Concepts and Calculations in Molecular Mass and Formula Mass, Study notes of Chemistry

Download Concepts and Calculations in Molecular Mass and Formula Mass and more Study notes Chemistry in PDF only on Docsity! Molecular Mass synonymous with molar mass and molecular weight Molecular Mass is the sum of the atomic masses of all the atoms in a molecule the mass in grams of one mole of a compound pg. 323 problem 27 What is the mass in grams of 3.25 mol sulfuric acid ( H2SO4 ). 3.25 mol H2SO4 98.07 g H2SO4 1 mol H2SO4 x = 319 g H2SO4 1 mol H2SO4 2 mol H = 2(1.00 g) 1 mol S = 32.07g 4 mol O = 4(16.00 g) = 98.07 g pg. 324 problem 30a Determine the number of moles present in 22.6 g AgNO3. 22.6 g AgNO3 1 mol AgNO3 169.87 g AgNO3 x = 0.133 mol AgNO3 1 mol AgNO3 1 mol N = 14.00 g 1 mol Ag = 107.87g 3 mol O = 3(16.00 g) = 169.87 g pg. 326 problem 31a A sample of silver chromate (Ag2CrO4 ) has a mass of 25.8 g. 25.8 g Ag2CrO4 1 mol Ag2CrO4 331.74 g Ag2CrO4 x = 9.36 x 1022 ions Ag+ 1 mol Ag2CrO4 1 mol Cr = 52.00 g 2 mol Ag = 107.87g 4 mol O = 4(16.00 g) = 331.74 g a. How many Ag+ ions are present? 2 mol Ag+ 1 mol Ag2CrO4 x 6.02 x 1023 ions Ag+ 1 mol Ag+ x Percent Composition of Compounds Percent composition is the percent by mass of each element the compound contains. Obtained by dividing the mass of each element in one mole of the compound by the molar mass of the compound and multiplying by 100% Example 45 Calculate the percent composition by mass of H,P and O for one mole of phosphoric acid (H3PO4) Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99 Determining Formula Elemental Composition Levels of Structure Empirical Formula Molecular Formula Constitution Configuration Conformation Examples: Elemental Composition Formaldehyde Glucose C: 40.00% C: 40.00% H: 6.73% H: 6.73% O: 53.27% O: 53.27% Examples: Formaldehyde and Glucose Empirical Formula Elemental Composition C: 40.00% H: 6.73% O: 53.27% assume a 100g sample calculate atom ratios by dividing by atomic weight 40.00 g 6.73 g 53.27 g 40.00 g x 1 mol 12.01 g = 3.33 mol 6.73 g x 1 mol 1.00 g = 6.73 mol 53.27 g x 1 mol 16.0 g = 3.33 mol 100 g Calculating Empirical Formula C: H: O: Examples: Formaldehyde and Glucose Empirical Formula Elemental Composition C: 40.00% H: 6.73% O: 53.27% assume a 100g sample calculate atom ratios by dividing by atomic weight 40.00 g 6.73 g 53.27 g 3.33 mol 6.73 mol 3.33 mol determine the smallest whole number ratio by dividing by the smallest molar value A 1.723 g sample of aluminum oxide (which consists of aluminum and oxygen only) contains 0.912g of Al. Determine the empirical formula of the compound. Example 0.912 g Al = 0.811 g O 1.723 g sample - 0.912 g Al x 1 mol Al 26.98 g Al 0.811 g O x 1 mol O 16.0 g O = 0.0338 mol = 0.0507 mol 0.0338 mol 0.0338 mol = 1.0 1.5= x 2 = x 2 = Al2O3 2 3 Elemental Composition Empirical Formula Molecular Formula Constitution Configuration Conformation Levels of Structure √ √ determined from empirical formula and experimentally determined molecular mass Molecular Formula Compound Empirical Formula Molar mass formaldehyde CH2O glucose CH2O 30 180 Example: Elemental Composition Lysine C: 49.20% H: 9.66% O: 21.94% N: 19.20% Elemental Composition Empirical Formula Molecular Formula Constitution Configuration Conformation √ Levels of Structure Example: Elemental Formula Lysine C: 49.20% H: 9.66% O: 21.94% N: 19.20% assume a 100-g sample 49.20 g 9.66 g 19.20 g 21.94 g calculate atom ratios by dividing by atomic weight determine smallest whole-number ratio by dividing by smallest number Elemental Composition Empirical Formula Molecular Formula Constitution Configuration Conformation √ √ Levels of Structure determined from empirical formula and molar mass Molecular Formula Compound Empirical Formula Molar mass Molecular formula lysine C3H7ON ~150 C6H14O2N2 73g 150g = 2 Caffeine contains 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2g. Determine the molecular formula formula of caffeine Example First determine the mass of each element in one mole of caffeine 100g caffeine 49.48g C 5.15g H 100g caffeine 28.87g N 100g caffeine 16.49g O 100g caffeine Example x 194.2g caffeine 1 mol x 194.2g caffeine 1 mol x 194.2g caffeine 1 mol x 194.2g caffeine 1 mol = 1 mol caffeine 96.09g C = 1 mol caffeine 10.0g H = 1 mol caffeine 56.07g N = 1 mol caffeine 32.02g O then convert to moles Example 1 mol caffeine 96.09g C 1 mol caffeine 10.0g H 1 mol caffeine 56.07g N 1 mol caffeine 32.02g O x x x x 12.011g C 1 mol C 1.008g H 1 mol H 14.01g N 1 mol N 16.00g O 1 mol O 1 mol caffeine 8.00 mol C = 1 mol caffeine 4.00 mol N = 1 mol caffeine 2.00 mol O = 1 mol caffeine 9.92 mol H = C8H10N4O2