1. Percent Composition and Molecular Vs. Empirical Formulas
Videodisk Unit 4 1 1

2. Percent Composition (always by mass)
Divide mass of each element by total compound mass and multiply by 100
Can be found experimentally or from the formula itself
Can be used to find the empirical formula of a compound 2

3. Sample Problem
Glucose has a molecular formula of C6H12O6. What is the percent composition of each element in this compound?
Whole:
C6H12O6 = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = g/mol
(Part/Whole) x 100:
C: 6(12.01 g/mol) x 100 = % C
g/mol
H: 12(1.008 g/mol) x 100 = % H
O: 6(16.00 g/mol) x = % O 3

4. Sample Problem C: 0.09480 g/0.2370 g x 100 = 40.00 %
A g sample of an unknown compound is extracted from the roots of a plant. Decomposition of the sample produces g of carbon, g of oxygen, and g of hydrogen. What is the percent composition of the compound?
C: g/ g x 100 = %
O: g/ g x 100 = %
H: / g x 100 = 6.67 % 4

5. Empirical & Molecular Formulas
Empirical formula is the simplest whole number ratio of a compound.
Molecular formula shows the actual number of atoms in the molecule.
The molecular formula is some multiple of the empirical formula.
Ex. C6H12O6 is glucose’s molecular formula
CH2O is glucose’s empirical formula 5

6. Empirical & Molecular Formulas
C6H12O6 is glucose’s molecular formula
It means in one mole of glucose you have 6 moles of Carbon, 12 moles of hydrogen, and 6 moles of oxygen
CH2O is glucose’s empirical formula
It means glucose has a ratio of
1 mole C: 2 moles H: 1 mole O 6

7. What is the empirical formula of:
Molecular formula : C25H10O5
Empirical Formula: C5H2O
Molecular Formula: N2O6
Empirical Formula: NO3 7

8. Steps for Determining Empirical Formulas from Percents or Masses
Example: a compound is 27.3% C and 72.7% O
1. Turn percent to mass (if needed). Assume 100 g if a mass isn’t given
Assuming 100g, 27.3g C and 72.7g O
2. Turn mass into moles
27.3g C (1mole/12.01g) = 2.27 moles C
72.7g O (1 mole/16.00g) = 4.54 moles O 8

9. Steps for Determining Empirical Formulas from Percents or Masses
3. Write a formula using the number of moles as subscripts.
C2.27O4.54
4. Divide both by the smallest number of moles (remember the empirical formula is a ratio)
C2.27/2.27O4.54/2.27
C1O2 or CO2 9

10. Steps for Determining Empirical Formulas from Percents or Masses
5. If needed, multiply until all values are whole numbers
Often you will be close to whole numbers, rarely will it be exact
Example: If after dividing you had something like X1Z1.47 you would multiply everything by 2 , ending up with X2Z2.94, meaning your empirical formula would be X2Z3 10

11. Sample Problem A compound was analyzed and found to contain (by mass):
54.1% Ca, 43.2% O, and 2.7% H. What is the empirical formula for the compound?
Ca: 54.1 g Ca x 1 mol Ca = 1.35 mol Ca
40.08 g Ca
O: 43.2 g O x 1 mol O = 2.70 mol O
16.00 g O
H: 2.7 g H x 1 mol H = 2.7 mol H
1.008 g H 11

12. Divide each mole value by the smallest number of moles and write formula.
Ca: 1.35 mol Ca / 1.35 =1
O: 2.70 mol O / 1.35 = 2
H: 2.7 mol H / 1.35 = 2
CaO2H2 12

13. Determine the empirical formula for a compound containing 2.128 g Cl
Sample Problem
Determine the empirical formula for a compound containing g Cl
and g Ca.
1.203 g Ca x 1 mol/40.08 g = mol Ca
2.128 g Cl x 1 mol/35.45 g = mol Cl
Ca / Cl /
CaCl2 13

14. What is the empirical formula if a compound consists of:
29.1% Na 40.5% S 30.4% O
Assuming a 100 g sample:
29.1 g Na x 1 mol/22.99 g = 1.27 mol Na
40.5 g S x 1 mol/32.07 g = 1.26 mol S
30.4 g O x 1 mol/16.00g = 1.90 mol O
Na 1.27/1.26 S 1.26/1.26 O 1.90/1.26 = NaSO1.5
(NaSO1.5) x 2 = Na2S2O3 14

15. Determining Molecular Formulas
Find the empirical formula
Find molar mass of empirical formula
Divide molecular mass by empirical molar mass
Multiply empirical formula by answer.
You must be given molecular mass in the problem. 15

16. Sample Problem CH2O = 30.03 g 150.0 g / 30.03 g = 4.99500= 4.995
Ribose is an important sugar that is found in RNA. Ribose has a molecular mass of g and an empirical formula of CH2O.
CH2O = g
150.0 g / g = = 4.995
5 (CH2O) = C5H10O5 16

17. Sample Problem Empirical to Molecular
Find the molecular formula for a compound that contains 30.4% Nitrogen and 69.6% Oxygen. The molecular mass of the compound is 92.0 g/mol.
N: 30.4 g N x 1 mol N = 2.17 mol N
14.01 g N
O: 69.6 g O x 1 mol O = 4.34 mol O
16.00 g O
N 2.17/2.17 O 4.35/2.17 = NO2 (empirical formula) 17

18. Sample Problem Empirical to Molecular
NO2 = (16.00) = 46.01g/mol
92.0 g/mol/46.01 g/mol = 2
2 x (NO2) = N2O4 (molecular formula) 18