Answer:
An interesting way to approach this problem is to determine the oxide's molecular formula first, then backtrack and find its empirical formula.
So, you know that this oxide of phosphorus has a molar mass of
284 g mol
−
1
. This tells you that one mole of this oxide has a mass of
284 g
.
You also know that the oxide has a percent composition of
43.66
%
phosphorus and
56.34%
oxygen.
This means that every
100 g
of oxide will contain
43.66 g
→
phosphorus,
P
56.34 g
→
oxygen,
O
Pick a
284-g
sample of this oxide, which will be equivalent to one mole, and determine how many moles of each constituent element it contains. This will get you the compound's molecular formula.
So, you can say that this sample will contain
284
g oxide
⋅
43.66 g P
100
g oxide
=
123.994 g P
284
g oxide
⋅
56.34 g O
100
g oxide
=
160.006 g O
Use the molar masses of the two elements to convert the masses to moles
For P:
123.994
g
⋅
1 mole P
30.9738
g
=
4.003
≈
4 moles P
For O:
160.006
g
⋅
1 mole O
15.9994
g
=
10.001
≈
10 moles O
Since this is how many moles of phosphorus and oxygen you get in one mole of the oxide, it follows that its molecular formula will be
molecular formula
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
P
4
O
10
a
a
∣
∣
−−−−−−−−−−
→
phosphorus pentoxide
To find the empirical formula, all you have to do is rewrite the
4
:
10
mole ratio that exists between the two elements in its smallest whole number form.
In this case, you can say that
4
:
10
⇔
2
:
5
This means that the empirical formula will be
empirical formula
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
P
2
O
5
a
a
∣
∣
−−−−−−−−−
