for practice:

1) 7.09L of a gas has a mass of 4.34 what is the molar mass at STP?

7.09L x 1mol / 22.4 = 0.3165mol
molar mass = 4.34g / 0.3165mol = 13.71g/mol

2) How many moles are in 25.0 mL of HCN (g) at STP?
# of moles = 0.0250 L x 1 mol/22.4 L = 1.12 x 10 the power of negative 3 moles

3) What is the density of gold if 0.02780 mol of gold has a volume of 0.2836 mL?
Density = 0.02780 mol/0.2836 x 10 to the power of negative 3 litres x 197.0 g/ 1 mol = 1.931 x to the power of 4 g/L

4) How many oxygen atoms are contained in 75.0L of SO3 (g) at STP?
# of O atoms = 75.0L x 1 mol/ 22.4 L x 6.02 x 10 to the power of 23 molecules/ 1 mol x 3 atoms/ 1 molecule = 6.05 x 10 the power of 24 O atoms

5) What is the mass of 2.50 L of PF5(g) at STP?
Mass = 2.50 L x 1 mol/22.4 L x 126. 0 g/1 mol = 1.74 mol


Getting tired of the mole yet? Here's a comic that someone made on the mole (Avogadro's number).

Chem Class- November 5, 2009

We went over more of our unit 2 homework questions and then we got started on a new topic:

 Density and Moles.

_____________________________________________

Density----->Mass per unit volume

D = m/v

m=Dv

v=m/D

Density of a gas at STP- 1 mole of gas

Dstp = m/v = Molar Mass g/mol/22.4l/mol

Ex: Calculate the density of CO2 at STP
Density of CO2 (g) = 44.0 g/22.4 L =1.96 g/L

Density of air is approximately 1.5 g/L.

_____________________________________________________________

The following chart shows the conversions between mass, particles and volume.

Summary of the conversions we've learned so far and their conversion factors:

Moles<----->Mass: (molar mass) g/ 1 mol or 1 mol/(molar mass) g

Moles<----->Volume (gases at STP) 22.4 L/1 mol or 1 mol/22.4 L

Moles<-----> Number of Particles (Atoms or Molecules):

6.02 x 10 to the power of 23 particles/ 1 mol or 1 mol/6.02 x 10 to the power of 23 partilces

Molecules<----->Atoms: # of atoms/ 1 molecule or 1 molecule/# of atoms

_____________________________________________________________

More Examples:

1) The density of CCl4(l) is 1.59 g/mL. How many molecules of CCl4 are there in 2.59 L of CCL4?

# molecules = 2.50 L x 1.59 g/ 10 to the power of negative 3 L x 1 mol/154.0 g x 6.02 x 10 the power of 23 molecules/ 1 mol = 1.55 x 10 to the power of 25 molecules

2) What is the mass of 250.0 mL of C3H8?

Mass = 0.2500 L x 1 mol/22.4 L x 42.0 g/1 mol = 0.469 g

3)How many moles are there in 6.00 L of NO3F(g) at STP?

# of moles= 6.00 L x 1 mol/22.4 L = 0.268 mol

4)What is the volume occupied by 10.0 g of H2S(g)?

Volume = 10.0 g x 1mol/34.1 g x 22.4 L/1 mol = 6.57 L

Ahaha! Here's a song that kind of sums up a lot of what we've learned for this unit so far:

Chem Class- November 3

Today we marked a few questions from our previous assignment. Here's an example of one:

Find the mass in grams of 1 million CO molecules: 1 million Co molecules x 1 mol/6.02 x 10 to the power of 23 molecules x 28 g /1 mol = 4.65 x 10 the power of -17.

_______________________________________________________________

We also did the Molar Volume Lab. We explained a bit about how it would work in the last entry, but here it is anyway.

Problem: To experimentally determine the molar volume of a gas.


Materials: Butane, sink, water, 100 mL Graduated Cylinder, Weight scale

Hypothesis: What is the expected value for the molar volume of Butane at STP?

Prelab:
1) The chemical formula for Butane: CH3CH2CH2CH3 or C4H10

1) The molar mass of Butane: 4 C = 4(12)/ 10 H = 58 g/ mol


Procedure:
1) Fill the sink full of water


2) Weigh the butane lighter to the nearest 0.01 g


3) Submerge the 100 mL Graduated Cylinder into the sink, making sure it fills completely with water and then invert it (there shouldn’t be any air bubbles).


