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The Mole

The Mole. 1 mole = 6.022045*10 23 particles Avogadro’s number N A. AW of O = 16.0 amu/atom = 16.0 g/mole FW of NaCl = 58.4 amu/formula unit = 58.4 g/mole. 6.022 *10 23 NaCl formula units 6.022 *10 23 Na + ions 6.022 *10 23 Cl - ions.

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The Mole

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  1. The Mole 1 mole = 6.022045*1023 particles Avogadro’s number NA AW of O = 16.0 amu/atom = 16.0 g/mole FW of NaCl = 58.4 amu/formula unit = 58.4 g/mole 6.022 *1023 NaCl formula units 6.022 *1023 Na+ ions 6.022 *1023 Cl- ions

  2. What is the mass of one formula unit of NaCl in grams? AW Na=23.0g/mole; AW Cl=35.5g/mole FW NaCl=58.5 g/mole 1mole=6.022*1023 NaCl formula units g 1 mole g mole 6.022*1023NaCl NaCl 58.5 * = 9.71*10-23 0.0000000000000000000000971g/NaCl

  3. How many moles of Flourine atoms are there in 73.4g of Magnesium Flouride? How many Flourine atoms? MgF2 ; 1 mole MgF2 has 2 moles F atoms ; FW(MgF2)=62.3g/mole 1mol MgF2 62.3 g MgF2 2 mol F 1 mol MgF2 = 2.36 mol F 73.4g MgF2 * * 2.36 mol F atoms*(6.022*1023 atoms/mol) = 1.42*1024 F atoms FW grams of MgF2 moles MgF2 moles F atoms F Composition Stoichiometry Unit Factor How many moles of Magnesium in 73.4g of Magnesium Flouride? Composition Stoichiometry 1 mol Mg 2 mol F 2.36 mol F * =1.18 mol Mg

  4. % Composition Parts of A (by mass) 100 parts mixture (by mass) %A (mass)= Parts of A=atom of interest Parts Mixture=caffeine molecule 8C 10H 2O 4N 49.5% C 5.15% H 16.5% O 28.9% N 49.5 grams of Carbon 100 grams of caffeine MW=8*AW(C)+10*AW(H)+2*AW(O)+4*AW(N) =8*(12.0g/mole)+10*(1.01g/mole)+2*(16.0g/mole)+4*(14.0g/mole) =194g/mole of caffeine 8*12.0 grams of Carbon g C 100 g caffeine 194 grams of caffeine g caffeine 100 g caffeine %C = = 0.495 * =49.5%

  5. AW Carbon 100 g Caffeine MW Caffeine 100 g Caffeine % C in caffeine = * Same thing as: AW Carbon MW Caffeine % C in caffeine = * 100% because 100% means parts per 100

  6. Empirical (Simplest) Formula - The smallest whole-number ratio of atoms present in a compound. xZn + yS ZnxSy 67.1% Zn & 32.9% S

  7. 67.1% Zn & 32.9% S xZn + yS ZnxSy 67.1g Zn mole1.03mol Zn 100 g ZnxSy 65.4g 100 g ZnxSy * = 32.9g S mole 1.02mol S 100 g ZnxSy 32.1g 100 g ZnxSy 1.03mol Zn 1.02 1.02mol S 1.02 = 1 mol Zn = 1 mol S The empirical formula for Zinc Sulfide is ZnS & EW=100g/mol To determine the molecular formula, we need the MW MW EW 100g/mol 100g/mol n= if MW=100.0g/mol then n= = 1 (experimentally determined)

  8. Combustion Train for the analysis of Carbon, Hydrogen and Oxygen containing compounds 0.400 g sample 0.163 g H2O 0.600 g CO2

  9. 0.400 g sample 0.163 g H2O & 0.600 g CO2 1 mol H2O 2 mol H 18.0 g H2O 1 mol H2O 0.163 g H2O * * = 0.0181 mol H 1 mol CO2 1 mol C 44.0 g CO2 1 mol CO2 0.600 g CO2 * * = 0.0136 mol C mass O = mass of sample - mass H - mass C 1.01 g H 1 mol H 0.0181 mol H* = 0.0183 g H C1H1.33O1 C3H4O3 12.0 g C 1 mol C EW=88.0g/mol 0.0136 mol C * = 0.163 g C MW=176 g/mol 0.400g-0.0183g-0.163g=0.218g O 176 g/mol 88.0 g/mol n= = 2 1 mol O 16.0 g O 0.218g O * = 0.0136 mol O C6H8O6 0.0181 mol H/0.0136 = 1.33

  10. Law of Multiple Proportions When two elements, A and B, form more than one compound, the ratio of the masses of element B than combine with a given mass of element A in each of the compounds can be expressed by small whole numbers. Find the ratio of masses of oxygen that combine with a given mass of hydrogen in H2O2 and H2O AW O=16.0g/mol AW H=1.01g/mol H2O2 32.0gO/2.02gH = 15.8gO/gH H2O 16.0gO/2.02gH = 7.92gO/gH ratio H2O2/H2O = 15.8/7.92 = 2/1 2:1 H2O2:H2O

  11. Purity of a Sample The purity of a sample is given by it’s % Composition A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO3 and impurities in 250 g of this sample? 98.3g Na3PO4 1.7g impurity 100 g sample 100 g sample ; 98.3g Na3PO4 100 g sample * 250 g sample = 2.46*102g Na3PO4 1.7g impurity 100 g sample * 250 g sample = 4.3g impurities

  12. A 245g sample is found to contain 225g of Calcium Carbonate. What is the purity of Calcium Carbonate in the sample? 225g CaCO3 100g sample 245g sample 100g sample % CaCO3= * = 91.8% CaCO3 What is the % impurities? 20g impurity100g sample 245g sample 100g sample % impurity = * = 8.2% impurity

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