1. C H What is the Empirical mass if it has 65.5% C, 5.5% H AND 29.0% O
STEP 1: assume a 100 gram sample.
STEP 2: express the % of each element as grams.
STEP 3: follow the chart below.
ELEMENT
MASS
MOLES
(raw ratio)
REDUCE
(divide by smallest)
ROUNDED RATIO C
65.5 g H
5.5 g O
29.0 g

2. C H What is the Empirical mass if it has 65.5% C, 5.5% H AND 29.0% O
Mol = MASS/MOLAR MASS
Mol C = g = mol C
12.01g/mol
STEP 1: assume a 100 gram sample.
STEP 2: express the % of each element as grams.
STEP 3: follow the chart below.
ELEMENT
MASS
MOLES
(raw ratio)
REDUCE
(divide by smallest)
ROUNDED RATIO C
65.5 g
5.45 mol H
5.5 g
5.44 mol O
29.0 g
1.81 mol

3. C H What is the Empirical mass if it has 65.5% C, 5.5% H AND 29.0% O
STEP 1: assume a 100 gram sample.
STEP 2: express the % of each element as grams.
STEP 3: follow the chart below.
ELEMENT
MASS
MOLES
(raw ratio)
REDUCE
(divide by smallest)
ROUNDED RATIO C
65.5 g
5.45 mol
5.45/1.81=3.011 H
5.5 g
5.44 mol
5.44/1.81=3.001 O
29.0 g
1.81 mol
1.81/1.81=1.00

4. What is the Empirical mass if it has 65.5% C, 5.5% H AND 29.0% O
The final mole ratio is
C3H3O
STEP 1: assume a 100 gram sample.
STEP 2: express the % of each element as grams.
STEP 3: follow the chart below.
ELEMENT
MASS
MOLES
(raw ratio)
REDUCE
(divide by smallest)
ROUNDED RATIO
(moles) C
65.5 g
5.45 mol
5.45/1.81=3.011 3 H
5.5 g
5.44 mol
5.44/1.81=3.001 O
29.0 g
1.81 mol
1.81/1.81=1.00 1