CHM 101 GENERAL CHEMISTRY

FALL QUARTER 2008

Section 2

 

Lecture Notes Ė 10/15/2008

(last revised: 10/15/08, 4:30 PM)

 

 

3.3          The Mole: The mole (abbreviated mol) is a unit of measure that greatly facilitates our ability to count atoms by weighing them.

n = 6.02214 x 1023

n = 6.022 x 1023

o       What is the mass in grams of 6 atoms of americium (Am)? (3.2, p. 83)

o       How many moles are there in 10.0 g of aluminum? How many atoms? (3.3, pp. 84-5)

o       How many silicon atoms are there in a silicon chip whose mass is 5.68 g? (3.4, p. 85)

o       How many moles are there in a sample of cobalt containing 5.00 x 1020 atoms? What is the mass in grams? (3.5, p. 85)

3.4          Molar Mass: We can use the mole to help us count molecules by weighing them.

Mass of 1 mol C

=

 

=

12.01 g

Mass of 4 mol H

=

4 x 1.008 g

=

4.03 g

Mass of 1 mol CH4

=

 

=

16.04 g

 

o       The molecular formula for juglone is C10H6O3. (3.6, p. 86)

o       a) Calculate the molar mass of juglone.

o       b) How many moles of juglone are there in a 1.56 x 10-2 g sample of juglone?

o       The formula for calcium carbonate is CaCO3. (3.7, p. 86-8)

o       a) Calculate the molar mass of calcium carbonate.

o       b) What is the mass (in grams) of 4.86 mols of calcium carbonate?

o       c) What is the mass of the CO32- ions in 4.86 mols of calcium carbonate?

o       Isopentyl acetate (C7H14O2) is the molecule responsible for the odor of bananas. Bees release about 1 μg (1 x 10-6 g) of isopentyl acetate when they sting, in order to attract other bees to the attack. (3.8, p. 88)

o       a) How many molecules of isopentyl acetate are released in a typical bee sting?

o       b) How many carbon atoms are there in this amount of isopentyl acetate?

3.5          Percent Composition

o       By the numbers of each atom present in the molecular formula.

o       By the mass percent of each element in the compound. We can obtain these mass percentages from the molecular formula by the following procedure.

o       First we compute the molar mass (in grams/mol) of the compound. This also gives us the molar masses of each element present in the compound.

o       Then we compute the ratios of the masses of each element to the total molar mass.

o       When we express these ratios as percentages, we are done.

The example in the text is ethanol (C2H5OH). Letís work it out.

Mass of 2 mol C

=

2 x 12.01 g

=

24.02 g

Mass of 6 mol H

=

6 x 1.008 g

=

6.05 g

Mass of 1 mol O

=

1 x 16.00 g

=

16.00 g

Mass of 1 mol C2H5OH

=

 

=

46.07 g

 

We can check our arithmetic by adding the 3 percentages together. They should (and do) sum to 100.00%.

o       Calculate the mass percent of each element in carvone (C10H14O) (3.9, pp. 89-90)

o       Calculate the mass percent of each element in penicillin F (C14H20N2SO2) (3.10, pp. 90-1)

3.6          Determining the Formula of a Compound:

o       First we calculate the masses (in grams) of each element in the 100 g sample.

o       Second we calculate the number of moles of each element in the sample by dividing the masses by the atomic masses.

o       Finally, we reduce the result to a set of small whole numbers. This gives us the empirical formula of the compound.

Letís work out the empirical formula for ethanol.

Element

C

H

O

% Composition

52.14 %

13.13%

34.73%

g/100 g of compound

52.14 g

13.13 g

34.73 g

molar mass element (g/mol)

12.01

1.008

16.00

mols/100 g compound

52.14/12.01

13.13/1.008

34.73/16.00

mols/100 g compound

4.3413

13.025

2.1706

(above values)

2.000

6.000

1.000

2.1706

 

The result is C2H6O. If we compare this result against the known formula for ethanol, (C2H5OH), we see we have the correct numbers of atoms, but no information about the structure. For all we can determine from the % composition, the true formula could be CH3OCH3 (dimethyl ether), or C4H12O2, or any of an infinite number of possibilities where the C:H:O ratio is 2:6:1.

