### Empirical and Molecular Formulas

Combustion analysis can only determine the empirical formula of a compound; it cannot determine the molecular formula. However, other techniques can determine the molecular weight. Once we know this value, coupled with the empirical formulas, we can easily calculate what the molecular formula is.

Consequently, a full combustion analysis problem might look like this:

Example #1: A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO2 and 2.70 g of H2O. What is the empirical formula of this compound? In addition, its molecular weight has been determined to be about 78. What is the molecular formula?

Solution Step #1: The empirical formula was determined in the "Combustion Analysis" tutorial to be CH. From that we can determine the "empirical formula weight" to be 13 (one carbon plus one hydrogen). This term (empirical formula weight, abbreviation = "EFW") IS NOT a standard chemical term, so be alert to how others describe it.

Solution Step #2: Divide the molecular weight (a standard term in chemistry) by the "empirical formula weight" (a nonstandard term):

78 / 13 = 6

Solution Step #3: Multiply the empirical formula (CH in this example) by the answer to step #2. The result is the molecular formula:

CH x 6 = C6H6

Example #2: Many compounds have the empirical formula of CH2O. Here are the molecular weights of three:

1) 30.0
2) 60.0
3) 180.0

Determine the molecular formula for each. Go to the answers.

Example #3: Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol. What is its molecular formula?

Solution: (1) calculate the empirical formula, (2) compare "EFW" to molecular weight, (3) multiply empirical formula by proper scaling factor.

1) Calculate the empirical formula:

carbon: 49.98 g ÷ 12.011 g/mol = 4.16
hydrogen: 5.19 g ÷ 1.008 g/mol = 5.15
nitrogen: 28.85 g ÷ 14.007 g/mol = 2.06
oxygen: 16.48 g ÷ 15.999 g/mol = 1.03

carbon: 4.16 ÷ 1.03 = 4.04 = 4
hydrogen: 5.15 ÷ 1.03 = 5
nitrogen: 2.06 ÷ 1.03 = 2
oxygen: 1.03 ÷ 1.03 = 1

2) Empirical formula is C4H5N2O. The "empirical formula weight" is about 97.1, which gives a scaling factor of two.

3) The molecular formula is:

C4H5N2O times 2 = C8H10N4O2 <--- that's the molecular formula

There is another technique which reverses the calculation order. In the above examples the empirical formula was calculated first, then the molecular formula. In the technique below, the molecular formula will be calculated first.

Here is the basic technique:

(1) For each element, multiply the molecular weight by the percentage composition (expressed as a decimal).
(2) Divide each element's answer from (1) by its atomic weight.
(3) Round off to the best whole number ratio using the values obtained from (2). Do not remove the common factor This answer is the molecular formula.
(4) Remove the fractor that is common to whole number ratio in (3). This is the empirical formula.

Example #4: Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol. What is its molecular formula? What is the empirical formula? (By the way, this is repeat of example #3, with the addition of the last question.)

Now for the solution using the new technique:

1) Multiply the molecular weight by the percent composition:

 carbon: 194.19 x 0.4948 = 96.0852 hydrogen: 194.19 x 0.0519 = 10.07846 oxygen: 194.19 x 0.1648 = 32.0025 nitrogen: 194.19 x 0.2885 = 56.0238

2) Divide each answer by the atomic weight:

 carbon: 96.0852 ÷ 12.011 = 7.9997 hydrogen: 10.07846 ÷ 1.008 = 9.998 oxygen: 32.0025 ÷ 15.9994 = 2.000 nitrogen: 56.0238 ÷ 14.0067 = 3.9997

3) Round off to closest whole number ratio

 carbon: 8 hydrogen: 10 oxygen: 2 nitrogen: 4

The molecular formula is C8H10N4O2.

4) Remove common factor to get the empirical formula.

The common factor between 8, 10, 4 and 2 is 2. The empirical formula is C4H5N2O.

Example #5: What are the empirical and molecular formulas for a compound with 86.88% carbon and 13.12% hydrogen and a molecular weight of about 345?

Example #6: What are the empirical and molecular formulas for a compound with 83.625% carbon and 16.375% hydrogen and a molecular weight of 388.78?

Example #5 will be solved step-by-step and only the answer for example #6 will be given. Work #6 in parallel as you study #5.

Step One:

 carbon: 345 x 0.8688 = 299.736 hydrogen: 345 x 0.1312 = 45.264

Step Two:

 carbon: 299.736 ÷ 12.011 = 24.955 hydrogen: 45.264 ÷ 1.008 = 44.91

Step Three: the molecular formula is C25H45.

Step Four: The common factor is 5, the empirical formula is C5H9.

Example #7: A compound composed of sulfur and fluorine is found to contain 25.24% by mass of sulfur. If the molar mass of the compound is 254.11 g/mol, what is the molecular formula?

Solution:

1) Assume 100.0 g of the compound is present. That means:

S ---> 25.24 g
F ---> 74.76 g

2) Convert mass to moles:

S ---> 25.24 g / 32.065 g/mol = 0.787 mol
F ---> 74.76 g / 18.998 g/mol = 3.935 mol

3) Determine smallest whole-number molar ratio:

S ---> 0.787 / 0.787 = 1
F ---> 3.935 mol / 0.787 = 5

4) The empirical formula is SF5 and weighs 127.055. Determine the molecular formula:

254.11 / 127.055 = 2

2 times SF5 is S2F10 <--- that's the molecular formula

Example #8: A compound has the empirical formula CHO. If 0.0500 moles of the compound weighs 5.804 g, what is the molecular formula?