4) Place the lighter at the bottom of the sink (make sure it is completely beneath the water), underneath the graduated cylinder


5) Hold down the button of the lighter to release butane and fill the graduated cylinder between 10 and 20 mL (It will displace the water)


6) Remove the lighter, wrap a paper towel around it and shake it so any water that got inside the lighter will come out


7) Weigh the butane lighter once more. If the mass is heavier than it started, it means that there is still water inside and you will still need to try and shake it out.

Observations: We first filled the graduated cylinder completely with water in a sink, inverted it and made sure that there were no visible bubbles. When we placed the lighter underneath the graduated cylinder and held down the button, we saw bubbles beginning to rise inside and about 20 mL of the water was displaced. After we removed the lighter, placed it inside the paper towel and shook it a bit, we took it to the electronic scale to be measured. We noticed that the mass of the lighter was greater than when it started, so that meant that there was still water inside the lighter. So, we shook the lighter more vigorously and then measured it to find that the total amount of butane used in the experiment was 0.04 g. We had time to try the experiment once more and came up with similar results.

_______________________________________________________________

We're all part of different groups, but here's one sample of the results:

Analysis:
1) Mass of the butane lighter before the experiment: 16.48 g

2) Mass of the butane lighter after the experiment: 16.44 g

3) Mass of butane used in the experiment: (Mass of butane lighter before experiment – mass of butane lighter after the experiment) 0.04 g

4) Number of moles of Butane: 0.04 g x 1 mol/58 g = 6.9 x 10 to the power of -4 mol

5) Molar volume of Butane: L/1 mol = 20 mL/6.9 x 10 to the power of -4 mol = 28985.5 mL/mol.

28985.5 mL/1000 = 28.99 L/mol

Conclusion:
1) Percent Error (Measured-Accepted/Accepted x 100): 28.99 L – 22.4 L/22.4 L = 29.4 %.

2) A number of factors may have affected our results. First of all, perhaps there were some air bubbles inside the graduated cylinder underneath the water. It’s also possible that we didn’t release enough or released too much butane into the cylinder. In addition, we may have not have shook out enough water. Finally, there could be slight miscalculations in recording the mass of the butane lighter before and after the experiment.

Chem Class- October 30, 2009

Another lab is coming up, so today Mr. Doktor showed us what we will have to do. First we are going to fill the sink full of water. Then we will submerge a 100 mL graduated cylinder with water in the sink and then invert it, making sure there aren't any air bubbles. After that, we will place the lighter at the bottom of the sink underneath the water and graduated cylinder. We will also release butane from it so the graduated cylinder will fill up with gas. The purpose is to experimentally determine the molar volume of a gas. We will also find out what the mass of the Butane lighter is before the experiment, after the experiment, how much is used in the experiment and calculate the amount of moles of the Butane. After that, as usual, we are to calculate the percent error in our calculation of molar volume and list any factors that may have affected our results.


Notes on Atoms and Molecules.

For Monoatomic elements: A molecule = an element

Molecules and compounds: H2O = 2 "H" atoms in 1 molecule/ 1 "O" atom in 1 molecule

Conversion fator for moles to molecules and vice versa:
6.20 x 10 to the power of 23 molecules/ 1 mol or 1 mol/ 6.20 x 10 to the power of 23 molecules

Eg: How many molecues are there in 0.125 mol of molecules?
# of molecules = 0.125 mol x 6.02 x 10 to the power of 23 molecules/1 mol = 7.53 x 10 the power of 22 molecules.

Oh, and Mr. Doktor broke a wooden board in half!!!!!

Chem Class - October 28, 2009

Today in class we did the ‘Mole Ratio Lab Experiment’. It wasn’t too hard, but it took a while. It was a simple experiment but if you didn’t pay attention to how you were handling the chemicals you could have gotten hurt. The purpose was to the determine the number of moles of copper produced in the reaction of iron and copper (II) chloride, the number of moles of iron used up in the reactio of iron and copper (II) chloride, the ratio of moles of iron to moles of copper, and the number of atoms and formula units involved in the reaction.



When doing the experiment you had to be careful with the chemicals. Copper (II) Chloride and 1Mhydrochloric acid.

