Today is 8 March 2008 -- Why is carbon able to enter into chemical reactions with so many other elements? And other questions.

Carbon has four electrons in its outer shell.  It needs to gain four to complete the shell, or to lose four so the underlying shell is seen as complete.  The energy to remove the four electrons is enormous so that doesn't happen (on this planet, normally, anyhow).  Gaining electrons would give it a high negative charge (four electrons, remember) and the charge imbalance between the four positive protons and four more electrons results in a very unstable ion.  So THAT doesn't happen.  Instead, the carbon shares electrons covalently with other atoms.  If it is in a salt like calcium carbonate, CaCO3, it is still covalently bound to the oxygen.    Further, since there are four bonds to be formed, the number of combinations are large.  Carbon can form single, double and triple bonds.  When it bonds to other carbon atoms, it forms a crystal or lattice as in diamond or Buckyballs.

2.  Why is water a liquid under normal atmospheric conditions?

Because although the hydrogen atoms in a molecule of water are covalently bound to an oxygen atom, they are not completely neutral in the sense that they are able to be attracted to the oxygen of an adjacent water molecule.  This "hydrogen bonding" is strong enough to keep the molecules from drifting apart under normal conditions.  In the atmosphere, when the air pressure is low, the molecules have enough energy to break free and form water vapor, or if heated, they gain enough energy to overcome the strength of the hydrogen bonds.  If our plant had a lower atmospheric pressure, or was hotter, there would be more water in the vapor phase and less in the liquid.

3.  Write and balance the decomposition of calcium carbonate:

This was discussed over and over in class.  I'm really disappointed that more of you did not write the correct answer.  When calcium carbonate is heated, it forms carbon dioxide and calcium oxide.  

CaCO3 --> CaO + CO2 .

4.  Write and balance the following reaction:  Complete oxidation of sugar (C6H12O6) in the body to produce CO2 and H2O.  Again, we went over this in class.

C6H12O6 + 6O2 --> 6CO2 + 6H2O

5.  Show the Lewis dot diagram for CaCl2

You know that this means one calcium and two chlorine.  You should know that chlorine has 7 electrons in its outer orbital and wants to gain one.  You also should know that calcium has two electrons in its outer orbital and can lose them in forming a bond.  Draw the dot diagram with chlorine on either side of the calcium like this:  Cl : Ca : Cl   Then, add two dots over each Cl, two under them and two between each of the atoms.  Now, check:  There are 16 electrons in all, seven from each of two chlorine atoms and two from the one calcium atom.  That adds up to 16 electrons.  Two between the Cl and the Ca (on each side) uses 4, leaving 12.  Since we need the same number of electrons for each of two Cl, divide 12 electrons by 2 atoms and get 6 electrons for each Cl.  Arrange the dots around the Cl.  Then, stand back and admire your artwook.  

Today is 15 July 2009 -- We learned how to write a lab report

We made lab notebooklets in which to write up the report on the first experiment, "Combustion of Magnesium and associated reactions."  Students were provided with a data page that identified what they were to look for.  The data page is:

COMBUSTION OF MAGNESIUM RIBBON AND ALLIED EXPERIMENTS

1.  Weigh a portion of magnesium ribbon.  About 0.5g is enough.  Record the weight.

2.  Weigh a clean beaker.  Record the weight.

3.  Hold the magnesium with tongs.

4.  Ignite the ribbon over a burner; hold it over the beaker to catch the residue of combustion.

5.  What is the chemical equation that describes the combustion?  Write and balance the equation.

5a. How many moles of magnesium did you start with and how many moles of magnesium oxide resulted? 

6.  Weigh the beaker and combustion product.  This is the "gross weight."  Subtract the weight of the beaker (the "tare weight") from the gross.  This gives the "net weight."  The net weight is mass of the magnesium oxide collected.

7.  Based on the reaction equation, how many grams of magnesium oxide did you expect?  How much did you actually obtain?  What is the percentage recovery?

8.  Write and balance the equation showing the reaction of magnesium oxide with hydrochloric acid to produce magnesium chloride and water. 

8a.  What is the mole ratio of magnesium oxide to hydrochloric acid?

