1. CHEMICAL COMPOUNDS AND THE MOLE Chapter 7

2. Formula Mass  Mass of H 2 O? H 2(1.01) + O 16.00_ 18.02 amu  Formula Mass: mass of molecule, formula unit, or ion is sum of masses of all atoms represented (amu)  Ca(NO 3 ) 2 Ca 40.08 N 2(14.01) + O 6 (16.00) 164.10 amu

3. Molar Mass  Definition: mass of 1 mole of compound – use molar masses of elements (g/mol)  MgCl 2 24.31 g/mol + 2(35.45 g/mol) = 95.21 g/mol  (NH 4 ) 2 CrO 4 2(14.01 g/mol) + 8(1.01 g/mol) + 52.00 g/mol + 4(16.00 g/mol) = 152.10 g/mol  CuSO 4 * 5H 2 O 63.55 g/mol + 32.07 g/mol + 4(16.00 g/mol) + 5(18.02 g/mol) = 249.72 g/mol

4. Molar Mass in Conversions  Remember flow chart from chapter 3?  What is mass (g) of 3.04 mol of ammonia vapor, NH 3 ?  ? g = 3.04 mol X 17.04 g = 51.8 grams NH 3 1 mol  How many moles of sodium chloride are present in 100.0 grams? ? Moles NaCl = 100.0 g X 1 mol = 1.711 mol NaCl 58.44 g

5. PRACTICE MOLAR MASS Work as a group of 4.  1 st group member: 1,5,9,13,17,21 GroupII:1  2 nd group member: 2,6,10,14,18,22 Group II:2  3 rd group member: 3,7,11,15,19,23 Group II:3  4 th group member:4,8,12,16,20,24 Group II:4 Show work for each problem you complete. Explain your work to the other group members and write in their answers. HW: complete your set and 33-40.

6. “Super Mole” Conversions  How many molecules are in 4.15 x 10 -5 g C 6 H 12 O 6 ? ? Molec.= 4.15 x 10 -5 g X 1 mol X 6.02 x 10 23 molec. = 1.39 x 10 17 molecules 180.18 g 1 mole  How many H atoms are in 7.1 moles of C 6 H 12 O 6 ? ? Atoms = 7.1 mol X 12 mol H X NA = 5.1 x 10 25 atoms H 1 mol C 6 H 12 O 6 1 mol H  How many formula units are in 4.5 kg Ca(OH) 2 ? ?f.un = 4.5 kg X 10 3 g X 1 mol X NA X 1 f. un = 3.7 x 10 25 formula units 1 kg 74.10 g 1 mol 1 molecule **NA = 6.02x10 23 molecules

7. More “Super Mole” Conversions  What is the mass of H 2 SO 4, if you have 1.53 x 10 23 sulfate ions your compound? 1.53x10 23 ions x 1 mol SO 4 x 1 mol H 2 SO 4 x 98.09 g = 24.9 g NA 1mol SO 4 1 mol  How many water molecules are present in in a 5.00 g sample of copper (II) sulfate pentahydrate? 5.00 g CuSO 4 * 5H 2 O x 1 mol x 5mol H 2 O x NA 249.72 g 1mol CuSO 4 * 5H 2 O 1 mol H 2 O = 6.03x10 22 molecules H 2 O

8. Percent Composition  The percent by mass of each element in a compound.  % = mass due to 1 element x 100 mass of whole compound

9. Percent Composition  What is the percent composition by mass of each element in (NH 4 ) 2 O? [MM (NH 4 ) 2 O = 52.10 g/mol] %N = 2(14.01) x 100 = 53.78% 52.10 % H = 8(1.01) x 100 = 15.5% 52.10 % O = 16.00 x 100 = 30.71% 52.10  What percentage by mass of Al 2 (SO 4 ) 3  6H 2 O is water? % H 2 O = 6(18.02) x 100 = 24.01% 450.29  Given a 25.0 gram sample of aluminum sulfate hexahydrate, how much water (g) could be driven off? g H 2 O= (%H 2 O) (total sample mass) (24.01%) (25.00 g) = 6.00g

