1. Empirical Formula and Molecular Formula

2. Empirical Formula  Empirical formula is the simplest whole number ratio of atoms in a formula  Empirical information is always based on lab data

3. Molecular Formula  Molecular formula is the actual ratio of the atoms that form a molecule of a substance.  Sometimes the empirical and molecular are the same but they do not have to be!

4. More molecular….  Example H 2 O for water empirical and molecular formula are the same.  H 2 : O 1 is the simplest whole number ratio of atoms in the formula and it is also how the molecule actually forms.  Example glucose C 6 H 12 O 6 for glucose the molecule is 6:12:6 This is not the simplest whole number ratio of atoms in the formula.

5. Molecular and Empirical Formulas  If glucose were analyzed in the laboratory its empirical ratio would be 1:2:1. Additional test would have to be run to determine the mass of the substance

6. Calculations Step one= List individual elements in the formula. Change the given information from grams or percent into moles Step two= Decide which mole value is the smallest on the list. Step three= Divide all mole values by that smallest value

7. more Calculations……  Step four= If a decimal of.5 or.3 appear in the ratio you obtained do not round up or off! Ratios with.5 must be multiplied by two. Ratios with.3 must be multiplied by three.

8. Formula of Hydrates  The three basic steps are the same for hydrates. But with hydrates……  The two parts being compared are:  Moles of anhydrous in the sample vs.  Moles of water in the sample  Elements are not listed separately!

9. Example Empirical /Molecular  An organic compound was found to contain 92.25 % carbon and 7.75 % hydrogen. If the molecular mass is 78.0 grams, please calculate both the empirical and molecular formulas for this substance.

10. Step 1  C 92.25g/12.01 g = 7.681 moles  H 7.75g/1.01 g= 7.67 moles  Divide through by the smaller value!  1 : 1 ratio C 1 H 1 is empirical formula  Now we must check for molecular formula!

11. Step 2  C 1 H 1 add up the empirical mass of this substance.  12.01+ 1.01g = 13.02grams  Divide the given molecular mass by the empirical mass.  78.0/ 13.02 = 6  Molecular formula is C 6 H 6

12. Final Answer  Empirical formula C 1 H 1  Molecular formula C 6 H 6

13. Modern Chemistry example p.246  Quantitative analysis shows that a compound contains 32.38% sodium, 22.65%sulfur, and 44.99% oxygen. Please find the empirical formula for this compound.

14. Hydrate Example  Strontium chloride hexahydrate  158.52g plus 108.12g = 266.64g  108.12/266.64 x 100 = 40.5 % H 2 O

15. Hydrate Calculation  CuSO 4 * ? H 2 O  Crucible&cover111.55g  Crucible& cov& hydrate131.53g  Crucible/cov& anhydrous124.32g

16. Hydrates  Subtract C&C from anhydrous (after heating) this will equal grams of CuSO4.  Change grams to moles.

17. Hydrates  If you subtract hydrate from anhydrous you will find the mass of the water removed from the sample.  Change grams water to moles.

18. Hydrates  Look at mole values.  Decide which mole value is smaller  Divide both values by the smaller to get a one to something ratio  Taa! Daa! you are done.

19. The End!  Go forth and calculate!