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:

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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

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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

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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.

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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.

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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

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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.

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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.

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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.