VSEPR: Molecular Yoga
The Goal: To help students derive and visualize molecular geometries from electron
domain geometries.
The Method:
- Act out the key shapes (especially tetrahedral) to understand how electron
domains are maximally separated in each—this is the “yoga” part. I make the
kids get up and act out the shapes with me.
- Show these shapes using a building toy (“flexoam”)
(http://www.amazon.com/FLEXOAM-42-PIECE-SETMindscope/dp/B001CJQFXO)
- Have them build the three confusing shapes (tetrahedral, trigonal
bipyramidal and octahedral) and say the names of the electron domain
geometries and the molecular geometries
Follow up:
We do a model building activity (worksheet based, predict polarities of bonds and of
molecules built)
I refer back to molecular yoga throughout our study of intermolecular forces, boiling
points and solubility.
The Woofer Quiz
28.7 mL of ethanol was poured into the water bottle, which was stoppered and left
overnight. The density of ethanol is 0.80g/ml.
Please SHOW the steps of your calculations in your work below
1. Write the formula for ethanol: _____________________
2. Calculate the molar mass of ethanol.
3. How many moles of ethanol were poured in the bottle yesterday?
4. Write the balanced chemical reaction for the reaction of ethanol and oxygen
to make water and carbon dioxide.
5. Assuming that all of the ethanol reacts, what is the theoretical yield of water
(in mL, remembering that water’s density is 1 g/mL) from this reaction?
6. Now we will pour out the liquid left in the bottle. The volume is: _____________
Assuming it is water, calculate the percent yield of this reaction.
7. Based on the flame test of the liquid left in the bottle, was it water? Explain
Woofer Quiz Answers and How To
1. C2H5OH
2. (2 x 12.01) + 6 (1.01) + 16.0 = 46.08 g/mol
3. (28.7mL) (.80 g/mL) = 22.96 g
22.96g/46.08g/mol = .498 mol  .50 mol
4. C2H5OH + 3 O2  2 CO2 + 3 H2O
5. 3 moles of water from each mole of ethanol so 3(.5mol) = 1.5 mol water
(1.5 mol)(18.02 g/mol) = 27.03 g water = 27 mL of water
6. This will vary depending on how much liquid is left in the woofer. Let’s say
it’s 24ml of liquid: (24/27) x 100% = 89% yield
7. No, it wasn’t water because it was flammable. It must have been ethanol.
There must have been more ethanol in the bottle than oxygen, so the fire
went out before all of the ethanol was able to be burned up. (I often do this
reaction after introducing stoichiometry but before talking about limiting
reagents)
How to set up and light a “woofer”
Use between 10ml and 30ml of alcohol (isopropanol or ethanol, NOT methanol).
Pour it in the night before or several hours in advance, stopper well and
shake/roll the bottle around to help start the vaporization. Tape the bottle with
clear packing tape. (This is to limit the pieces flying if the bottle cracks during
combustion.)
To light: Put the bottle behind a safety shield. Light a match and hold it with
tongs. Remove the stopper, stand behind the safety shield, reach around and
drop the match in the bottle. I sometimes do this with a colleague and one of us
removes the stopper and steps away while the other drops the match. Best to
have the lights out for this reaction.
After the explosion is done and the fire is out: I stopper the bottle while the kids
do some math. Then, for #6 of the quiz, I unstopper the bottle (noting the inrush of air...why is that?) and pour its contents into a graduated cylinder. Then I
call out the volume of liquid for them to write down. Then I pour the liquid into
a metal bowl and drop a lit match in the bowl—it burns for quite awhile, making
the point that the liquid is not water.
Other notes:
The explosion can be loud, and the flame can go on for a while. If the flame
persists at the mouth of the bottle, you may have to put it out (bottle mouth with
stopper). This makes an interesting talking point—why does the flame persist
only at the mouth of the bottle?
Check the bottle after using it. If it starts to look fatigued, use a new one. Flinn
recommends no more than 20 times.
Fire and Math: The Woofer
It’s always fun to do a fiery reaction—but can you predict the “boom”? In this lab
you’ll be trying out two different fuels—but first you have to predict which trial will
yield the most spectacular results and back up your predictions with calculations.
The fuels:
ethanol (95%), isopropanol (91%)
You may assume:
The reactants will be at 25oC at the start of the reaction and will have volatilized to
the greatest extent possible (they will have been sitting overnight or for several
hours).
The pressure in the room will be 1.00 atm
The reactants will be confined (until the stopper is removed) to a 5 gallon water
cooler bottle
Scenario: 30mL of each fuel will have been placed into each bottle, the bottle
stoppered and left for hours. At reaction time the stopper will be removed and a lit
match dropped into the bottle. The bottle will be re-stoppered promptly after the
reaction is complete.
Making Predictions
When you make your predictions, consider the amounts of reactants (in moles, of
course), the stoichiometric ratios of the balanced combustion equations, the vapor
pressures of the different alcohols, and the delta H of each reaction. Make
predictions about relative heats of reactions, as well as other kinds of energy (light?
sound?) that might be released. Once you’ve done the calculations for ethanol and
isopropanol, try them for methanol (it’s too dangerous to use here, but interesting to
calculate!).
What information do you need to calculate?
