1. Empirical and Molecular Formulas

2. Definition: the lowest whole number ratio of elements in a compound.
Steps to determine empirical formulas:
1) find moles of each element [find the moles of H20 if compound is hydrated].
2) divide by smallest number of moles to get a whole number.
3) multiply by a common whole number, if necessary.

3. Calculate EF of a compound with 25.9% N and 74.1% O.
%=assume mass mass mols
25.9g N x 1 mol N = mols N
14.007g N
74.1g O x 1 mol O = 4.63 mols O
16 g O
2) N: = 1
1.85
O: = 2.5
3) Multiply by common whole number
N: 1 x 2 =2
O: 2.5 x 2 = = N2O5

4. Just remember this!!
Percents to grams Grams to moles Divide by smallest Multiply ‘til whole

5. Given: 36.5g Na, 25.4g S, and 38.1g O 25.4g S x 1 mol S = .79 mols S
36.5g Na x 1 mol Na = 1.59 mols Na
22.989g Na
25.4g S x 1 mol S = .79 mols S
32g S
38.1g O x 1 mol O = 2.4 mols O
16g O
Na: = S: .79 = O: 2.4 = 3
= Na2SO3

6. An oxide of aluminum is formed by the reaction of 4
An oxide of aluminum is formed by the reaction of 4.151g of aluminum with 3.692g of oxygen. Find EF.
4.151g Al x 1 mol Al = mol Al atoms
26.98 g Al
3.692g O x 1 mol O = mol O atoms
16 g O
mol Al = mol Al atoms
0.1539
mol O = mol O atoms
1.500 O x 2 = = 3 O atoms
1.000 Al x 2 = 2.000= 2 Al atoms
= Al2O3

7. Multiple = molecular mass x empirical formula (whole number)
Molecular formulas show the number of atoms in a compound. In order to determine the molecular, you must have the empirical formula first.
The molecular mass of molecule will always be given.
Multiple = molecular mass x
empirical formula (whole number)
The molecular formula is always an integer multiple of the empirical formula.

8. Determine the molecular formula of a compound whose EF is CH2O and molecular mass is 120 g/mol
-Distribute that 4 throughout the empirical formula =
C4H8O4

9. A compound is 64. 9% carbon, 13. 5% hydrogen, and 21. 6% oxygen
A compound is 64.9% carbon, 13.5% hydrogen, and 21.6% oxygen. Its molc mass is 74 g/mol. What is its MF?
64.9g C x 1 mol C = 5.40 mols C = 4 C
12.01 g C
13.5g H x 1 mol H = mols H = 10 H
1.01 g H
21.6g O x 1 mol O = 1.35 mols O = 1 O
16.0 g O
74 g/mol = 1
74 g/mol = C4H10O

10. Ex) a compound is 54. 5% carbon, 9. 1% hydrogen, and 36. 4% oxygen
Ex) a compound is 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen. It’s molc mass is 88 g/mol. What is its molecular formula?
54.5 C x 1 mol C = 4.54 mols C
12.01 g C
9.1 g H x 1 mol H = 9 mols H
1.01 g H
36.4 g O x 1 mol O = 2.28 mols O
16 g O
4.54 mols C = 2 C
2.28 mols
9 mols H = 4H
2.28 mols O = 1O
= C4H8O2

11. Ex) A compound has an empirical formula of ClCH2 and a molecular weight of g/mol. What is it’s molecular formula?
Mass Cl + mass C + 2(mass H)
Mass of empirical unit= (2.008) = g/mol
98.96 g/mol = 2.000
49.47 g/mol
= Cl2C2H4