Empirical and Molecular Formulas

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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 C8H10N4O2.


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:

Step One - 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

Step Two - 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

Step Three - round off to closest whole number ratio

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

The molecular formula is C8H10N4O2.

Step Four - 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.

The answer to example #6.


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