This is the answer key to the problems from the powerpoint slides on empirical
and molecular formulas
Example 1:
A sample of ethylene shows that it contains 85.7g of carbon and 14.3 g of hydrogen. Determine
the empirical formula of ethylene.
Element
C
H
Mass (g)
85.7
14.3
Mole
7.14
14.3
Molar Ratio
1
2.00
Formula Ratio
1
2
Empirical Formula of ethylene: CH2
Example 2:
A sample of a pure compound is found to be made up of 1.61 g of phosphorus and 2.98 g of
fluorine. Find the empirical formula of the compound.
Element
P
F
Mass (g)
1.61
2.98
Mole
.052
.157
Molar Ratio
1
3.02
Formula Ratio
1
3
Empirical Formula of compound: PF3
Example 3:
An unknown compound is found in the lab to contain 91.8 g of Si and 8.2 g of H. Determine the
empirical formula of the unknown compound.
Element
Mass
Mole
Molar Ratio
Formula Ratio
Si
91.8
3.28
1
2
H
8.2
8.2
2.5
5
 Because the molar ratio is not a whole number, we need to multiply both the molar
ratio by a number that makes BOTH numbers whole. Do this with the simplest number
possible. In this case, we can multiply the molar ratio by 2. This leaves us with a
formula ratio of 2:5
 Empirical Formula of unknown compound: Si2H5
Example 3 continued:
Another experiment indicates that the molar mass of the unknown compound is 122 g/mol.
Determine the molecular formula of the compound.
In order to determine the molecular formula, you must do the following:
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
Therefore,
122 𝑔/𝑚𝑜𝑙
61 𝑔/𝑚𝑜𝑙
=2
This number “2” tells us how many empirical formula units can go into the molecular formula.
Therefore, multiply all the subscripts in the empirical formula by 2 in order to get the molecular
formula.
(Si2H5) x 2 = Si4H10 = molecular formula of the unknown compound
Example 4:
A compound contains 52.2% C, 13.0% H, and 34.8% O. Calculate the empirical formula of the
compound.

If we assume, which we can, that the total mass of the compound is 100 g, then it makes
sense that we are able to make the claim that, in this 100g sample, there are 52.2 g C,
13 g H, and 34.8 g O. We will use these values for our masses.
Element
Mass
Mole
Molar Ratio
Formula Ratio
C
52.2
4.35
1.99
2
H
13
13
5.96
6
O
34.8
2.18
1
1
 When presented with 3 or more elements, the molar ratio is determined by dividing all
the moles of each element by the smallest amount of moles (in this case, oxygen since it
only contains 2.18 moles). Therefore, the molar ratio is all RELATIVE to oxygen since it is
the least amount.

Empirical Formula of compound: C2H6O
The empirical formula cannot be reduced any further. You must remember that there is
technically a subscript of “1” after the oxygen and that “1” cannot be reduced any
further.