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.

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 C2H2 (ecetylene), or C6H6

(benzene), or C8H8 (styrene)

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

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 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

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

3.220 mole C 16.09 mole H 3.219 mole N

C1H5N1 is the empirical formula

• C3.22H16.09N3.219If we divide all of these by the smallest

number (3.22), it will give us whole numbers and therefore the empirical formula

Compound Molecular Formula Empirical Formula

Water H2O H2O

Hydrogen Peroxide H2O2 HO

Glucose C6H12O6 CH2O

Methane CH4 CH4

Ethane C2H6 CH3

Octane C8H18

C4H9

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– CH2O, however, many molecules can have that ratio, each having a different molecular formula

C2H4O2 is the molecular formula of ethanoic acid CH2O is the molecular formula of methanal C6H12O6 is the molecular formula of glucose

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 CH4N experimental molar mass --- 60.0 g

empirical molar mass --------- 30.0 g 2 (CH4N) = C2H8N2

= 2

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

Cl2.02C2.02H4.07

divide by lowest (2.02 mol ) Cl1C1H2 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, Cl2C2H4 is the molecular formula