empirical formula

1. Empirical and Molecular Formulas

2. Review
 We learned how to calculate the molar
mass of compounds.
 Calculate the molar mass of Ca(CN)2.
 1 x Ca = 1 x 40.08 g/mol = 40.08 g/mol
 2 x C = 2 x 12.01 g/mol = 24.02 g/mol
 2 x N = 2 x 14.01 g/mol = 28.02 g/mol
 TOTAL = 92.12 g/mol

3. Review
 We also learned how to determine the
percentage composition of a compound.
 Calculate the % composition of Ca(CN)2.
 %Ca = (40.08)/(92.12) x 100% = 43.51% Ca
 %C = (24.02)/(92.12) x 100% = 26.07% C
 %N = (28.02)/(92.12) x 100% = 30.42% N

4. Empirical Formulas
 Molecular Formula (MF) - shows how
many atoms are actually in a molecule.
 EXAMPLE: Glucose has the MF C6H12O6.
 EXAMPLE: Water has the MF H2O.
O
H H

5. Empirical Formulas
 Empirical Formula (EF) - shows the
lowest whole-number ratio of atoms in a
compound.
 EXAMPLE: Glucose has the EF CH2O.
 EXAMPLE: Water has the EF H2O.
 Different cmpds can have different MFs but
have the same EF.
 EXAMPLE: NO2 and N2O4 have different MFs
but the same EF (NO2).

6. Empirical Formulas
 You can discover the empirical formula
of a compound if you know the %
composition.
 Will NOT tell you which molecular
formula is correct!

7. Empirical Formulas
 An unknown compound is analyzed:
 15.77% carbon
 84.23% sulfur
 Calculate the EF.

8. Empirical Formulas
 First, assume you have exactly 100
grams of the sample.
 Why 100 grams?
 Because percents become grams.
 In 100 grams of this compound you
would have:
 15.77 g C
 84.23 g S

9. Empirical Formulas
 Next, change grams to moles.
 15.77 g C
 84.23 g S
x
12.01 g C
1 mol C
= 1.313 mol C
x
32.07 g S
1 mol S
= 2.626 mol S

10. Empirical Formulas
 The formula so far:
 C1.313S2.626
 Divide all subscripts by the lowest one.
 CS2
 This is the empirical formula of our
mystery compound.
 We don’t know if it’s the correct molecular
formula.
 Could be CS2, C2S4, C3S6, C4S8, etc...

11. Empirical Formulas
 A mystery compound has the following
composition:
 3.0856% hydrogen
 31.604% phosphorus
 65.310% oxygen
 What is this compound’s empirical
formula?

12. Empirical Formulas
 Convert grams to moles.
 3.0856 g H x = 3.0613 mol H
 31.604 g P x = 1.0203 mol P
 65.310 g O x = 4.0820 mol O
 Formula so far: H3.0613P1.0203O4.0820
 Reduced: H3PO4
H
g
1.00794
H
mol
1
P
g
30.9738
P
mol
1
O
g
15.9994
O
mol
1

13. Empirical Formulas
 A mystery compound has the following
composition:
 25.940% nitrogen
 74.060% oxygen
 What is this compound’s empirical
formula?

14. Empirical Formulas
 Convert grams to moles.
 25.940 g N x = 1.8520 mol N
 74.060 g O x = 4.6289 mol O
 Formula so far: N1.8520O4.6289
 Reduced: NO2.5
 Double subscripts to eliminate fractions.
 N2O5
N
g
14.0067
N
mol
1
O
g
15.9994
O
mol
1

15. Molecular Formulas
 If we know a compound’s empirical formula
and its molar mass, we can work out its
molecular formula.
 Benzene, a common non-polar solvent, has the
empirical formula CH and a molar mass of 78.12
g/mol
 Formula mass of CH: 13.02 g/mol
 How many times does this go into 78.12 g/mol?
 78.12 g/mol  13.02 g/mol = 6 times
 Molecular formula of benzene = (CH)6 = C6H6

16. Molecular Formulas
 The empirical formula of uracil (a base found in
RNA) is C2H2NO. If the molar mass of uracil is
122.09 g/mol, what is the molecular formula of
uracil?
 Formula mass of C2H2NO: 56.05 g/mol
 How many times does this go into 122.09 g/mol?
 122.09 g/mol  56.05 g/mol ≈ 2
 Molecular formula of uracil = (C2H2NO)2 = C4H4N2O2