10. Empirical Formula  Definition: formula showing smallest whole-number mole ratio of atoms in a compound  Ex: B 2 H 6  Molecular Formula BH 3  Empirical Formula  Given (CH 2 O) x as the empirical, determine possible molecular formulas. x=2 x=1 x=6 C 2 H 4 O 2 CH 2 O C 6 H 12 O 6 Formaldehyde acetic acid glucose (all have different molar masses)

11. Empirical Formula Calculation  Given the molecular formula: reduce  Given % compostion data  Find grams of each element  Find moles of each element  Find mole ratio of atoms by dividing through by the smallest number of moles  If the ratio yields a 0.33, 0.50 multiply the entire formula through to clear fractional mole amounts.

12. Finding Empirical Formula  Determine the empirical formula of the compound with 17.15% C, 1.44% H, and 81.41% F.  (CHF 3 )x Assume 100 g sample. C 1.427 H 1.43 F 4.285 17.15g C x 1mol = 1.427mol C 1.427 1.427 1.427 12.01g 1.44g H x 1mol = 1.43mol H =(CHF 3 ) x 1.01 g 81.41g F x 1mol = 4.285mol F 19.00g

13. Finding Empirical Formula  Find empirical formula of 26.56% K, 35.41% Cr, and rest O.  (K 2 Cr 2 O 7 )x Assume 100 g sample. 26.56 g K x 1 mol =.6793 moles K K.6793 Cr.6810 O 2.377 39.10 g.6793.6793.6793 35.41 g Cr x 1 mol =.6810 moles Cr 52.00 g = (KCrO 3.5 ) x 2 38.03 g O x 1 mol = 2.377 moles O = (K 2 Cr 2 O 7 ) x 16.00 g

14. Finding Molecular Formula x(empirical formula) = molecular formula x(emp.form mass) = molec.form mass x = Molecular formula Mass Empirical formula Mass

15. Finding Molecular Formula  Determine molecular formula of compound with empirical formula CH and formula mass of 78.110 amu. x = 78.110 amu = 6 (CH) 6 = C 6 H 6 (12.01+1.01) amu

16. Finding Molecular Formula  Sample has formula mass of 34.00 amu has 0.44 g H and 6.92 g O. Find its molecular formula. %H =.44g x 100 = 6.0% 7.36g %O = 6.92g x 100 = 94.0% 7.36 g Assume 100 g sample. 6.0g H x 1 mol = 5.9 mol 94.0 g O x 1mol = 5.88 mol 1.01g 16.00g H 5.9 O 5.88 = (HO) x x = 34.00 amu = 2 H 2 O 2 5.88 5.88 (1.01 + 16.00) amu hydrogen peroxide

17. Combustion Analysis  A compound contains only carbon, hydrogen, and oxygen. Combustion of the compound yields.01068 grams of carbon dioxide and.00437 grams of water. The molar mass of the compound is 180.1 g/ mol. The sample has a total mass of.0100 grams. What are the empirical and molecular formulas of the compound? CHO + O 2  H 2 0 + CO 2.01068g CO 2 x 1 mol CO 2 x 1 mol C = 2.427 x 10 -4 mol C x 12.01 g =.002915 g C 44.01gCO 2 1 mol CO 2 1 mol.00437g H 2 0 x 1mol H 2 0 x 2mol H = 4.85 x 10 -4 mol H x 1.01 g = 4.90 x 10 -4 g H 18.02 g 1mol H 2 0 1 mol C+H+O = CHO.002915 g + 4.90 x 10 -4 g + O =.0100 g gO =.00660 g = 4.12 x 10 -4 mol O

18. Combustion Analysis (continued) C 2.427 x 10 -4 H 4.85 x 10 -4 O 4.12 x 10 -4 = (CH 2 O 2 )x 2.427 x 10 -4 X = molecular mass = 180.1 g = 4 empirical mass 46.03 g (CH 2 O 2 ) 4 = C 4 H 8 O 8