8b.  Based on the amount of magnesium oxide recovered, how much hydrochloric acid do you need to react with it?

9.  Hydrochloric acid is hydrogen chloride dissolved in water.  The concentration of hydrochloric acid that you will work with is 3.7%, the rest being water.  How many milliliters of the dilute acid will you need to react with the magnesium oxide?

10.  Add the acid to the magnesium oxide and stir or swirl to react.  Use a glass stirring rod, if necessary.

11.  When the reaction is complete, the solution should be clear.  If you have undissolved residue, carefully add more dilute acid.  Measure EXACTLY how much you are adding. 

12.  Using a hot plate or burner with wire gauze, warm the solution and evaporate to dryness.   The result should be magnesium chloride crystals. 

13.  Cool and reweigh.  Subtract the beaker tare from the gross weight to find the number of grams of magnesium chloride.

14.  How many moles of magnesium chloride were recovered?  How much was expected? How much magnesium was present in the material recovered?  Compare the amount of magnesium recovered to the amount you started with as magnesium ribbon.  

Today is 17 July 2009 -- We are learning to balance chemical equations

After discussing reaction yield, I taught you the algebraic method to balance equations.  (Refer to "CHEM Class Notes")  I gave you some problems to solve in class and I promised to give you more to practice at home.  Here are four more:

1.  Al + CuO --> Al(sub2)O(sub3) + Cu

2.  Fe(sub2)(SO(sub4))(sub3) + KSCN --> K(sub3)Fe(SCN)(sub6) + K(sub2)SO(sub4)

NOTE:  In the first term, there is a sulfate group (SO(sub4)) taken 3 times. 

3.  CaCl(sub2) + AgNO(sub3) --> AgCl + Ca(NO(sub3))(sub2)

4.  C(sub6)H(sub5)COOH + O(sub2) --> CO(sub2) + H(sub2)O

Today is 16 July 2009 -- We are learning about reaction "yield"

"Yield" from a chemical reaction means the amount of product you actually got compared to what is predicted from chemical theory.  For example, we burned a piece of magnesium to produce magnesium oxide.  The reaction is:

2Mg + O(sub2) --> 2MgO

If we started with 0.5 gram of magnesium, how much MgO should we get?

Use the mole equation to determine:  moles = mass / atomic mass.

The mass is stated.  Get the atomic mass from the periodic table:  24.

Then the number of moles is 0.5g / 24 = 0.0208

You can see from the equation that for every atom of magnesium reacted, one molecule of MgO is formed.  Thus, 0.0208 moles of magnesium should produce 0.0208 moles of MgO.  To find the number of grams of MgO, rearrange the mole equation to solve for mass:

mass = (moles)(atomic mass)

Find the atomic mass of MgO by adding the atomic weight of magnesium and of oxygen, or 24+ 16 = 40.   Then, multiply 0.0208 x 40 = 0.833 gram.

But is that really how much MgO was recovered?  No, because some was lost as a coating on the tongs I used to hold the magnesium strip and more was lost as a fume.  (I showed you what happened if we inverted a beaker over the burning strip of magnesium:  the fume coated the walls of the beaker so we collected more than if we just burned it in the open air.  You saw that the burning it beneath the mouth of the beaker starved it for oxygen and the combustion stopped.)

I told you to weigh a beaker before the experiment.  This gave you a "tare" weight.  Then, when you burned the magnesium strip, you were to collect the resulting MgO in the beaker as it fell off the burning strip.  Then you weighed the beaker with the MgO in it.  This gave you a "gross" weight.  You subtracted the tare from the gross to get the "net" weight of the MgO.  This weight would be less than 0.833 gram because of the losses I discussed above.  Let us say the gross weight was 108.52 grams and the tare weight was 108.1 grams.  Then the amount of MeO you recovered would have been 0.42 grams.  The yield, or "percent recovery" is the amount recovered divided by the expected amount times 100% or 0.42/0.833 x 100 = 50.4%.