Copper (II) Chloride                            1 Mhydrocholirc acid

If you came in contact with either of them you would have to rinse the affected area with water. The chemicals should never be touched, tasted, consumed and inhaled directly in any way! We wore our safety goggles and lab aprons.

The main point of the experiment was to find the difference of ratio of moles of Iron to copper. We mixed the Copper (II) Chloride with distilled water, and it turned into a light blue color.

 Copper (II) Chloride mixed with distilled water

After that we added two Iron nails and waited 20 minutes for the results.

After waiting 20 minutes we took out the nails from the solution and examined them. We saw that they were covered in copper. We then scrapped off the copped back into the solution then placed the nails to dry.

We then separated the Copper (II) Chloride solution and the leftover copper, then rinsed the left over copper with 1Mhydrochloric acid. We dried the left over copper and weighed the nail and dry copper. After recording out results we dumped the left over chemicals in a chemical container that would be disposed of later on and cleaned up our lab materials. Then we washed our hands.

You can find the lab on page 55 of the Heath Chemistry (Laboratory Experiments-Canadian Edition) text book. Give it a go yourself!

Chem Class- October 26, 2009

We've added on to what we've learned today with GASES AND MOLES.
The volume of a baloon occupied by a certain gas depends on the temperature and pressure.

Standard Temperature and Pressure (STP)
O degrees Celcius and 101.3 kPa (or 273 k)

Recall Standard Ambient Temperature and Pressure (SATP)
25 degrees Celcius and 100 kPa (or 298 k)

The volume of 1.0 mole of any gas at STP is 22.4 L
The Molar Volume at STP is 22.4

___________________________________________________________

Conversion Factors: 22.4 L/1 mol or 1 mol/22.4 L (At STP only)

____________________________________________________________

Example: Find the volume at STP occupied by 12.5 mol of NH3 (g)
Volume= 12.5mol x 22.4 L/1 mol =2.80 x 10 to the power of 2 L.

Example: Find the number of moles of 375 mL of SO3 (g)
# of moles = 375 mL x 10 the power of negative 3 L/ 1 mL x 1 mol/22.4 L = 0.0167 mol

Chem Class- October 21, 2009

We've now moved on to calculations of molar mass and moles. First though, Mr. D checked our homework that was assigned last class and then we went over number 5. The question for that asked what the purpose of using a satndard term to describe a certain number was. It's basically created to abbreviate something, making it more convenient, faster and easier.
Also: " Mr. D stole my noodles :(.... but I got dem back :)"

Ok, back to Chemistry. First of all, we started with the Atomic Mass.

The atomic mass is the mass of one mole of atoms in an element. It is also the mass of 1.0 mole of 'C' atoms in 12.0 g  (Eg: The mass of 1.0 mol of 'Ca' atoms is 40.1 g).

Molecular Mass is the mass of 1.0 mole of molecules of an element or a compound.
It's good to remember those that are diatomic: N2, O2, F2, CL2, Br2, I2, Hr (the special seven)
It's also good to remember those that are polyatomic: P4 and S8
You can assume, now, that all the rest are monoatomic

Finding the Molar Mass of Compounds
Eg: Find the molar mass of Ammonium Phosphate

NH4(+) and PO43(-) is (NH4)3PO4
3 N = 3(14.0)
12 H = 12(1.0)
1 P = 1 (31.0)
4 O = 4 (16.0)
          _______
          149.0 g/mol

Converting Mass<----->Moles
Conversion Factor g/mol or mol/g

Eg: Find the mass of 2.5 mol of water
1mol/18.0g x 1/2.5 mol = 1/45 g =45 g

Eg: Find the number of moles in a 391 g sample of Nitrogen dioxide
NO2
1 N = 1(14.0)
2 O = 2 (16.0)
         _______
          46.0 g

391 g x 1 mol/46 g = 8.5 mol

Chem Class- October 19, 2009

Last class we had our test for Unit 2 and today we got our test back. We hope everyone did well!
After that, Mr.D did a small balloon experiment:

Now we're on to something new!

_______________________________________________________

Today we were introduced to the Mole.

