1. 5.5 Molecular Formula Percent Composition

2. Review of Empirical Formula

3. A compound is found to contain 77.9% I, 22.1 % O. Calculate the empirical formula. = = = mol 77.9 g I x 1 mole g = mol 22.1 g O x 1 mole g

5. Molecular Formula Actual number of atoms of each element in a molecule. Compare it to empirical formula, which is the simplest whole number ratio of atoms in a molecule.

6. Empirical FormulaMolar MassMolecular FormulaMolecular Mass C 4 H 8 NO86.08 g/molC 12 H 24 N 3 O 3 258.24 g/mol 192.24 g/mol C 14 H 24 96.12 g/molC 7 H 12 360.4 g/mol C 20 H 40 O 5 72.08 g/molC4H8OC4H8O 78.06 g/mol C6H6C6H6 13.01 g/molCH 192.24 96.12 = 2 360.4 72.08 = 5 78.06 13.01 = 6

7. A compound is found to contain 30.8 % C, 4.27 % H, 23.9 % N, and 41.0 % O. The molecular mass is 468.0 g/mole. Determine the empirical and molecular formulas. = 2 = 1.504 = 5 117.0 g/mole468.0 g/moleC3H5N2O3C3H5N2O3 C 12 H 20 N 8 O 12 30.8 g C x 1 mol= 2.567 mol 12.0 g 4.27 g H x 1 mol= 4.227 mol 1.01 g 23.9 g N x 1 mol= 1.707 mol 14.0 g 41.0 g O x 1 mol= 2.563 mol 16.0 g 1.707 mol = 2.476 = 1 = 1.501 x 4 = x 2 = = 3= 3 = 3= 3

8. How to find the molecular formula 1- Find the empirical formula as you always do. 2- Calculate the molar mass of the empirical formula you got. 3- Divide the molecular mass by the molar mass: Molecular mass Molar mass 4- Multiply the subscripts in the EF by this number in step(3) This new formula will be your molecular formula!

9. TRY:A compound is found to contain 66.35 g C, 17.54 g H, 33.17 g N, and 63.22 g O. The molecular mass is 456.44 g/mole. Determine the empirical and molecular formulas. Hint: use 1.01 g/mol for H. = 3 = 2.334 = 22 228.22 g/mole 456.44 g/mole C 7 H 22 N 3 O 5 C 14 H 44 N 6 O 10 66.35 g C x 1 mol= 5.529 mol 12.0 g 17.54 g H x 1 mol= 17.37 mol 1.01 g 33.17 g N x 1 mol= 2.369 mol 14.0 g 63.22 g O x 1 mol= 3.9513 mol 16.0 g 2.369 mol = 7.332 = 1 = 1.668 x 2 = x 3 = = 7= 7 = 5= 5

10. Another type of questions You are not given the molecular mass. What to do?! Usually these questions are only about gases at STP.

11. A gas has the empirical formula CH 2. If 0.85 L of the gas at STP has a mass of 1.59 g, what is the molecular formula?

12. What did we just learn? We use 22.4 L/mol to help us find the molecular mass –For gases only –At STP We use the density (g/L) to find molecular mass (g/mole)

13. Homework Try in this order! Page 95 #54, 50, 52.