7.3 Molecular Formulas 3/8/2016

Empirical Vs. Molecular Formulas A molecular formula tells exactly how many of each atom are in the molecule. C O H An empirical formula tells the ratio of atoms in a molecule C H O empirical formula of sugar = CH2O molecular formula of sugar = C6H12O6

A molecular formula is a whole-number (n) multiple of the empirical formula. Compound Empirical Formula Molecular formula Molar Mass (grams/mole) Formaldehyde CH2O 30.03 Acetic acid C2H4O2 n = 2 60.06 Glucose C6H12O6 n = 6 180.18

Example: The empirical formula for a compound is CH2O Example: The empirical formula for a compound is CH2O. Its experimental molar mass is 120.12 g/mol. Determine the molecular formula of the compound. 1st Find the molar mass of the empirical formula CH2O. 1 x 12.01 g = 12.01 g/mol 2 x 1.01 g = 2.02 g/mol + 1 x 16.00 g = 16.00 g/mol molar mass of CH2O = 30.03 g/mol

2nd Solve for n, the factor multiplying the empirical formula n = experimental molar mass of compound molar mass of empirical formula n = 120.12 g/ mol = 4 30.03 g/mol Molecular Formula = n(E.F) 4(CH2O) = C4H8O4

Example: Isobutylene is a raw material for making synthetic rubber Example: Isobutylene is a raw material for making synthetic rubber. It contains 85.7% carbon and 14.3% hydrogen by mass. The experimental molar mass is 56.12 g/mol. What are the empirical and molecular formulas of isobutylene?

85.7 g C / 12.01 g C = 14.3 g H / 1.01 g H = 7.13 mol C 14.2 mol H Find E.F first…. 85.7 g C / 12.01 g C = 14.3 g H / 1.01 g H = 7.13 mol C 14.2 mol H C 7.13 H 14.2 7.13 7.13 CH2

Find n for M.F 1 x 12.01 g C = 2 x 1.01 g H = 12.01 g C 2.02 g H 14.03 g/mol CH2 n= 56.12 g/ mol = 4 14.03 g/ mol 4(CH2)  C4H8

Empirical formula = CH2 Molecular formula = C4H8