Now let's get this straight. We're not talking about the kind of burrowing animal or the so called 'beauty spot', but the SI unit for the amount of substance . It is equal to the number of atoms in exactly 12.0 grams of carbon-12 (6.02 x 10 to the power of 23)

1 mole is equal to 602 000 000 000 000 000 000 000 or 6.02 x 10 to the power of 23. This is a really big number and it is known as Avogadro's number. Let's say we had that number equal to dollars. If we divided that number up among the six billion people on Earth, every person would have $100 000 000 000 000 each.

__________________________________________________________

Many people tried to see how gases could combine. John Dalton, for example, looked at masses of gasses but he could see no pattern:

• 11.1g of H2 reacts with 88.9g of O2
• 46.7g of N2 reacts with 53.3g of O2
• 42.9g of C reacts with 57.1g of O2

The reason he couldn't see a pattern is because 1 molecule of H2 doesn't have the same mass as 1 molecule of O2.

Joseph Gay-Lussac tried combining gases on volume:

• 1 L of H2 reacts with 1 L of Cl2----->2 L of HCl
• 1 L of N2 reacts with 3 L of H2----->2 L of NH3
• 2L of CO reacts with 1 L of O2----->2 L of CO2

As you can see, gases combine in simple whole number ratios.

_____________________________________________________

AVOGADRO'S HYPOTHESIS
Equal volumes of any gas at a constant temperature and pressure contain equal numbers of molecules.

Chem Class- October 13, 2009

Today we did a lab on hydrates. We weighed our empty test tube, then our teacher put a hydrate inside. We then used our bunsen burner to evaporate all the water out of the hydrate. It went from a fushia, to a light blue. About 45% of the water in the hydrate evaporated, which was the accepted value. Finally, we calculated our percent error (measured-accepted/accepted). The purpose of the experiment was to determine the empiracle formula of a hydrate.

When it comes to lab safety, we need to be aware that we are using a bunsen burner. This means that long hair should be tied and sleeves should be rolled up so that it doesn't catch on fire. We are also dealing with glassware so it needs to be handled with care and broken glassware should not be used. In addition, we are  electrical equipment is being used (electrical scale) and so it should not be unplugged by pulling on the chord. Finally, we should also wear safetly goggles and a lab apron.

For our lab, the mass of the hydrate (calculate by subtracting the mass of the test tube) was 1.0 g. The mass after heating was 0.6 g which meant that 0.4g of water was realsed during the heating and 40% of the hydrate waswater. Our percent error was 11.1%.

We also have our test coming up for this unit.
60% of the test will be mainly naming chemicals (Chemical Nomenclature):

• Ionic
• Polyatomic
• Multivalent (IUPAC and Classical-Latin)
• Molecular
• Acids/Bases
• Hydrates
• Ions

The other portion of the exam will consist of the classification of matter (Heterogeneous and Homogeneous substances) and separating mixtures (Chromatography, Crystalization, Distillation, Filtration, by hand...)

Chem Class- October 8, 2009

Let's get started for the day. Today we:

- Handed in homework

- Got our marks for our blogs for the first unit

- Mr. D did a demonstration lab that involved combining sugar (C12H22O11) and Sulfuric Acid (H2SO4). Smoke came out and the substance inside turned black (carbon) and slowly rose out of the beaker. A very strong smell of burnt sugar or raisins was produced.

Equation for experiment: C12H22O11 + H2SO4 -----> C(s) + H2O +  S

Actually, you can observe the experiment yourself without having to smell the results. Here's a youtube video:

Down to the notes now. Acids and Bases. What are acids and bases? Let's start with acids.

SATD stands for standard ambient temperature and pressure which is 25 degrees Celcius and 100 kPa.

NAMING ACIDS

• 
  
  It's also good to know that acids are aqueous (dissolved in water). Hydrogen compound are acidic (Eg: HCL (aq)-----> Hydrochloric Acid and H2SO4(aq)----->Sulfuric Acid)
  
  
  
• 
  
  Hydrogen appears first in the formula unless it is part of a polyatomic group (Eg: CH3COO----->CH3COOH/Acetate becomes Acetic Acid).
  
  
  
• 
  
  Classical rules use the suffix -ic and/or the prefix hydro (EG: HBr = Hydrobromic acid and HI = Hydrochoric Acid).
  
