1. More than one Type of Formula????

2. Empirical and Molecular Formulas  Believe it or not, there is more than one kind of chemical formula  Judging from the heading of this section, they would be empirical and molecular formulas.

3. Empirical and Molecular Formulas  Empirical formulas give the lowest whole number ratio of the atoms in a compound  Molecular formulas give the exact composition of one molecule of a compound.

4. Empirical and Molecular Formulas  Examples  HO H 2 O 2  CHC 2 H 2  CHC 6 H 6  NO NO  NO 2 N 2 O 4

5. Calculating Empirical Formulas Empirical formulas can be calculated using experimental data based on a ratio of the weights of the elements in the compound Empirical formulas can be calculated using experimental data based on a ratio of the weights of the elements in the compound Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.4333 g Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.4333 g Calculate the empirical formula of this compound. Calculate the empirical formula of this compound.

6. Calculating Empirical Formulas The mass of oxygen can be found by subtracting the mass of phosphorus from the mass of the compoundThe mass of oxygen can be found by subtracting the mass of phosphorus from the mass of the compound 10.150 g sample -4.333 g phosphorus 5.717 g oxygen 5.717 g oxygen

7. Calculating Empirical Formulas  Convert the mass of each element to moles of that element 4.333g P x 1 mol P/30.97 g P = 0.1431 mol P4.333g P x 1 mol P/30.97 g P = 0.1431 mol P 5.717 g O x 1 mol O/16.00 g O = 0.3573 mol O5.717 g O x 1 mol O/16.00 g O = 0.3573 mol O  Divide through each value by the smallest number of moles to get a 1 : 2.497 ratio 1 : 2.497 ratio

8. Calculating Empirical Formulas  Divide through each value by the smallest number of moles to get a 1 : 2.497 ratio 1 : 2.497 ratio  The number of oxygen atoms cannot be rounded off to a whole number  Recognize that this is a 1:2½ ratio  Multiply by the denominator 2  Formula for the compound is P 2 O 5

9. Calculating Empirical Formulas Empirical formulas can be calculated using percentage composition: Empirical formulas can be calculated using percentage composition: Given that a certain compound is 69.58% Ba, 6.090% C and 24.32% O Given that a certain compound is 69.58% Ba, 6.090% C and 24.32% O calculate the empirical formula of this compound. calculate the empirical formula of this compound.

10. Calculating Empirical Formulas  Assume that you have 100.00 g of the compound: Therefore: 69.58% Ba = 69.58 g Ba 6.090%C = 6.090 g C 24.32% O = 24.32 g O

11. Calculating Empirical Formulas Convert the mass of each element to moles of that element: Ba = 69.58 g Ba C = 6.090 g C O = 24.32 g O

12. Calculating Empirical Formulas  Divide through each value by the smallest number of moles to get a 1 : 1.001 : 2.999 ratio  This rounds of nicely to give the formula BaCO 3

13. Calculating Molecular Formulas  Just figure out what you would need to multiply the empirical formula by to get that molecular weight  OR  Add up the weights of A,B,C,D and figure out which one matches the molar mass given!