1. Finding the Empirical Formula of a compound

2. “The Ultimate Dog Tease” isn’t near as entertaining as Empirical formula! Trust me!!

3. The empirical formula for a compound can only be found through experimentation where repeated synthesis/ decomposition reactions are performed.

4. The empirical formula represents the smallest whole number ratio of atoms of each element in a molecule of a compound
e.g. The chemical formula for hydrogen peroxide is H2O2 but its empirical formula is HO. Similarly for glucose, whose empirical formula is CH2O while its chemical formula is C6H12O6.

5. Therefore, the empirical formula is H2O.
If you know the percent mass composition of a compound after it has been decomposed (of each part it broke into), then you can find its empirical formula using:
Element
%Mass
Mass in
100g
Moles
Mole Ratio H
11.2
11.2g O
88.8
88.8g
Therefore, the empirical formula is H2O.

6. How’d we do that??
Write down the symbol for each element in the first column then the % mass composition for each element. e.g. H=11.2% and O=88.8%
Express each % as a mass out of 100 g.
Express each as a quantity in moles (Mass Moles)
Express each as a mole ratio by comparing each to the element with the least number of moles.
Write the empirical formula using the mole ratios as subscripts for their respective elements.

7. So really…

8. Therefore, the empirical formula starts as NO2.5 but becomes N2O5.
Find the empirical formula for a compound where: N=25.9% and O=74.1% by mass. (What do you do with mole ratios that are not whole numbers?)
Element
%Mass
Mass in
100g
Moles
Mole Ratio N
25.9
25.9g O
74.1
74.1g
Therefore, the empirical formula starts as NO but becomes N2O5.
*Notice atoms/ subscripts in a formula must be whole numbers!

9. Since the data collected for these questions are very accurate, then fractional mole ratios can be changed to whole numbers by multiplying by the appropriate whole number. e.g x 2 = x 4 = x 5= x 3= 7.0
Note: All values in the empirical formula must be multiplied by the same whole number.

10. Sometimes a compound is decomposed and the masses of the various elements are found instead of % mass composition. e.g. When 120.0g of a compound containing lead and oxygen is decomposed, the mass of lead left is 103.9g. What is the empirical formula of the compound.
This problem is done the same way as above except the masses of the elements are used instead of the % mass composition.

11. The rest of the procedure is the same.
5 (cont).
The rest of the procedure is the same.
Element
Mass
Moles
Mole Ratio Pb
103.9g O
120.0g-103.9g
= 16.1g
Therefore, the empirical formula is PbO2.

12. Empirical formula is Ca5O11H11 or Ca5(OH)11
A compound was analyzed and found to contain 12.5g of Ca, 10.8g of O, and 0.675g of H. What is the Empirical Formula?
Element
Mass
Moles
Mole Ratio Ca
12.5g O
10.8g H
0.675g
2.2 x 5 = 11… so everything else is multiplied by 5 as well!
Empirical formula is Ca5O11H11 or Ca5(OH)11

13. Let’s Try some Practise!! Hebden Pg. 93 #46 (a,c,i,k,m)

14. Finding “Molecular” Formulas
The chemical or molecular formula can be found if the molecular or molar mass of the compound is known and either the % mass composition or the mass of each element involved in the synthesis or decomposition.
The molecular mass is the "true ratio" of the elements in a compound (ie C6H12O6), whereas the empirical is the lowest whole number ratio (ie CH2O)
Finding “Molecular” Formulas

15. Find the molecular formula for a compound whose molar mass is 26
Find the molecular formula for a compound whose molar mass is 26.0g and where the % mass composition is H=7.7% and C=92.3%.
Element
%Mass
Mass in
100g
Moles
Mole Ratio H
7.7
7.7g C
92.3
92.3g
So the empirical formula is CH.
But the empirical formula is the simplified version (it has a molar mass of 12.01g+1.01g= 13.01g) but the real molar mass is 26.0g… So to find the molecular mass…

16. 6. Since the ratio of molar mass to empirical molar mass is
. Then the empirical formula is multiplied by two (2) to give a Molecular formula of C2H2
Molecular Formula  True molar mass / empirical molar mass = # you multiply the E.F by!!

17. Try another… %Mass Moles Mole Ratio
Ribose is an important sugar that is found in DNA and RNA. Ribose has a molecular mass of 150g/mol, and a chemicla composition of 40.0% Carbon, 6.67% Hydrogen, and 53.3% Oxygen. What is the molecular formula of Ribose?
Element
%Mass
Mass in 100g
Moles
Mole Ratio C
40.0%
40.0g H
6.67%
6.67g O
53.3%
53.3g
Empirical formula is CH2O (molar mass = 30.0g/mol )

18. 6. Since the ratio of molar mass to empirical molar mass is
. Then the empirical formula is multiplied by five (5) to give a Molecular formula of C5H10O5

19. Let’s Try Some Practise !!! Hebden Pg. 95 # 49,51,54,556.
Let’s Try Some Practise !!!
Hebden Pg. 95 # 49,51,54,55