  
  
• 
  
  IUPAC uses the aquesous hydrogen compound (EG: HCL(aq)----->Aqueous Hydrogen Chloride)
  

Now, what are bases?

Naming Bases

• 
  
  For now, all bases will be aqueous solutions of ionic hydroxides
  
  NH3 is an exception. Even though it does have OH at the end, it is a base.
  
• 
  
  You Use the cation name followed by hydroxide
  
  
  
• 
  
  (NaOH----->Sodium Hydroxide and Ba(OH)2----->Barium Hydroxide).
  
  
  

Chem Class- October 6, 2009

Today in class Mr. D showed us a demonstration. There were two test tubes and each of them had something in them. One of them was white and one of them was purple. Mr. D used the Bunsen burner to heat up the purple one and it turned blue with a bit of moisture. Cool or what?

After that, we went over our homework which was the work sheet for naming compounds. Then we did some notes. Luckily, they weren’t that long!

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More Chemical Nomeclature!

First of all though, we must learn about hydrates.
Hydrates are substances that contain water.
Some compounds can form lattices that bond to water molecules (Eg: Copper Sulfate, Sodium Sulfate).
These crystals contain water inside them which can be released by heating. Without water, the compound is often preceded by Anyhydrous (Eg: Copper (II) Sulphate) just to make it clear that it is not a hydrate.

To name Hydrates:

1. Write the name of the chemical formmula
2. Add a prefix indicating the number of water molecules
    3. Add hydrate after the prefix

Examples: Cu(SO4) . 5H20(s)----->Copper (II) Sulfate Penta Hydrate
                 Nickel (II) Sulfate hexahydrate-----> NiSO4 . 6H2O(s)

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Molecular Compounds

- Composed of two or more non-metals

- Have low melting points and boiling points

- Usually end in -gen or -ine (EG: Hydgrogen, Oxygen and Nitrogen)

- 7 molecules are diatomic (H2, N2, O2, F2,Cl2, Br2, I2) which means that they have two of the same element

- 2 molecules are polyatomic (S8 and P4) which means that they have more than 2 (many) of the same element

Examples

N2O4----->Dinitrogen Tetraoxide

P4O10----->Tetraphosphorous decaoxide

Nitrogen Trichloride----->NCl3

Sulphur Dibromide----->SBr2

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These molecular compounds are very common and have names that don't follow the same rules for naming:

Water is H2O

Hydrogen Peroxide is H2O2

Ammonia is NH3

Glucose is C6H12O6

Sucrose is C12H22O11

Methane is CH4

Propane is C3H8

Octane is C8H8

Methanol is CH3OH

Ethanol is C2H5OH

Chem Class- October 2, 2009

Ah yes. Chemical Nomenclature. This is all about naming chemical compounds. Different systems have been used through the enturies but the most common system today is IUPAC (International Union of Pure and Applied Chemistry) for most chemicals.

• Ions
• Binary Ions
• Polyatomic Ions
• Molecular Compounds
• Acids

Be careful when dealing with superscripts and subscripts. Superscripts refer to the number that is beside and above the symbol, representing the ion charge while subscripts refer to the number that is beside and below the symbol, representing the number of ions.

The first two columns in the table below contain symbols with superscripts while the last column represents symbols with subscripts (look at the numbers).

You probably noticed the two terms: Cation and Anion. Cations are ions with positive charches while Anions are ions with negative charges.

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When it comes to naming ions, use the name of the element and then add ion for metals (Al3+ becomes Aluminum Ion) and simply remove the original ending and add -ide for non metals (F- becomes Fluoride).

Below is a table of polyatomic ions which have special names:

Here's a youtube video on how to write the formula for Binary Ionic Compounds. It's pretty good on listing the steps and providing a visual example of a lewis diagram for why the compound is written the way it is:

Some elements can form more than one ion and they are called Multivanet Ions. The more common ion is the top one of the Periodic Table. IUPAC uses Roman numerals in parenthesis to show the charge (I, II, III, IV, V, VI, VII)

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Finally, Classical Systems use Latin names of elements and the suffixes -ic for larger charges (Eg: FeO----->smaller charge-----> Ferrous Oxide) and -ous for smaller charges (Eg: Fe2O3----->larger charge----->Ferric Oxide)

Other Classical Names

Ferr-Iron

Cupp-Copper

Mercur-Mercury

Stann-Tin

Aunn- Gold

Plumb-Lead

Chem Class- September 30, 2009

Mr. Doktor checked our homework from last class today. He then went over a powerpoint on the Classification of Matter.

