1. Empirical Formula & Molecular Formula

2. Empirical Formulas
Best way to identify an unknown compound is to determine its chemical formula.
Empirical Formula: the smallest whole number ratio of a compound. (Ionic compounds are already in its empirical formulas)

3. Empirical Vs. Molecular Formulas
Empirical Formula
a formula that shows the simplest whole-number ratio of elements in compounds
Molecular Formula
a formula that shows the element symbols & exact number of each type of atom in a molecular compound

4. can be reduced to lowest ratio can be reduced to lowest ratio
CH4
Vs.
C6H24
Vs.
C18H72
Lowest ratio
can be reduced to lowest ratio
can be reduced to lowest ratio
In empirical & molecular formula
18 can go into 18 & 72--> CH4
6 can go into 6 & 24--> CH4
same empirical formula. But not the same molecular formula

5. M= 12.01g/mol + (4 x 1.01g/mol) = 16.00 g/mol
If you know the chemical formula you can find percent composition.
CH4
C6H24
M= 12.01g/mol + (4 x 1.01g/mol) = g/mol
MC6H24= 96 g/mol
MC= 72 g/mol
MC= 12.01g/mol
MH= 1.01g/mol
(72 g/mol ) x
1mol x
100%
= 75%
96.00g
(12.00g/mol ) x
1mol x
100%
= 75%
CH4
&
C6H24
16.00g
- same empirical formula
- should be ratios of each other
-helps determine percent composition
(4.01g/mol) x
1mol x
100%
= 25%
16.00g

6. Determining Empirical Formula
Find the empirical formula of a compound with percentage composition 35.4% Na and the remainder nitrogen
Step 1: Assume your conveniently working with a mass of g.
Na= 35.4 g
N= 64.6 g
100.0g g = 64.6 g
Step 2: Calculate the amount of each element.
nNa = (35.4 g) (1 mol)
nN = (64.5 g) (1 mol)
22.99g
14.01g
= mol
= mol

7. Determining Empirical Formula
Find the empirical formula of a compound with percentage composition 35.4% Na and the remainder nitrogen
Step 3: To determine the simplest ratio of the elements in the compound, divide the amount of each element by the smallest amount
nNa nN
mol
1.540 mol = =
nNa
nNa
1.540 mol
1.540 mol
= 1
= 2.98

8. Determining Empirical Formula
Find the empirical formula of a compound with percentage composition 35.4% Na and the remainder nitrogen
Step 2: Calculate the amount of each element.
nNa
nNa
1.540 mol
mol = =
nNa nN
1.540 mol
1.540 mol
= 1
= 2.98
The ratio for Na & N is1:2.98. Since we cannot have a fraction of an element in a compound, the value of N is rounded off to the nearest whole number. As a result, the simplest whole-number ratio is 1: NaN3

9. If a number is within 0.05 of a whole number, you can round it up or down to the nearest whole number.
But what do you do if one of the values is not close to a whole number? You multiply all the subscripts by the fraction denominator gives whole numbers
C1 x 3 H1.333 x 3 = C3H4

10. Practice nc (52.2 g) (1 mol) = = 4.3464 mol 12.01g nH (6.15g) (1 mol)
Determine the empirical formula of a compound listed that contains 52.2% carbon, 6.15% hydrogen and 41.7% oxygen
Step 1: Assume your conveniently working with a mass of g. nc
(52.2 g) (1 mol) =
= mol
12.01g nH
(6.15g) (1 mol) =
= mol
1.01g n0
(41.7g) (1 mol) =
= mol
16.00g

11. Practice nc 4.3464 mol = = 1.67 n0 2.6063mol 6.0891mol nH = = 2.29 n0
Determine the empirical formula of a compound listed that contains 52.2% carbon, 6.15% hydrogen and 41.7% oxygen
Step 2: Divide the amount of each element by the smallest amount. nc
mol =
= 1.67 n0
2.6063mol
Step 3: These calculations give an empirical formula of C1.67H2.29O1
Multiplying each of the subscripts by 3 give C5H7O3
6.0891mol nH =
= 2.29 n0
2.6063mol n0
2.6063mol =
= 1 n0
2.6063mol

12. Practice
Write the empirical formula of compounds with the following percentage
20.2% Al; 79.8% Cl

13. Practice
Write the empirical formula of compounds with the following percentage
50.85% carbon; 8.47% hydrogen; and 40.68% oxygen

14. Molecular Formula
gives the exact number of atoms of each element present
“chemical formula” = molecular formula
A compound’s molar mass is always a whole number multiple of the molar mass of the empirical formula. You ca find this multiple x, by dividing the molar mass of the compound by the molar mass of the empirical formula

15. Molecular Formula
x = molar mass of compound
molar mass of empirical

16. Find a Molecular Formula Given its Empirical formula
Compound with empirical formula CH2 & the molar mass of g/mol
Step 1: Find the empirical molar mass by adding the molar mass of each element
MCH2 = 1( g/mol) + 1(1.01g/mol)
= g/mol

17. Find a Molecular Formula Given its Empirical formula
Compound with empirical formula CH2 & the molar mass of 84.18g/mol
Step 2: Solve for x,
x = g/mol
14.03 g/mol
x = 6

18. Find a Molecular Formula Given its Empirical formula
Compound with empirical formula CH2 & the molar mass of 84.18g/mol
Step 3: therefore, the molar mass of the compound is 6 times the molar mass of the empirical formula
CH2= C6H12
molecular formula

19. Practice
Determine the molecular formulas
KSO4; g/mol

20. Using Percentage Composition & Molar Mass Data
Determine the molecular formula of vitamin C (ascorbic acid). This compound contains 40.5% Carbon, 4.6% hydrogen, 54.5% oxygen. Its molar mass is g/mol