1. Aim # 7: How can we determine the empirical and molecular formulas of a compound?
H.W. # 7
Study pp
Ans. ques. p. 189 # 33, 34
p. 190 # 37,40,42
H.W. # 7a
Study pp
Ans. ques. p. 190 # 44,45,46,49,50

2. I Determining empirical formulas
A. Problem: Find the empirical formula of a compound, a sample of which consists of 11.2g of iron and 21.3g of chlorine.
1. Determine the number of moles of each element in the sample.
11.2g Fe x 1 mole Fe = mol Fe
55.9g Fe
21.3g Cl x 1 mole Cl = mol Cl g Cl

3. # of mole Fe atoms = 0.200 = 1 # of moles Cl atoms 0.600 3
2. From the number of moles, determine the smallest whole-number ratio of the atoms of each element in the compound.
# of mole Fe atoms = = 1
# of moles Cl atoms
3. Use the numbers from the whole-number ratio as the subscripts in the empirical formula.
1 atom Fe : 3 atoms Cl
FeCl3
Remember, the metal comes first!

4. 1. Assume that you have a 100-g sample of the compound.
B. Problem: Determine the empirical formula of the compound with a percentage composition of % carbon and 72.7% oxygen.
1. Assume that you have a 100-g sample of the compound.
27.3% of 100g = 27.3g C
72.7% of 100g = 72.7g O
The problem may now be solved as before.
g C x 1 mol C = mol C g C
72.7g O x 1 mol O = mol O
16.0g O

5. 3. # of moles C atoms = 2.28 = 1 ≈ 1 # of moles O atoms 4.54 1.99 2
1 atom C : 2 atoms O
CO2
Problem: Find the empirical formula of the compound that consists of 75.0% carbon and 25.0% hydrogen.

6. 75. 0 g C x 1 mol C = 6. 25 mol C 12. 0g C 25. 0 g H x 1 mol H = 25
75.0 g C x 1 mol C = mol C g C g H x 1 mol H = mol H g H
# moles C = = 1
# moles H
Formula: CH4

7. II To determine the molecular formula of a compound,
Derive the empirical formula.
Calculate gram-molecular mass gram-formula mass of empirical formula
Multiply the subscripts of the empirical formula by the number calculated in step 2, above.

8. the gram-formula mass of CH2 is 14g.
e.g. If the empirical formula of the compound is CH2 and the gram molecular mass is 56g,
the gram-formula mass of CH2 is 14g.
The ratio is 56g = 4 = g
The molecular formula is 4(CH2)
C4H8

9. Problem: A compound of phosphorous and oxygen contains 56. 4% oxygen
Problem: A compound of phosphorous and oxygen contains 56.4% oxygen. If the gram-molecular mass is 284g, what is the molecular formula of the compound?
Ans. A 100-g sample of the compound contains 56.4g O
and 43.6g P.
43.6g P x 1 mol P = mol P g P g O x 1 mol O = mol O g O
mol P = = = empirical formula: P2O5
gram-formula mass: 2(31.0g) + 5(16.0g) =
142.0g

10. Therefore, the molecular formula is 2(P2O5) = P4O10
284g = g
Therefore, the molecular formula is
2(P2O5) = P4O10
Problem: Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol.
a) What is the empirical formula of ethylene glycol?
b) What is its molecular formula?

11. Problem
Determine the empirical and molecular formulas of nicotine, a component of tobacco: 74.1% C, 8.6% H, and 17.3% N by mass. Nicotine’s molar mass is 160 ± 5g.