Here's a diagram I got from a Chemistry book on classifying matter:

As you can see, there are two types of pure substances: Homogenous and Heterogeneous.
Homogenous substances consist of only one visible component (Eg: distilled water, oxygen, graphite) and Heterogeneous substances contain more than one visible compent (Eg: chocolate chip cookie, granite).

You can also see in the above diagram that there are also two types of pure substances: Elements and Compounds.
Elements are substances that cannot be broken down into simpler substances by chemical reactions (Eg: oxygen, iron and magnesium)
Compounds are substances that are made up of two or more elements and can be changed into elements (or other compounds) by chemical reactions (Eg: water, sugar).
The differences between them are only visible at the atomic level

We also learned about Solutions.
A solution is a homogeneous mixture of 2 or more substances. The component present in the greater amount is the solvent and the component present in the smaller amount is the solute. The symbol (aq) is used when something is dissolved in water.

Finally, Mixtures:
Many mixtures are easy to identify, but others are easily confused as pure substances.
In Heterogenous mixtures the different parts are clearly visible (granite, sand, fog) and in Homogeneous mixtures the different parts are not visible (salt, water, air).
There are different methods to separate methods depending on the type of mixture. These include:

• By hand (Heterogeneous mixtures only)
• Filtration (Heterogeneous mixtures only)
• Distillation
• Crystalization
• Chromatography

The above are all physical changes.

Chem Class- September 28, 2009

Here we are now in Unit 2: Properties of Matter. Today in class, we took notes on matter and then watched a video (part of a MythBusters episode), trying to see how many chemical and phyiscal changes we could spot.
So just what is matter?

MATTER

- Anything that has mass and occupies space

- It can exist in many different states but the most common are: Solid,Liquid,Gas,Plasma,Aqueous,Amorphous

Solid: Holds one shape and has a definite volume
Liquid: Can change shape, but has a definite volume
Gas: Can change shape and colume
Aqueous: Something dissolved in water

In this picture, we see that solids have strong bonds, liquids have weak bonds, gases have no bonds and plasma involves ionization:

- Can undergo many changes: Physical, Chemical and Nuclear

 Physical Changes: Involves changing shape or state of matter and no new substances are formed
(Examples: Crushing, tearing, boiling water, cutting wood, smashing cars)
Changing from a solid to a gas can often be confused as a chemical change but remember that if that chemicals remain the same it is a chemical change.

Chemical Changes: Properties of matter change and new substances are formed
(Examples: Conductivity, acidity, colour, iron rusting, burning wood, digesting food)

So remember: Changes of state are considered phsical changes:

Also, during the melting process, chemicals usually follow this path:

New substances are formed in a chemical change:

One last point before we are through: In physical and chemical changes, matter is neither created or destroyed. This is known as: Conservation of Matter. The French chemist Antoine-Laurent Lavoisier is considered to have discovered this concept.

Chem Class- September 23, 2009

Today we had our very first chapter test. It included the very first things we learned up until this point. So, after reviewing a bit on the board, we proceeded with the writing of the test.

   The main topics were:

Lab Safety

Lab Equipment

Fundamental Units

Scientific Notation

The SI System

Significant figures

Error

And Dimensional Analyis

Chem Class- September 21, 2009

Today:
Mr. Doktor talked to us about our Unit Test. Ooooooooh scary!
If we had already completed the Lab, we were to work on our Lab write up as well as our review questions (In preparation for next class...Ah!). The people who either didn't finish their lab or didn't start their lab had the time to do that.

I'm sure you're probably wondering what the actual mass of salt is that can be dissolved in 200 mL of water. So, after everyone verified their results from the lab, Mr. Doktor told us just that: 11.5 g! We calculated the percent error by using this answer and the answer we were using. Remember how to calculate that?

Percent Error = (observed - theoretical/ theoretical) x 100.

I have to say, a lot of us were off by a lot!;)

Here's another youtube video. This one's on calculating percent error. One thing to note, however, is that they use the word 'actual' value. We learned in class that 'accepted' is probably the better word since there would still be error in the measurement.



Time to study now for that test!