1. Empirical and Molecular Formulas How to find out what an unknown compound is

2. A chemist obtains a new product  What is the formula for the compound?  First step – determine constituent elements and their amounts  This info can be used to determine chemical formula

3. Formula of a compound represents relative numbers of atoms present  E.g. “CO 2 ” tells us that in a molecule of this compound there is 1 carbon atom to every 2 oxygen atoms  To determine the formula of a substance we need to count the atoms – we can do this by weighing

4. An example  You have a compound that you know contains only Carbon, Hydrogen and Oxygen.  You have a 0.2015 g sample  Analysis shows you have 0.0806 g C, 0.01353 g H, and 0.1074 g O  We can convert these masses to moles, and then moles to atoms using dimensional analysis…..

5. Determining # of moles of elements in unknown compound

6. Determining number of atoms of unknown compound

7. To summarize up to this point  We have.00671 moles of Carbon  We have.01342 moles of Hydrogen  We have.006713 moles of Oxygen  Amount of Carbon = Amount of Oxygen .01342/.006713 = 2, so we have twice as much Hydrogen as we have of Oxygen and as we have of Carbon  We have a ratio of 1:2:1

8. We know the ratio of elements  1 Carbon : 2 Hydrogens : 1 Oxygen  We can write this as CH 2 O  Is this the molecular formula? Maybe…but the molecule might also have 2 Carbons, 4 Hydrogens and 2 Oxygens, or it might have 16 Carbons, 32 Hydrogens and 16 Oxygens  We have found the EMPIRICAL FORMULA – a formula that represents the ratio of elements in a compound.  This is also called “simplest formula” since it is smallest whole-number ratio of elements in the compound

9. Empirical Formula Vs Molecular Formula  Empirical formula gives relative numbers of atoms e.g. CH 2 O  Molecular formula gives the actual numbers of atoms e.g. C 6 H 12 O 6  C 6 H 12 O 6 = (CH 2 O) 6

10. Find the Empirical Formula  Empirical Formula of Benzene = CH  Empirical Formula of Dioxin = C 6 H 2 Cl 2 O

11. How to calculate empirical formula  We have 0.2636 grams of nickel. We heat it in the presence of oxygen to produce 0.3354 grams of a nickel oxide. What is the formula of the compound we made?  First - what is the mass of oxygen that reacted with the copper?  Mass of copper oxide – Mass of copper = mass of oxygen  0.3354 g copper oxide – 0.2636 g copper = 0.0718 g oxygen  Next – Find number atoms involved…

12. Find number of atoms in the compound Mole quantities represent number of atoms We have same number of moles of oxygen as of nickel Empirical formula will be NiO

13. Another example  You have a metal oxide made by reacting 4.151g Al with 3.692 g O. What is the empirical formula?  Atomic mass Al = 26.98 g/mol  Atomic mass O = 16.00 g/mol  Need to know relative numbers of atoms, so need to convert grams to moles and then find the whole number ratio of atoms

14. Find whole number ratios Do this by dividing both numbers by the smallest of the two. This converts the smallest number to 1 This is not a whole number ratio – to get to a whole number, all we have to do is multiply by 2

15. Finally we have found our empirical formula  Al 2 O 3

16. Summary

17. If you have percent composition your can also find the empirical formula  You have a compound that is 27% Carbon and 73% Oxygen by mass. What is the empirical formula?  Assume you have 100 grams. That means you would have 27 g C and 73 g O  Convert grams to moles:  27 g C (1 mol/12 g C) = 2.25 mol C  73 g O (1 mol/16 g O) = 4.6 mol O  Divide by smallest # of moles:  4.6 /2.25 = 2.04 (O)  2.25/2.25 = 1 (C)  Empirical formula = CO 2

18. A poem to help  Percent to mass  Mass to mole  Divide by small  Multiply ‘til whole

19. Find empirical formula from percent composition  Nylon-6 is a compound that is 63.68 C, 12.38% N, 9.80% H and 14.4% O. Find the empirical formula C 6 H 11 NO

20. One more step to find molecular formula  To find the molecular formula of your unknown compound, you need another piece of information – the molar mass of the compound, in addition to the percent composition

21. You have the empirical formula – what is the molecular formula?  Empirical formula = P 2 O 5  Molar mass of compound (determined experimentally) is 238.88 g/mol  P = 30.97 g/mol ; O = 16.00 g/mol  Find empirical formula mass - the molar mass that the compound would have based on the empirical formula:  2(30.97g/mol) + 5(16.00 g/mol) = 141.94 g/mol

22. Finding molecular formula  The molar mass of the compound will be some multiple of the empirical mass – find the multiple by dividing molar mass by empirical mass:  (238.88 g/mol)/(141.94 g/mol) = 1.7  1.7 ≈ 2  Molar mass of unknown is about twice empirical mass. We will multiply empirical formula subscripts by 2  Molecular formula = P 4 O 10

23. One last problem

24. Homework  Read Ch 6, section 3 (pp. 196 – 208)  Do empirical formula homework sheets  Do molecular formula homework sheets  Monday – we will go over homework and talk about hydrates  Test Review  Test on Wednesday!!!!  We have 12 class meetings to go until semester final