2.  shows the smallest whole-number ratio of the atoms in the compound  CH only means it is a 1:1 ration between carbon and hydrogen  could be C 2 H 2 (ecetylene), or C 6 H 6 (benzene), or C 8 H 8 (styrene)

3.  What is the empirical formula of a compound that’s mass is 25.9% N and 74.1% O?  if there were 100g of this compound then 25.9g would be nitrogen and 74.1g oxygen  change to moles in order to convert the % to a number  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

4.  a formula is not just the ratio of atoms, it is also the ratio of moles  this leaves a ratio of N 1.85 O 4.63 which is not a whole number  therefore, divide each by the lowest quantity (in this case, 1.85)  now is N 1 O 2.5  again, still not a whole number, but if you were to double it, it would be N 2 O 5  the empirical formula is N 2 O 5

5.  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

6.  3.220 mole C  16.09 mole H  3.219 mole N  C 1 H 5 N 1 is the empirical formula C 3.22 H 16.09 N 3.219 If we divide all of these by the smallest number (3.22), it will give us whole numbers and therefore the empirical formula

7. CompoundMolecular FormulaEmpirical Formula WaterH2OH2OH2OH2O Hydrogen PeroxideH2O2H2O2 HO GlucoseC 6 H 12 O 6 CH 2 O MethaneCH 4 EthaneC2H6C2H6 CH 3 OctaneC 8 H 18 C4H9 C4H9

8.  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  the following have the same ratio of elements and therefore the same empirical formula– CH 2 O, however, many molecules can have that ratio, each having a different molecular formula  C 2 H 4 O 2 is the molecular formula of ethanoic acid  CH 2 O is the molecular formula of methanal  C 6 H 12 O 6 is the molecular formula of glucose

9.  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  calculate the molecular formula of a compound whose molar mass is found to be 60.0g and has an empirical formula of CH 4 N  experimental molar mass --- 60.0 g empirical molar mass --------- 30.0 g  2 (CH 4 N) = C 2 H 8 N 2 = 2

10.  A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known 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

11.  Cl 2.02 C 2.02 H 4.07  divide by lowest (2.02 mol )  Cl 1 C 1 H 2 is the empirical formula  this would give an empirical formula mass of 48.5 g  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, Cl 2 C 2 H 4 is the molecular formula