Empirical Formulas and Molecular Formulas

Empirical Formulas shows the smallest whole-number ratio of the atoms in the compound CH only means it is a 1:1 ratio between carbon and hydrogen, NOT the actual number of atoms in the compound could be C2H2 (ecetylene), or C6H6 (benzene), or C8H8 (styrene)

Determining Empirical Formula of a Compound What is the empirical formula of a compound that’s mass is 25.9% N and 74.1% O? % is not really a unit of measurement we can use in chemistry at times if there was 100g of this compound then 25.9g would be nitrogen and 74.1g oxygen change to moles in order to convert the % to a # 25.9g N x 1 mol N = 1.85 mol N 14.0 g N 74.1g O x 1 mol O = 4.63 mol O 16.0 g O

What is the whole number ratio? a formula is not just the ratio of atoms, it is also the ratio of moles this leaves a ratio of N1.85O4.63 which is not a a possible formula/whole number therefore, divide each by the lowest quantity (in this case, 1.85) now is N1O2.5 again, still not a whole number, but if you were to double it, it would be N2O5 the empirical formula is N2O5

Example #2 Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so… 38.67 g C x 1mol C = 3.220 mole C 12.01 g C 16.22 g H x 1mol H = 16.09 mole H 1.01 g H 45.11 g N x 1mol N = 3.219 mole N 14.01 g N

If we divide all of these by the smallest 3.220 mole C 16.09 mole H 3.219 mole N C1H5N1 is the empirical formula C3.22H16.09N3.219 If we divide all of these by the smallest number (3.22), it will give us whole numbers and therefore the empirical formula

Example #3 (notice this is already in grams and not a percentage) A compound is found to contain 16g of C and 4g of H. What is the empirical formula? 16 g C x 1mol C = 1.33 mole C 12.01 g C 4 g H x 1mol H = 4 mole H 1.01 g H 1.33 is too far off from 1 divide both by 1.33 1.33/1.33 = 1 4/1.33 = 3 empirical formula is CH3

Comparing Empirical and Molecular Formulas   Comparing Empirical and Molecular Formulas Compound Molecular Formula Empirical Formula Water H2O Hydrogen Peroxide H2O2 HO Glucose C6H12O6 CH2O Methane CH4 Ethane C2H6 CH3 Octane C8H18 C4H9  

Molecular Formulas a molecular formula is the same as, or a multiple of, the empirical formula it is based on the actual number of atoms of each type in the compound, NOT just the ratio the following all have same empirical formula (CH2O at a 1-2-1 ratio) but different molecular formulas C2H4O2 is the molecular formula of ethanoic acid CH2O is the molecular formula of methanol C6H12O6 is the molecular formula of glucose

Determine the molecular formula from the empirical formula and the experimental molar mass divide the experimental molar mass by the mass of one mole of the empirical formula this results in the multiplier to convert to the molecular formula (this sounds confusing but it really isn’t) ex: calculate the molecular formula of a compound whose molar mass is found to be 60.0g and has an empirical formula of CH4N experimental molar mass of ? --- 60.0 g empirical molar mass of CH4N ---- 30.0 g 2 (CH4N) = C2H8N2 = 2

need to calculate the empirical formula first assume 100 g so… Determining the empirical formula and then molecular formula of a compound A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known to be 98.96 g. What is its molecular formula? need to calculate the empirical formula first assume 100 g so… 71.65 g Cl x 1mol Cl = 2.02 mole Cl 35.5 g Cl 24.27 g C x 1mol C = 2.02 mole C 12.0 g C 4.07 g H x 1mol H = 4.07 mole H 1.0 g H

Cl2.02C2.02H4.07 divide by lowest (2.02 mol ) ClCH2 is the empirical formula if ClCH2 was the molecular formula, this would give a mass of 48.5 g however, recall the problem asked for the molecular formula of a compound with a molar mass of 98.96 g which is twice that of 48.5 g therefore, Cl2C2H4 is the molecular formula