In this example we are told that we start with 0.1156 g of a purified compound containing only carbon, nitrogen, and hydrogen. When we burn it in the presence of enough oxygen to make it burn completely, we find that the apparatus has collected 0.1638 g of CO2 and 0.1676 g of H2O. From this information we are asked to determine the (empirical) formula of the compound. This is a complex problem, so we need to break it into a set of simple problems.

1.    How many grams of carbon are there in 0.1638 g of CO2? What percent is this of the total mass of the sample?

2.    How many grams of hydrogen are there in 0.1676 g of H2O?

3.    Since the only other element in the starting sample is nitrogen, this will be the 0.1156 g that we burned, less the masses of hydrogen and carbon that we determined in steps 1 & 2.

4.    What is the mass percent composition of the starting compound. We will calculate this from the masses of C, H, and N that we determined in steps 1-3 and from the mass of compound we started with.

5.    Now we can solve for the empirical formula in the same way we found the empirical formula for ethanol.

Letís work these simple problems in order:

1.    1 mol of CO2 contains 1 mol (12.01 g) of C plus 2 mols (32.00) of O, giving us a molar mass of 44.01 g/mol for CO2. Thus we can convert from g CO2 to g C:

And we can compute the % C in the starting compound:

2.    1 mol of H2O contains 2 mols (2.016 g) of H and 1 mol (16.00 g) of O, giving us a molar mass of 18.02 g/mol for H2O. Thus we can convert from g H2O to g H:

And we can compute the % H in the starting compound:

3.    Then the % N is:

4.    Now we can work out the mass percent composition of the compound.

Element

C

H

N

% Composition

38.67 %

16.22%

45.11%

g/100 g of compound

38.67 g

16.22 g

45.11 g

molar mass element (g/mol)

12.01

1.008

14.01

mols/100 g compound

38.67/12.01

16.22/1.008

45.11/14.01

mols/100 g compound

3.220

16.09

3.219

(above values)

1.000

4.998

1.000

3.219

 

5.    Thus the empirical formula (formula determined by experiment) is CH5N.

Suppose somehow that we know that the molar mass of the compound is 31.06 g/mol. How do we work out the true formula? Letís start by determining the molar mass of CH5N.

Mass of 1 mol C

=

1 x 12.01 g

=

12.01 g

Mass of 5 mol H

=

5 x 1.008 g

=

5.04 g

Mass of 1 mol N

=

1 x 14.01 g

=

14.01 g

Mass of 1 mol CH5N

=

 

=

31.06 g

Since we know that the molar mass of the compound is the same as the molar mass of CH5N, we can conclude that the true formula of the compound is (also) CH5N.

Element

Cl

C

H

% Composition

71.65 %

24.27 %

4.07 %

g/100 g of compound

71.65 g

24.27 g

4.07 g

molar mass element (g/mol)

35.45

12.01

1.008

mols/100 g compound

71.65/35.45

24.27/12.01

4.07/1.008

mols/100 g compound

2.021

2.021

4.04

(above values)

1

1

2

(smallest)

The empirical formula is ClCH2. Now we can determine the formula mass (i. e., the molar mass corresponding to the empirical formula):

Mass of 1 mol Cl

=

1 x 35.45 g

=

35.45 g

Mass of 1 mol C

=

1 x 12.01 g

=

12.01 g

Mass of 2 mol H

=

1 x 1.008 g

=

2.02 g

Mass of 1 mol ClCH2

=

 

=

49.48 g

If we divide the known molar mass (98.96 g/mol) by this formula mass (49.48 g/mol) we get 2. Thus the true formula of the compound is twice the empirical formula: (ClCH2)2 or Cl2C2H4. FYI: Two possible structures for this molecule are shown in Figure 2.7:

Element

P

O

% Composition

43.64 %

56.36 %

g/100 g of compound

43.64 g

56.36 g

molar mass element (g/mol)

30.97

16.00

mols/100 g compound

43.64/30.97

56.36/16.00

mols/100 g compound

1.409

3.523

(above values)

1

2.5 = 5/2

(smallest)

In order that our empirical formula contain only whole numbers, we multiply by 2 to get 2 P atoms and 5 O atoms. Thus the empirical formula is P2O5. Now we can determine the formula mass:

Mass of 2 mol P

=

2 x 30.97 g

=

61.94 g

Mass of 5 mol O

=

5 x 16.00 g

=

80.00 g

Mass of 1 mol P2O5

=

 

=

141.94 g

If we divide the known molar mass (283.88 g/mol) by this formula mass (141.94 g/mol) we get 2. Thus the true formula of the compound is twice the empirical formula: (P2O5)2 or P4O10. FYI: The structure of this interesting molecule is shown in Figure 2.8:

Element

P

O

% Composition

43.64 %

56.36 %

molar mass of compound (g/mol)

283.88

283.88

mass of element per mol of compound (g/mol)

0.4364 x 283.88

0.5636 x 283.88

mass of element per mol of compound (g/mol)

123.89

159.99

molar mass of element (g/mol)

30.97

16.00

mols element per mol compound

123.89/30.97

159.99/16.00

mols element per mol compound

4.000

10.00

Thus each mole of the compound contains 4 moles of P and 10 moles of O. This is just another way of saying that the true formula of the compound is P4O10.

Element

C

H

N

O

% Composition

49.48 %

5.15 %

28.87 %

16.49%

molar mass of caffeine (g/mol)

194.2

194.2

194.2

194.2

mass of element per mol of caffeine (g/mol)

0.4948 x 194.2

0.0515 x 194.2

0.2887 x 194.2

0.1649 x 194.2

mass of element per mol of caffeine (g/mol)

96.09

10.00

56.07

32.02

molar mass of element (g/mol)

12.01

1.008

14.01

16.00

mols element per mol caffeine

96.09/12.01

10.00/1.008

56.07/14.01

32.02/16.00

mols element per mol caffeine

8.001

9.92

4.002

2.001

We round these results to the nearest integer and get the result. Caffeine has a true formula of C8H10N4O2. The figure illustrates its structure.

 

3.7          Chemical Equations

CH4 + O2 = CO2 + H2O

In this equation, the reactants are CH4 and O2 and the products are CO2 and H2O. Notice that the carbon-hydrogen bonds from the methane and the oxygen-oxygen bond in the oxygen have all disappeared and that new carbon-oxygen and hydrogen-oxygen bonds have been formed. (If you have sharp eyes, you might notice that the text uses a right pointing arrow in its version of this equation, where I use an equals sign. Both ways are correct, but I use the equals sign to remind you that this is an equation whereas the text uses the arrow to indicate the direction of the reaction.)

CH4 + O2 = CO2 + 2H2O

Now there are 4 hydrogens on each side. And we now have 4 oxygens on the right, but we still have only 2 on the left. Letís change the number of oxygen molecules on the left from 1 to 2. This gives us:

CH4 + 2O2 = CO2 + 2H2O

Now there are 4 oxygens on each side. And we left the numbers of carbons and hydrogens unchanged. Each side still has 1 carbon and 2 hydrogens. The equation is now balanced.

o       gas (g),

o       liquid (l),

o       solid (s), or

o       dissolved in water (aq)

We do this by putting the appropriate abbreviation to the right of the chemical formula. Thus the methane/oxygen reaction can be written:

CH4 (g) + 2O2 (g) = CO2 (g) + 2H2O (g)

An example of a reaction involving all four of the states is the reaction of aqueous hydrochloric acid (HCl dissolved in water) with solid sodium hydrogen carbonate to generate carbon dioxide, water, and sodium chloride:

HCl (aq) + 2NaHCO3 (s) = CO2 (g) + H2O (l) + NaCl (aq)