Solution:

1) Determine the molecular weight of the compound:

5.804 g / 0.0500 mol = 116.08 g/mol

2) Determine the "empirical formula weight" of CHO:

12.011 + 1.008 + 15.994 = 29.013 g

3) Divide 1) by 2):

116.08 / 29.013 = 4

4) This means there are 4 "units" of CHO in the molecular formula:

CHO times 4 = C4H4O4 <--- the molecular formula

Example #9: A gaseous compound is 78.26% boron and 21.74% hydrogen by mass. A 0.600 g sample of it occupies 486 mL at STP. What is the molecular formula for the gas?

Solution:

1) Determine the empirical formula of the compound. Start by assuming 100 g of the compound is present:

B ---> 78.26 g
H ---> 21.74 g

B ---> 78.26 g / 10.811 gmol = 7.239 mol
H ---> 21.74 g / 1.008 g/mol = 21.567 mol

Do you see the 1:3 molar ratio?

B --->7.239 mol / 7.239 mol = 1
H --->21.567 mol / 7.239 mol = 2.98 = 3

The empirical formula is BH3.

2) Determine the molecular weight of the compound. Since we are at STP, we can use molar volume:

0.486 L / 22.414 L/mol = 0.0216829 mol

0.600 g / 0.0216829 mol = 27.67 g/mol

3) Determine the molecular formula:

BH3 weighs 13.835

27.67 13.835 = 2

BH3 times 2 = B2H6

The molecular formula is B2H6.

Example #10: An organic volatile compound was analyzed by combustion analysis and found to be 85.63% C and 14.37% H. In a Dumas bulb experiment, a 2.174 g sample of the compound's vapor occupied 1.00 L at 120.0 °C and 760.0 torr. Determine the (a) the molar mass and (b) the molecular formula of the compound.

Solution:

1) The Dumas experiment data provides us with the molar mass:

Ideal Gas Law ---> PV = nRT

(760.0 torr / 760.0 torr/atm) (1.00 L) = (n) (0.08206 L atm / mol K) (393 K)

n = 0.031008 mol

2.174 g / 0.031008 mol = 70.11 g/mol

2) The combustion analysis data will give us the empirical formula:

Assume 100 g of the compound is present. This means 85.63 grams of C and 14.37 grams of H are present.

Determine moles:

carbon ---> 85.63 g / 12.011 g/mol = 7.13 mol
hydrogen ---> 14.37 g / 1.008 g/mol = 14.256 mol

Divide through by smallest:

carbon ---> 7.13 mol / 7.13 mol = 1
hydrogen ---> 14.256 mol / 7.13 mol = 2

The empirical formula is CH2

3) Determine the molecular formula:

CH2 weighs 14.0268

70.11 / 14.0268 = 4.998

The molecular formula is C5H10

Bonus Example #1: An organic compound A contains C, H and N only. When a certain mass of the compound was subjected to combustion, CO2 and H2O ware obtained in the mole ratio 4.3. N2 gas was also evolved, but no measurements of it were taken. The relative molecular mass of the compound was determined to be 164. Determine the molecular formula.

Solution #1:

1) The carbon goes into the CO2 and the hydrogen into the H2O. Since the mole ratio is 4 mol of CO2 to 3 mol of H2O, there are 4 carbon atoms for every 6 hydrogen atoms.

2) There is also nitrogen, and the empirical mass has to be a whole-number factor of 164, the molecular mass. Let us try C4H6N. Is its weight a whole-number factor of 164?

C4H6N weighs 68.0984.

164 / 68.0984 = 2.41

3) One nitrogen doesn't work, so let us try C4H6N2:

C4H6N2 weighs 82.105.

164 / 82.105 = 1.997 = 2

We have a winner!

4) Since 82 is half of 164, the molecular formula is C4H6N2 doubled, giving this for the final answer:

C8H12N4

Solution #2:

1) I would like to start at this ratio:

4 carbon atoms for every 6 hydrogen atoms

2) Let's reduce that to:

2 carbon atoms for every 3 hydrogen atoms

C2H3

4) Let's add in one nitrogen and calculate its weight:

C2H3N

It weighs 41.0527

5) Is 41.0527 a factor of 164?

164/41.0527 = 3.995 = 4

We have a winner!

6) We need to multiply C2H3N by 4 to obtain:

C8H12N4

Bonus Example #2: A compound containing only sulfur and fluorine is 25.238% sulfur. 0.8902 g of the compound was found to occupy 115.5 mL at 715 mmHg and 105 °C. What is the compound's molecular formula?

Solution:

Plan: (a) Determine the empirical formula and its weight, (b) Determine the molecular weight, (c) Use the two weights to determine the molecular formula.

1) By subtraction, determine the percent fluorine in the compound to be 74.762%. Assume 100 g of the compound is present and convert to moles:

S ---> 25.238 g / 32.065 g/mol = 0.78709 mol
F ---> 74.762 g / 18.9984 g/mol = 3.935 mol

2) Determine lowest whole-number ratio:

S ---> 0.78709 mol / 0.78709 mol = 1
F ---> 3.935 mol / 0.78709 mol = 4.999 = 5

The empirical formula is SF5

The "empirical formula weight" is determined to be 127.057. <--- this number gets used below

3) Use PV = nRT to determine moles of the compound present:

(0.9408 atm) (0.1155 L) = (n) (0.08206 L atm mol¯11) (378 K)

n = 0.003503 mol

Note the mmHg and mL were converted to atm and L.

4) Determine the molecular weight:

0.8902 g / 0.003503 mol = 254.12 g/mol

5) Determine the "scaling factor:"

254 / 127 = 2

6) Determine the molecular formula:

SF5 times 2 = S2F10 <--- the answer