### 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:

Problem #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. This is the molecular formula:

CH x 6 = C6H6

Problem #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.

Problem #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.

Problem #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.

Problem #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?

Problem #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?

Problem #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.

Problem #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

Problem #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