Even if you had some difficulty with this, you could still use the amount of MgO actually recovered to determine how much MgCl(sub2) you would produce in the next step.  The reaction is:

MgO + 2HCl --> MgCl(sub2) + H(sub2)O

You know how much MgO you recovered.  From that determine the number of moles:  it is 0.42/40 = 0.0105 in this example.  Then, inspect the BALANCED equation.  You can see that for every mole of MgO, you produce one mole of MgCl(sub2).  Therefore, the number of grams of MgCl(sub2) is:

(0.0105) x [(24 + 2x35) = 0.987 grams.

After you evaporate the water from the beaker, let us say you reweigh it and find the gross weight is 108.6 grams.  This means the amount of MgCl(sub2) in the beaker is 108.6 - 108.1 = 0.5 grams.  What is the yield of MgCl(2)?  It is 0.5gram/0.987gram x 100% = 50.66%.

You need to find how much magnesium you have at the end of the reaction (in the MgCl(sub2)) compared to what you started with.  To do this, find the fractional amount of magnesium by dividing its atomic weight (24) by the total weight of the magnesium chloride (94).  Then multiply that fraction by the amount of MgCl(sub2) recovered:   24/94x0.5gram = 0.13 gram.  Compare that to the starting amount of magnesium, 0.5 gram.  What is the yield?  It is 0.13/0.5 x 100% = 26%

Today is 22 July 2010 - Magnesium mass balance experiment.

Following from our study of equilibrium constant and solubility product, we are conducting an experiment involving (1) burning magnesium, (2) collecting the magnesium oxide that results, (3) mixing it with water to form magnesium hydroxide, (4) adding acid to convert the magnesium hydroxide to a magnesium salt, (5) evaporating the liquid and (6) weighing the residue to determine what is in the salt.  Through it all, weighing and the mole concept are important concepts to be mastered.  The questions to be considered and the points involved in metacognitive analysis of the experiment are:

1.            What is an acid?

2.            What is a base?

3.            You react Mg with O2 in the air.  What kind of reaction is that?

4.            What is the product of the reaction?  Write it out and balance the reaction equation.

5.            You want to collect and quantify the product of the reaction.  How is this done?

6.            After you collect the reaction product, you try to dissolve it in water.  The solubility product is Ksp = 1.5 x 10-11.  What does this tell you about the solubility of the product of the reaction?

7.            You test the solution with a piece of pH paper.  The color indicates a pH between 8 and 9.  What is the best accuracy you can say for the pH?

8.            If you heat the solution, you think you have more reaction product dissolve.  What is a way to determine this?  What chemical principle would allow this way to work?

9.            To convert the reaction product into a sulfate or chloride salt, you can add drops of sulfuric or hydrochloric acid.   What is the formula for each of those acids?

10.            Write the reaction between the reaction product (in water) with the acid, and balance the chemical equation for the reaction.

11.            The acid solutions you have are 1 M.  What does this mean?

12.            What is your evidence that a reaction between the acid and the reaction product has occurred?

13.            You need to convert the number of drops of acid you added to your reaction product solution into milliliters.  How is this done?

14.            You want to calculate the “yield” of the oxidation of the Mg.  What did you do?

15.            You want to verify your work by calculating the amount of acid that reacted with the reaction product.  How did you do this?

16.            You can evaporate the liquid from the salt that formed when the acid reacted with the reaction product and weigh the residue.  How do you do this?  What measurements do you need to make?  What equipment do you need?

17.            What are the names, common or scientific, for the salt that results?

18.            What chemistry principles have you learned from this experiment? (Think of acids and bases, equilibrium, balancing equations, etc.)

DO A GREAT JOB ON YOUR LAB REPORT!!!

What we did in class 26 July 2010 -- We had an exam!

INSTRUCTIONS:  Read this exam carefully!  Section "A" with steps 1 - 10 is a practice problem worked out for you.  Understand it and then use your actual lab data from the experiment about mass balance and burning magnesium.  Apply the method to your ACTUAL lab data.  Show ALL work.  You will have 60 minutes.  The exam is worth 100 points.  

