1. WUP#22
A compound composed of 40.68% carbon, 5.08% hydrogen and 54.24% oxygen. Determine the empirical.
40.68 g C
1 mol C
= mol C
= 1.00
x 2
= 2 mol C
3.387
12.01 g C
5.08 g H
1 mol H
= 5.03 mol H
= 1.49
x 2
= 3 mol H
1.01 g H
3.387
54.24 g O
1 mol O
= mol O
= 1.00
x 2
= 2 mol O
16.00 g O
3.387
E.F. = C2H3O2

2. Molecular Formula

3. HCHO CH2O CH2O C6H12O6 CH2O 30.03 g/mol CH3COOH 30.03 g/mol
Compound
Empirical Formula
Molecular Formula
E.F.W
M.F.W
Formaldehyde
HCHO
Acetic Acid
CH3COOH
Glucose
C6H12O6
30.03
x 1
= g/mol
30.03 g/mol
CH2O
30.03
x 2
= g/mol
30.03 g/mol
CH2O
30.03
x 6 =
g/mol
30.03 g/mol
CH2O

4. Acetic Acid
Formaldehyde
Glucose

5. multiple empirical formula E.F. x 6 = M.F. CH2O x 6 = C6H12O6
Molecular Formula: a formula that tells the actual number of atoms in a compound
multiple
It is a __________
of its ______________________.
empirical formula
E.F. x 6 = M.F.
CH2O x 6 = C6H12O6

6. A compound composed of 40. 68% carbon, 5. 08% hydrogen and 54
A compound composed of 40.68% carbon, 5.08% hydrogen and 54.24% oxygen has a molar mass of g/mol. Determine the empirical and molecular formula.
40.68 g C
1 mol C
= mol C
= 1.00
x 2
= 2 mol C
3.387
12.01 g C
5.08 g H
1 mol H
= 5.03 mol H
= 1.49
x 2
= 3 mol H
1.01 g H
3.387
54.24 g O
1 mol O
= mol O
= 1.00
x 2
= 2 mol O
16.00 g O
3.387
E.F. = C2H3O2

7. E. F. = C2H3O2 = 59.05 g x  = 118.1 g  = 2 So your E.F. x 2 = M.F.
 = 2
Calculated molar mass of E.F.
given molar mass for the compound in the original problem
So your E.F. x 2 = M.F.
E. F. x 2 = C2H3O2 X 2 = M.F. = C4H6O4

8. 2) A compound has a molar mass of 462. 8 g/mol and contains 77
2) A compound has a molar mass of g/mol and contains 77.87% C, 11.76% H, and 10.37% O. Determine the empirical and molecular formulas.
77.87g C
1 mol C
=6.484 mol C
6.484 mol  10
mol
12.01g C
11.76g H
1 mol S
= mol H
11.64 mol  18
mol
1.01g S
10.37g O
1 mol O
mol = 1
mol
= mol O
16.00g O
E. F. = C10H18O

9. E. F. = C10H18O = 154.28g x  = 462.8 g  = 3 So your E.F. x 3 = M.F.
 = 3
Calculated molar mass of E.F.
given molar mass for the compound in the original problem
So your E.F. x 3 = M.F.
E. F. x 3 = C10H18O X 3 = M.F. = C30H54O3