Lesson 27 Molecular Formula

Objectives:-         The student will calculate molecular

formulas from an empirical formula and a molecular mass.

-         The student will calculate empirical and molecular formulas from

percentage composition data and a molecular mass. PA Science and Technology Standards: 3.4.10.A; 3.4.12.APA Mathematics Standards: 2.2.11.A, 2.4.11.E

I.    Empirical Formulasa.      Up to this point, we have represented formulas in

two different ways:

             i.      Lewis Structures

             ii.      Chemical formulas

b. Now, we will look at the difference between empirical formulas and molecular formulas

II. Empirical formulas are determined experimentally

a.      Many times when scientists discover a new compound, they determine the empirical formula from composition data.

b.      They can also use other methods to determine the molecular mass

of the compound present. This can be used to determine a molecular formula.

III. Molecular formulas

a.      An empirical formula just tells you how much of each element is present, but in lowest number ratios.b.      Molecular formula – gives type and actual number of elements in a compound.c.      Many different compounds can have the same empirical formula, but they may not have the same molecular formula.d. Also, many times the empirical formula for a compound is the same as the molecular formula – sometimes, it is not.

e.      Examples – all these have CH2O as their empirical formula

                            

        i.      Formaldehyde – CH2O

– poisonous

                     ii.      Acetic Acid (vinegar) – C2H4O2 – edible

                     iii.      Glucose – C6H12O6

– sugar

IV. Molecular formulas are determined from molar mass

a.      Molecular formulas are multiples of the empirical formulas

b.      X (empirical formula) = molecular formula

c.      X will be a whole number.

d.      To determine X, divide the molecular formula mass by the empirical

formula mass

Molecular Formula Calculations:

In calculations of molecular formula, you will be given percent compositions and the molecular weight of the entire compound.

 1.      Calculate the empirical formula as shown earlier.2.      Calculate the empirical mass of the empirical formula.3.      Divide the given molecular mass by the empirical mass.4.      The answer should be a whole number, multiply this through the subscripts of the empirical formula to get the molecular formula.

Example:

Problem 1 level 1 on page 213 26% C = 26g C (1 mole/ 12.011) =

2.16468moles C

2.2% H= 2.2g H(1 mole/ 1.0079) = 2.182756moles H

71.1%O=71.1g O(1 mole/15.999)= 4.444 moles O

2.16468 is the smallest so divide all by this 

C 2.16468/2.16468 = 1

H 2.182756/2.16468 = 1.008 = about 1

O 4.444/ 2.16468 = 2.053 = about 2 Use these numbers as the subscripts, giving you

CHO2

Now take this formula to calculate the empirical mass

 C- 1 x 12.011 = 12.011H- 1 x 1.0079 = 1.0079O- 2 x 15.999 = 31.998

45.169  empirical

mass

Now divide the molecular mass you were given in the problem, by the empirical mass that you calculated:

90/45.0169 = 2 take this number and multiply all the

subscripts in the empirical formula by it to get the molecular formula

 CHO2 becomes C2H2O4

Examples:Aspirin contains 60.8% C, 4.48% H and 35.5% O. It

has a molecular mass of 276μ. What are its empirical and molecular formula?

1. % grams2. Grams moles3. Divide by smallest number of moles4. Determine empirical formula5. Find empirical mass6. Divide molecular mass by empirical mass7. Multiply subscripts of empirical formula by the

answer in step 6

Level 1 1. A compound has the following percentage composition: 26.7%

carbon, 2.2% hydrogen, 71.1% oxygen. The molecular weight of this compound is 90. What is the compound’s true formula?

2. A certain compound was analyzed and found to have the following composition: 54.6% carbon, 9.0% hydrogen, 36.4% oxygen. The true molecular weight for the compound is 176. What is the molecular formula of the compound?

3. The percentage composition of ethane gas is 80.0% carbon and 20.0% hydrogen. The molecular weight for ethane is 30. What is the correct formula for this compound?

4. Analysis of a compound shows that it consists of 24.3% carbon, 4.1% hydrogen, and 71.6% chlorine. The molecular weight of the compound is determined to be 89.8. What molecular formula corresponds to these data?

5. An unknown compound is analyzed and found to consist of 49.0% carbon, 2.7% hydrogen, and 48.2% chlorine. Boiling point data suggest that the molecular weight of this compound is around 150. What molecular formula would you predict for this compound?

6. A gaseous compound is found to have the following composition data: 30.5% nitrogen and 69.5% oxygen. The molecular weight of the gas is found to be 91.8. What molecular formula corresponds to these data?