1. Empirical and Molecular Formulas

2. Empirical Formula Molecular Formula Empirical Formula H2O2 HO C6H12O6
A formula that gives the simplest whole-number ratio of the atoms of each element in a compound.
Molecular Formula
Empirical Formula
H2O2 HO
C6H12O6
CH2O
CH3O
CH3O
CH3OOCH = C2H4O2
CH2O

3. 2. Divide each mole by the smallest mole.
Determine the empirical formula for a compound containing g Cl and g Ca.
Steps
1. Find mole amounts.
2. Divide each mole by the smallest mole.

4. 1. Find mole amounts.
2.128 g Cl x 1 mol Cl = mol Cl
35.45 g Cl
1.203 g Ca x 1 mol Ca = mol Ca
40.08 g Ca

5. 2. Divide each mole by the smallest mole.
Cl = mol Cl = 2.00 mol Cl
0.0300
Ca = mol Ca = 1.00 mol Ca
Ratio – 1 Ca: 2 Cl
Empirical Formula = CaCl2

6. A compound weighing 298. 12 g consists of 72. 2% magnesium and 27
A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?
Hint
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”

7. A compound weighing 298. 12 g consists of 72. 2% magnesium and 27
A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?
Percent to mass: Mg – (72.2%/100)* g = g
N – (27.8%/100)* g = g
Mass to mole: Mg – g * ( 1 mole ) = 8.86 mole
24.3 g
N – g * ( 1 mole ) = 5.92 mole
14.01 g
Divide by small: Mg mole/5.92 mole = 1.50
N mole/5.92 mole = 1.00 mole
Multiply ‘til whole: Mg – 1.50 x 2 = 3.00
Mg3N2
N – 1.00 x 2 = 2.00

8. Molecular Formula
The molecular formula gives the actual number of atoms of each element in a molecular compound.
Steps
1. Find the empirical formula.
2. Calculate the Empirical Formula Mass.
3. Divide the molar mass by the “EFM”.
4. Multiply empirical formula by factor.
Find the molecular formula for a compound whose molar mass is ~ and empirical formula is CH2O3.
2. “EFM” = g
/62.03 = 2
4. 2(CH2O3) = C2H4O6

9. 1. Find the empirical formula.
Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.
Steps
1. Find the empirical formula.
2. Calculate the Empirical Formula Mass.
3. Divide the molar mass by the “EFM”.
4. Multiply empirical formula by factor.

10. Empirical formula. A. Find mole amounts.
4.90 g N x 1 mol N = mol N
14.01 g N
11.2 g O x 1 mol O = mol O
16.00 g O

11. B. Divide each mole by the smallest mole.
N = = 1.00 mol N
0.350
O = = 2.00 mol O
Empirical Formula = NO2
Empirical Formula Mass = g/mol

12. Molecular formula Molar Mass = 92.0 g/mol = 2.00
Emp. Formula Mass g/mol
Molecular Formula = 2 x Emp. Formula = N2O4

13. A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula?

14. A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula?
g C – (48.38/100)* g = g
g H – (8.12/100)* g = g
g O – (43.5/100)* g = g
mole C g * ( 1 mole ) = mol
12.01 g
mole H – g * ( 1 mole ) = mol
1.01 g
mole O – g * ( 1 mole ) = mol
16.00 g

15. From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O
A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula?
From last slide: mol C, mol H, mol O
C – 21.29/14.27 = 1.49
H – 42.49/14.27 = 2.98 (esentially 3)
O – 14.27/14.27 = 1.00
C – 1.49 x 2 = 3
H – 3 x 2 = 6
O – 1 x 2 = 2
C3H6O2

16. From last slide: Empirical formula = C3H6O2
A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula?
From last slide: Empirical formula = C3H6O2
“EFM” = 74.09
Molar mass = = ~3
EFM
3(C3H6O2) = C9H18O6