3.8          Rules for Balancing Chemical Equations: An unbalanced chemical equation is not very useful. You should get in the habit of checking any chemical equation you encounter to see if it is balanced. There are two fundamental rules to observe and a systematic trial-and-error procedure to follow when balancing an equation.

o       Balancing a unbalanced chemical equation is a little like solving a Sudoku puzzle. You start by looking for an empty cell where only one of the nine possible digits can fit, i. e., an empty cell with a uniquely determined number. By filling the cell with this number, you eliminate the number as a choice for several other cells, and hopefully, one of these cells now has become uniquely determined. You keep repeating the process until the puzzle is done.

o       Now back to equation balancing. We start with the most complicated molecule among the reactants and products. (It helps if that molecule also contains all the atoms of a given type to be found on one of the sides of the equation. We can call these unique occurrence atoms.) We will use the example from p. 99 of your text: ethanol (C2H5OH) reacting with oxygen (O2). We write out the unbalanced reaction. Then we choose ethanol as being the most complicated molecule, and we observe that it contains all the carbon atoms and all the hydrogen atoms to be found among the reactants:

C2H5OH (l) + O2 (g) = CO2 (g) + H2O (g)

o       From the chosen molecule (ethanol in this case), we select a unique occurrence atom. In this example, we have our choice between C and H. Weíll do them one at a time, starting with carbon. We have 2 carbon atoms on the left, both of them in the ethanol molecule, but only 1 on the right (in CO2). We fix this by changing the coefficient of CO2 from 1 to 2:

C2H5OH (l) + O2 (g) = 2CO2 (g) + H2O (g)

o       This balances carbon, and it increases oxygen on the right hand side, but it does not affect hydrogen. Now we can balance hydrogen. We have 6 hydrogen atoms on the left, all in the ethanol molecule, but only 2 on the right (in H2O). We fix this by changing the coefficient of H2O from 1 to 3:

C2H5OH (l) + O2 (g) = 2CO2 (g) + 3H2O (g)

o       Now we have balanced both of the unique occurrence type atoms from ethanol, the most complicated molecule in the reaction. What types of atom does this leave us to balance? In this case, there is only oxygen. The equation as it is now written has 7 oxygen atoms on the right, all of them in molecules that are balanced with respect to their other atoms. We have only 3 oxygens on the left, but if we try to increase the number to 7 by changing the coefficient of ethanol from 1 to 4, we lose our carbon and hydrogen balance. The only coefficient we can adjust is the 1 on the O2. However, if we change it to 3, we gain the 4 oxygen atoms we need to finish balancing the equation:

C2H5OH (l) + 3O2 (g) = 2CO2 (g) + 3H2O (g)

o       Now we can check our final balance by tallying the atoms on each side:

C2H5OH (l) + 3O2 (g)

=

2CO2 (g) + 3H2O (g)

2 C atoms

 

2 C atoms

5 + 1 = 6 H atoms

 

6 H atoms

1 + 6 = 7 O atoms

 

4 + 3 = 7 O atoms

 

o       Here is a summary of the process:

o       Determine the reaction that occurs, i. e., identify the reactants, the products, and their physical states.

o       Write the unbalanced equation.

o       Balance the equation by inspection. Select the most complex of the molecules in the reaction and leave its coefficient set at 1. (However, you may need to adjust it later.) Starting with any unique occurrence type atoms in the selected molecule, balance each atom, type by type, reviewing each molecular coefficient for possible adjustment.

o       Exercise 3.14 (p. 100): Solid ammonium dichromate, (NH4)2Cr2O7, reacts spectacularly when it is ignited. Assuming the reaction products are solid chromium (iii) oxide, nitrogen gas (N2) and water vapor, balance the reaction.

o       Exercise 3.15 (p. 101): At 1000 įC, ammonia gas (NH3) reacts with oxygen gas (O2) to form gaseous nitric oxide (nitrogen monoxide, NO) and water vapor. Balance the reaction.

3.9           

3.10