A.  Reactions of Magnesium

2Mg + O2 ⇒ 2MgO

MgO + H2O ⇒ Mg(OH)2

Ksp = 1.5 x 10 -11

Mg(OH)2  + H2SO4 ⇒ MgSO4 + 2H2O

Mg(OH)2  + 2HCl ⇒ MgCl2 + 2H2O

1.            Assume 0.010 g Mg.  How many moles of Mg is that? 

(moles = mass/At. Wt. = 0.010 g/24 amu = 4.16 x 10-4)

2.            Mole ratio of Mg to MgO is 1:1.  Therefore, 4.16 x 10-4mole of MgO is formed.

3.            Formula weight of MgO = 40.  Mass of MgO = (4.16 x 10-4)(40) = 0.0167 g.

4.            Assume recovery of 0.008 g or 48%.  (0.008 g/0.0167 = 0.48 = 48%)

5.            When added to water, solution pH is 8.5.  pH = - log [H+] so,

8.5 = - log [H+]

- 8.5 = log [H+]

antilog -8.5 = antilog (log [H+])

10-8.5 = [H+] = 10-9 x 10.5 = 3.16 x 10-9

Since Kw = [H+][OH-] = 1.0 x 10 -14

[OH-] = 1.0 x 10-14 / 3.16 x 10-9 = 0.316 x 10 (-14 – (-9)) = 3.16 x 10-1 x 10-5

[OH-] = 3.16 x 10-6 M = 3.16 x 10-6 moles per liter

6.            Since formula is Mg(OH)2, and 3.16 x 10-6 moles per liter is the TOTAL [OH-] per liter, the concentration of Mg(OH)2 must be 3.16 x 10-6 moles/2 = 1.58 x 10-6 moles per liter.   The mass of Mg(OH)2 is  (1.58 x 10-6 )(24 + (2)(17)) = 9.16 x 10-5 g.

7.            Acid (H2SO4 or HCl) was added.   The concentration is 1 M (1 Molar or 1 mole per liter). It reacted with the dissolved Mg(OH)2 to form MgSO4 and water (or MgCl2 and water).  As it reacted, more of the MgO was able to react with the water to form more Mg(OH)2  which then reacted with the acid, and so on util the MgO was all used up. 

8.            If the only Mg(OH)2 was the amount that originally dissolved, and it was in 50 ml water, not 1 liter, the amount of Mg(OH)2 that would have reacted would have been (50 ml / 1,000 ml per liter) (9.16 x 10-5 g) = 4.58 x 10-6 g.

9.            However, since 0.008 g MgO was recovered and eventually reacted, the amount of MgSO4 possible for recovery is in a 1.0 : 1.0 ratio.  (What would it be for MgCl2?) Thus, [(0.008g/(24+16) amu MgO per mole] x (24 + 32 + 64) amu MgSO4 = 0.024 g.

10.            But you don’t know how much water there was.  It was not measured.  You do know the final weight.  If the beaker weighed 45.57 grams, then the gross weight of MgSO4 with beaker would be 45.59 g.

B.  Now, using the model above, solve the following:  

1.  You know how much Mg you started with.  From your lab report, how much was that?  _____________ 

2.  You know how much MgO you recovered. From your lab report, how much was that?  _____________  

3.  You know what the final weight was of either the MgCl2 or MgSO4 that you recovered at the end of the experiment. From your lab report, how much was that?  _____________   

4.  You also know how many drops of acid you added.  From your lab report, how much was that?  _____________   

5.  How many ml of acid is that, based on your finding of the number of drops to make a ml? _____________ 

6.  Now, calculate the following:

a.            Moles Mg at start.

b.            Weight of MgO theoretically possible.

c.            Percentage of MgO actually recovered.

d.            Weight of MgSO4 or MgCl2 theoretically possible.

e.            Percentage of MgSO4 or MgCl2 actually recovered.

f.            Moles of acid needed to react with the Mg(OH)2. 

g.            Moles of acid you actually used.

h.            When you heated the MgO, some students reported that the pH increased to 9.  What is the [OH-] concentration when the pH is 9?

C.  In the next section of this exam,  write and balance the following reactions using the algebraic method that I taught to you:

1.  K4Fe(CN)6 + H2SO4 + H2O = K2SO4 + FeSO4 + (NH4)2SO4 + CO

2.  KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2