Woofer: Potentially useful information and assumptions:
Woofer bottle volume is 5 gal
1 gal = 3.75 L
Air is 21% oxygen
R = .0821 L atm/mol K
T = 25 oC = 298 K
1 atm = 760 mmHg
Information about the alcohols:
Alcohol
Chemical
formula
Boiling point
oC
Density g/mL
methanol
ethanol
isopropanol
CH3OH
C2H5OH
C3H7OH
64.7
78.4
82.5
.7913
.7892
.7854
H formations in kJ/mol
CH3OH (g)
C2H5OH (g)
C3H7OH (g)
CO2 (g)
H2O (g)
-201.2
-235.1
-261.1
-393.5
-241.82
Average bond enthalpies in kJ/mol
C-H
C-C
C-O
O-H
C=O
O=O
413
348
358
463
799
495
Vap. Pres. in
mmHg at 25oC,
1atm
97
68
18
Woofer: some answers
(typos/errors possible!)
Amount of oxygen gas in woofer?
n = PV/RT
vol woofer = 3.75L/gal x 5 gal = 18.75L
nO2 = (.21 atm)(18.75L)/(.0821 L atm/mol K)(298K) = .161mol O2 for all reactions
Sample calculation for woofer if methanol was used: (I sometimes give these to the
kids if they need example calculations to get them going)
max amount of methanol that can vaporize into woofer:
nCH3OH = (97mmHg/760mmHg/atm)(18.75L)/(.0821 L atm/mol K)(298K) =
.0978 mol methanol possible
Combustion equation for methanol:
2CH3OH + 3 O2  2 CO2 + 4 H2O
If have .161 mol O2 then need .161 (2/3) = .107 mol CH3OH for complete reaction.
Only have .0978 mol in vapor, so CH3OH is LIMITING
Methanol’s heat of combustion calculated via H formation data is -675.94 kJ/mol:
So for our sample : (.107mol)(-675.94kJ/mol) = - 66.1 kJ
Using bond enthalpies it’s -647.5kJ/mol, so for our sample: -63.3 kJ
Summary
for all of the alcohols (using 30mL of each alcohol in the woofer)
Alcohol
D
g/mL
Mass
(g)of
30
mL
CH3OH .7913 23.74
C2H5OH .7892 23.68
C3H7OH .7854 23.56
MM
mol
g/mol
32.05
46.08
60.11
Mol req. to
react
w/.161mol
O2
.741 .107
.51
.054
.392 .036
Max
Limiting H
vaporizable reagent (kJ)
mol
(“LR”)
based
on LR
.0978
CH3OH
-66.1
.0686
O2
-68.6
.0182
C3H7OH -34.0
Notes:
It is interesting that the alcohols are limiting in two cases, while the oxygen is
limiting in the case of ethanol.
The Hs in the table above are calculated from H of formation data.
The Wax Fireball
(sound track: Great Balls of Fire)
The idea that heat is added to cause phase changes from solid to liquid to gas makes
intuitive sense to most people, but the reverse idea (that energy is given off during
phase changes from gas to liquid to solid) is less evident. In this experiment a phase
change from liquid to solid releases enough heat to ignite a fireball.
Set up:
- Make a small amount (2-7 g) of shavings of pure paraffin wax. You can use a knife,
scupula or potato peeler to do this. Smaller shavings work better.
- Put the wax shavings into a test tube. I like to use a larger/shorter one (about 2cm
diam x 10cm long) but many sizes work.
- Mount the test tube onto a clamp at the end of a meter stick
- Set up an ice bath of about 2L volume
- Set up a Bunsen burner about two feet away from the ice bath
- use goggles, tie your hair back etc
Doing the demo:
- Heat the test tube over the burner until the wax reaches a rolling boil. (You don’t
have to hold it from the far end of the meter stick until you move it to the ice bath in
the next step,) For drama, turn off the lights before the next step.
- Now make sure that you are holding the meter stick at the far end. Move the test
tube quickly to the ice bath, submerge the test tube about halfway in the ice bath.
Disposal:
- the test tube usually breaks (or cracks)—put it in the glass disposal.
Other considerations and tips:
- Plan the location of the fireball—it will appear about a foot above the mouth of the
test tube and depending on the amount of wax you use may be 2 feet in diameter.
- You may want to cover the floor below the location of the fireball with a sheet of
plastic because there will be wax splatter below the fireball region
- I angle the test tube about 45 degrees when heating and a little less when putting it
in the ice bath
- I have the students wear goggles and be at least 10 feet back.
- You have to do the transfer from Bunsen burner to ice bath quickly. Try not to
have drafts in the room. Make sure the ice bath is really cold. Pure paraffin works
great, but other wax (such as from some commercial candles) does not.
What’s going on:
As the hot wax cools and changes phase from liquid to solid it releases heat. The
heat is given away to the ice bath, but also leaves via the open end of the test tube as
a “jet” of hot gas. This heat ignites the paraffin vapor above the test tube. Paraffin
vapor is very flammable...that’s what it burning when you see a candle burning.
I precede this demonstration with a discussion (or demonstration) of the heating
curve of water from ice through steam.
Calculations....
My colleagues and I did some calculations to look at how much heat can be made by
this reaction. For 7 g of paraffin (estimating the molar mass of paraffin as 436
g/mol) and assuming that the wax changes temperature from 370oC to 10oC, we
calculated 8.59 kJ of heat produced.
(information about paraffin needed for these calculations: formula C31H64 ; melting
point 55oC ; boiling point 370oC ; specific heats solid 2.3 J/goC ; liquid 2.9 J/goC ;
Hfus 210 J/g ; Hvap 2260 J/g )
This is enough to raise 41L of room temperature air to the flash point of paraffin
vapor or enough to raise 32 L of air to the auto ignition temperature of paraffin.
(paraffin flash point: 199 oC ; autoignition point: 245 oC; density of air assumed to
be 1.2g/L and specific heat of air assumed to be 1005 J/kg oC)
Of course, much of the heat lost goes into the ice water, but clearly enough is
directed upwards to light the paraffin vapor.