1. Due: HW 7C, Lab Reports Today: Determining Chemical Formulas Empirical Formulas Molecular Formulas Friday: Magnesium Oxide Lab Bring your own goggles if you don’t want to wear the ones in the classroom set! Weds March 5, 2014

2. Multiplication Division Convert mass to moles Necessary skills:

3. A Molecular formula includes the symbol of elements and the number of atoms of each element in that molecule Empirical formula includes symbols of elements in compounds with subscripts that show the smallest possible whole numbers that describe the atomic ratio Calculation of Chemical Formulae

4. Ionic compounds – formula unit IS ALREADY the smallest whole- number ratio Molecular compounds (covalent bonds) – molecule is not always the smallest whole number ratio! Determining Chemical Formula

5. Molecular formula: CH 4 Empirical formula: CH 4 Methane

6. Molecular formula: C 2 H 6 Empirical formula: CH 3 Ethane

7. Molecular formula: H 2 O Empirical formula: H 2 O Water

8. Molecular formula: C 6 H 6 Empirical formula: CH Benzene

9. 1. Determine Mass Composition If given percentages: Use percentage composition to convert to a mass composition (assume 100 g sample so percent is equal to the mass in g) If given mass: Skip this step 2. Convert mass to moles for each element 3. Find the smallest whole-number ratio by dividing each number of moles by the smallest number 4. These whole numbers are the subscripts in your compound Calculation of Empirical Formula

10. A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula. 1. Use % composition to get to mass composition 32.38% Na = 32.38 g Na 22.65% S = 22.65 g S 44.99% O = 44.99 g O Example 1 – Step 1

11. A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula. 2. Convert mass to moles 32.38 g Na / (22.989 g/mol) = 1.408 mol Na 22.65 g S / (32.065 g/mol) = 0.706 mol S smallest 44.99 g O / (15.999 g/mol) = 2.812 mol O Example 1 – Step 2

12. A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula. 3. Find smallest ratio by diving by smallest number of moles Na: 1.408 mol / 0.706 = 1.99 = 2 S: 0.706 mol / 0.706 = 1 O: 2.812 mol / 0.706 = 3.98 = 4 4. Write empirical formula Empirical formula = Na 2 SO 4 Example 1 – Step 3 These numbers become the subscripts in the empirical formula.

13. A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula. 1. Use % composition to get to mass composition (you are already there!) 4.43 g P 5.72 g O Example 2 – Step 1

14. A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula. 2. Convert mass to moles 4.43 g P / (30.974 g/mol) = 0.143 mol P smallest 5.72 g O / (15.999 g/mol) = 0.358 mol O Example 2 – Step 2

15. A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula. 3. Find smallest ratio by diving by smallest number of moles P: 0.143 mol / 0.143 = 1 x 2 = 2 O: 0.358 mol O / 0.143 = 2.5 x 2 = 5 THEY MUST BE WHOLE NUBMERS! YOU CAN’T HAVE HALF AN ATOM! Multiply by 2 (or 3, if the decimal is.333 or.666) to get whole numbers Example 2 – Step 3

16. ROUNDING HELPS If the number ends in.98,.99,.01, or.02  Round to nearest whole number If the number ends in.50,.51,.52,.49,.48, etc  Round to.5 then multiply by 2 If the number ends in.33,.32,.34, etc  Round to.3 then multiply by 3 If the number ends in.66,.65,.64, etc  Round to.6 then multiply by 3

17. A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula. 4. Write empirical formula P = 2 O = 5 Empirical formula: P 2 O 5 Example 2 – Step 4 These numbers become the subscripts in the empirical formula.

18. If you know the empirical formula and the molecular mass of a compound, you can determine the molecular formula. There are four pieces of information. If you know three, you can solve for the fourth. 1. Empirical formula 2. Molecular formula 3. Empirical mass (is the mass of the empirical formula) 4. Molecular mass (is the mass of the molecular formula) Calculation of Molecular Formula

19. Molecular Formula Molecular Mass Empirical Formula Empirical Mass Relationship between Masses (x) MethaneCH 4 EthaneC2H6C2H6 CH 3 WaterH2OH2OH2OH2O BenzeneC6H6C6H6 CH Calculation of Molecular Formula 16.05 30.08 18.02 78.12 16.05 15.04 18.02 13.01 1 2 1 6

20. The relationship between masses (x) will tell you the relationship between formulas. Find x by: x = molecular mass empirical mass What is x?

21. 1. Calculate the empirical mass (like you are used to doing.) 2. Solve for x using molecular mass (given in the problem) and empirical mass. 3. Multiply the empirical formula by x x(empirical formula) = molecular formula Steps

22. Determine the molecular formula of the compound with an empirical formula of CH and a molecular mass of 78.110 amu. Empirical formula= CH Molecular formula= ?? Empirical mass= mass C + mass H = 12.01 + 1.01 = 13.02 amu Molecular mass= 78.110 amu x = molecular mass = 78.110 = 5.999 formula mass 13.02 6(CH) = C 6 H 6 Example 1

23. A sample of a compound with a molar mass of 34.00 g/mol consists of 0.44 g H and 6.93 g O. Find its molecular formula. Empirical formula: Must determine from data given in the problem (See p.6 of notes) Molecular formula: ?? Empirical mass: Can determine from empirical formula Molecular mass: 34.00 H: 0.44 g H x (I mol H) = 0.436/0.433 = 1 (1.01 g H)  HO (empirical formula) O: 6.93 g O x (I mol O) = 0.433/0.433 = 1 (16.00 g O) x = molecular mass = 34.00 = 1.99 empirical mass HO (16.00 + 1.01) 2(HO) =H 2 O 2 Example 2