1. Empirical Formulas  Empirical formula – gives the lowest whole-number ratio of the atoms (or moles of atoms) of the elements in a compound

2. The empirical formula may or may not be the same as the molecular formula  Ex. Hydrogen peroxide  Molecular formula: H 2 O 2  Empirical formula: HO  Molecular and Empirical formulas are different  Ex. Carbon dioxide  Molecular formula: CO 2  Empirical formula: CO 2  Molecular and Empirical formulas are the same.

3.  Dinitrogen Tetroxide  Molecular formula: N 2 O 4  Empirical formula: NO 2  Ex. C 2 H 2 (Ethyne) and C 8 H 8 (Styrene)  Both have Empirical formula of CH

4. What is the empirical formula for a compound that is 25.9% N and 74.1% O?  Known:  25.9%N(same thing as saying 25.9g N/100.0g compound)  74.1%O (same as saying 74.1g O/100.0 g compound)  Empirical formula = N ? O ?  Step #1: To make this easy on ourselves: assume we have 100.0g of the compound. That means the mass is the same value as the percentage.  This would give us 25.9 g N and 74.1 g O  Step #2: Convert these mass values to moles  25.9 g N (1.00 mol N / 14.01 g N) = 1.85 mol N  74.1 g O (1.00 mol O / 16.00 g O) = 4.63 mol O

5.  This would give us a mole ratio of Nitrogen to Oxygen of: N 1.85 O 4.63  This is not an empirical formula (not the lowest whole-number ratio  Step #3: Divide both molar quantities by the smaller # of moles  1.85 mol N / 1.85 = 1 mol N  4.63 mol O / 1.85 = 2.50 mol O  This would give us NO 2.50  We’re still not there…

6.  Step #4, needed sometimes: We need to multiply each part of this ratio by the lowest number that would convert this fraction into a whole number.  In this case, the lowest # is 2  1 mol N x 2 = 2 mol N  2.5 mol O x 2 = 5 mol O  Lets’ try this again…The new formula is:  N 2 O 5  Now that looks like an empirical formula!

7. Review the steps:  To find the empirical formula, we:  Converted the known % of each element in our compound into grams by assuming 100.0 g of the compound  Converted the amount of each element from grams to moles  We did not have a whole # ratio, so we divided the # of moles of each compound by the smallest of the # of moles to get a whole number  Since we still did not have a whole # ratio, we had to multiply each part of the ratio by the lowest whole number that would give us a whole # ratio.

8. Calculating Molecular Formulas  Different compounds can have the same empirical formulas, but the will have different molecular formulas, and molar masses  Ex. Molecular Formulas: Ethyne (C 2 H 2 ), Benzene (C 6 H 6 )  Empirical formula CH  Molar Mass of CH = (12.0gC) + (1.0gH) = 13.0g CH This is the Empirical Formula Mass (efm)  Molar Mass of: C 2 H 2 = 2(12.0gC) + 2(1.0gH) = 26.0 g C 2 H 2 Same as (efm) x 2 13.0 g x 2 = 26.0 g C 6 H 6 = 6(12.0gC) + 6(1.0gH) = 78.0 g C 6 H 6 Same as (efm) x 6 13.0 g x 6 = 78.0 g

9.  The molar mass of a compound is a simple whole number multiple of the empirical formula mass (efm)  To find the molecular formula from the empirical formula and molar mass: 1. Calculate the efm from the empirical formula 2. Divide the known molar mass from the efm 3. Multiply your empirical formula by this value to give you your molecular formula

10. Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH 4 N 1. Calculate the efm from the empirical formula Empirical Formula: CH 4 N efm = 1(12.0gC) + 4(1.0gH) + 1(14.0gN) = 30.0g 2. Divide the known molar mass from the efm 60.0 g / 30.0g = 2 3. Multiply your empirical formula by this value to give you your molecular formula Molecular Formula: (CH 4 N) 2 = C 2 H 8 N 2