CHEMICAL COMPOUNDS and THE MOLE - PowerPoint PPT Presentation

chemical compounds and the mole n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
CHEMICAL COMPOUNDS and THE MOLE PowerPoint Presentation
Download Presentation
CHEMICAL COMPOUNDS and THE MOLE

play fullscreen
1 / 18
CHEMICAL COMPOUNDS and THE MOLE
256 Views
Download Presentation
feng
Download Presentation

CHEMICAL COMPOUNDS and THE MOLE

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. CHEMICAL COMPOUNDS and THE MOLE Chapter 7

  2. Formula Mass • Mass of H2O? 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(NO3)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) • MgCl224.31 g/mol + 2(35.45 g/mol) = 95.21 g/mol • (NH4)2CrO4 2(14.01 g/mol) + 8(1.01 g/mol) + 52.00 g/mol + 4(16.00 g/mol) = 152.10 g/mol • CuSO4 * 5H2O 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, NH3? • ? g = 3.04 mol X 17.04 g = 51.8 grams NH3 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. • 1st group member: 1,5,9,13,17,21 GroupII:1 • 2nd group member: 2,6,10,14,18,22 Group II:2 • 3rd group member: 3,7,11,15,19,23 Group II:3 • 4th 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 **NA = 6.02x1023 molecules • How many molecules are in 4.15 x 10-5g C6H12O6? ? Molec.= 4.15 x 10-5 g X 1 mol X 6.02 x 1023 molec. = 1.39 x 1017 molecules 180.18 g 1 mole • How many H atoms are in 7.1 moles of C6H12O6? ? Atoms = 7.1 mol X 12 mol H X NA = 5.1 x 1025 atoms H 1 mol C6H12O6 1 mol H • How many formula units are in 4.5 kg Ca(OH)2? ?f.un = 4.5 kg X 103 g X 1 mol X NA X 1 f. un = 3.7 x 1025 formula units 1 kg 74.10 g 1 mol 1 molecule

  7. More “Super Mole” Conversions • What is the mass of H2SO4, if you have 1.53 x 1023 sulfate ions your compound? 1.53x1023 ions x 1 mol SO4 x 1 mol H2SO4 x 98.09 g = 24.9 g NA 1mol SO41 mol • How many water molecules are present in in a 5.00 g sample of copper (II) sulfate pentahydrate? 5.00 g CuSO4 * 5H2O x 1 mol x 5mol H2O x NA 249.72 g 1mol CuSO4 * 5H2O 1 mol H2O = 6.03x1022 molecules H2O

  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 (NH4)2O? [MM (NH4)2O = 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 Al2(SO4)36H2O is water? % H2O = 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 H2O= (%H2O) (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: B2H6 Molecular Formula BH3 Empirical Formula • Given (CH2O)x as the empirical, determine possible molecular formulas. x=2 x=1 x=6 C2H4O2 CH2O C6H12O6 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. • (CHF3)x • Assume 100 g sample. C1.427H1.43F4.285 17.15g C x 1mol = 1.427mol C 1.427 1.427 1.427 12.01g 1.44g H x 1mol = 1.43mol H =(CHF3)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. • (K2Cr2O7)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 = (KCrO3.5 ) x 2 38.03 g O x 1 mol = 2.377 moles O = (K2Cr2O7 )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 = C6H6 (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 H5.9O5.88 = (HO)x x = 34.00 amu = 2 H2O2 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 + O2  H20 + CO2 .01068g CO2 x 1 mol CO2 x 1 mol C = 2.427 x 10-4 mol C x 12.01 g = .002915 g C 44.01gCO2 1 mol CO2 1 mol .00437g H20 x 1mol H20 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 H20 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 = (CH2O2)x 2.427 x 10 -4 X = molecular mass = 180.1 g = 4 empirical mass 46.03 g (CH2O2)4 = C4H8O8