Formula Mass. not all compounds are molecular formula mass calculated exactly the same way as molecular mass. Solid structure of NaCl
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1 Molecular Mass
2 Molecular Mass synonymous with molar mass and molecular weight is the sum of the atomic masses of all the atoms in a molecule the mass in grams of one mole of a compound
3 Formula Mass not all compounds are molecular formula mass calculated exactly the same way as molecular mass Solid structure of NaCl lowest whole number ratio
4 Example Calculate the number of moles of chloroform (CHCl 3 ) in 198 g of chloroform. Molecular mass of chloroform: 1 mol C = g 1 mol H = g 3 mol Cl = 3(35.46 g) = g 1 mol CHCl 3 = g 198 g CHCl 3 x 1 mol CHCl g CHCl 3 = 1.66 mol CHCl 3
5 Percent Composition of Compounds
6 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%
7 Example A sample of a compound containing carbon and oxygen had a mass of 88g. Of this sample 24g was carbon, 64g was oxygen. What is the percent composition of this compound. %carbon = 24g 88g x 100% = 27% %oxygen = 64g 88g x 100% = 73%
8 Example Calculate the percent composition by mass of H,P and O for one mole of phosphoric acid (H 3 PO 4 ) Molar mass = 3(1.008g) g + 4(16.00) = 97.99
9 Molar mass = 3(1.008g) g + 4(16.00) = %H = 3(1.008g) 97.99g x 100% = 3.086% 30.97g %P = x 97.99g %O = x 4(16.00) 97.99g 100% = 100% = 31.61% 65.31%
10 Determining Formula
11 Determining the Formula O 2 CO 2, H 2 O, O 2, and N 2 O 2, and other gases g C,H,N compound H 2 O absorber CO 2 absorber Burning the sample completely compound + O 2 H 2 O + CO 2 + other gasses
12 Determining the Formula Reacting the C,H,N sample with O 2, we assume all of the hydrogen and carbon present in the g sample is converted to H 2 O and CO 2 respectively grams collected g CO 2 x 12.01g C g CO 2 = g C g H 2 O x 2.016g H = g H 2 O g H
13 Determining the Formula Remembering the original mass of the sample the percentages of the components can be determined g C x 100% g sample = 38.67% C g H g sample x 100% = 16.23% H 38.67% C % H + % N = 100% % N = 45.10%
14 Levels of Structure Elemental Composition Empirical Formula Molecular Formula Constitution Configuration Conformation
15 Elemental Composition Examples: Formaldehyde Glucose C: 40.00% C: 40.00% H: 6.73% H: 6.73% O: 53.27% O: 53.27%
16 Levels of Structure Elemental Composition! Empirical Formula Molecular Formula Constitution Configuration Conformation
17 Empirical Formula The empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms.
18 Empirical Formula Examples: Formaldehyde and Glucose C: 40.00% H: 6.73% O: 53.27% Elemental Composition g 6.73 g g assume a 100g sample calculate atom ratios by dividing by atomic weight
19 Calculating Empirical Formula C: g x 1 mol g = 3.33 mol H: 6.73 g x 1 mol 1.00 g = 6.73 mol O: g x 1 mol 16.0 g = 3.33 mol 100 g
20 Empirical Formula Examples: Formaldehyde and Glucose Elemental Composition C: 40.00% H: 6.73% O: 53.27% assume a 100g sample g 6.73 g g 3.33 mol 6.73 mol 3.33 mol calculate atom ratios by dividing by atomic weight determine the smallest whole number ratio by dividing by the smallest molar value
21 Calculating Empirical Formula C: H: O: g x 6.73 g x g x 100 g 1 mol g 1 mol 1.00 g 1 mol 16.0 g = 3.33 mol 3.33 mol = 6.73 mol 3.33 mol = 3.33 mol 3.33 mol = 1.00 = 2.02 = 1.00
22 Empirical Formula Examples: Formaldehyde and Glucose Elemental Composition C: 40.00% H: 6.73% O: 53.27% g 6.73 g g Empirical Formula: CH 2 O 3.33 mol 6.73 mol 3.33 mol 1 2 1
23 Example A g sample of aluminum oxide (which consists of aluminum and oxygen only) contains 0.912g of Al. Determine the empirical formula of the compound g sample g Al = g O Al 2 O g Al x 1 mol Al g Al = mol mol = 1.0 x 2 = g O x 1 mol O 16.0 g O = mol = mol 1.5 x 2 = 3
24 Example Write the empirical formulas for the following molecules: (a) acetylene (C 2 H 2 ), (b) dinitrogen tetroxide (N 2 O 4 ), (c) glucose (C 6 H 12 O 6 ), diiodine pentoxide (I 2 O 5 ). This problem is not realistic. Molecular formulas are derived from empirical formulas, not vice versa. Empirical formulas come from experiment.
25 Levels of Structure Elemental Composition Empirical Formula!! Molecular Formula Constitution Configuration Conformation
26 Molecular Formula determined from empirical formula and experimentally determined molecular mass Compound formaldehyde glucose Empirical Formula CH 2 O CH 2 O Molar mass
27 Calculation of empirical mass 1 mol C = g 2 mol H x 2 = g 1 mol O = 16.00g glucose formaldehyde g 180g 30g = 6 30g 30g = 1 Molecular mass Empirical mass
28 Molecular Formula determined from empirical formula and experimentally determined molecular mass Compound Empirical Formula Molar mass Molecular formula formaldehyde glucose CH 2 O CH 2 O CH 2 O C 6 H 12 O 6
29 Elemental Composition Example: Lysine C: 49.20% H: 9.66% N: 19.20% O: 21.94%
30 Levels of Structure Elemental Composition! Empirical Formula Molecular Formula Constitution Configuration Conformation
31 Elemental Formula Example: Lysine C: 49.20% g H: 9.66% 9.66 g N: 19.20% g O: 21.94% g assume a 100-g sample 4.10 mol 9.58 mol 1.37 mol 1.37 mol calculate atom ratios by dividing by atomic weight determine smallest whole-number ratio by dividing by smallest number
32 Elemental Formula (cont d) Example: Lysine C: 4.10 mol C atoms H: 9.58 mol H atoms N: 1.37 mol N atoms O: 1.37 mol O atoms determine smallest whole-number ratio by dividing by smallest number (1.37 mol) C 3 H 7 ON
33 Levels of Structure Elemental Composition Empirical Formula Molecular Formula Constitution Configuration Conformation!!
34 Molecular Formula determined from empirical formula and molar mass Compound Empirical Formula Molar mass Molecular formula lysine C 3 H 7 ON ~150 C 6 H 14 O 2 N 2
35 Alternative Method for determining Molecular Formula the molar mass and the percentages by mass of each element present can be used to compute the moles of each element present in one mole of compound.
36 Example 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
37 Example First determine the mass of each element in one mole of caffeine 49.48g C 194.2g caffeine 96.09g C x = 100g caffeine 1 mol caffeine 1 mol 5.15g H 194.2g caffeine 10.0g H x = 100g caffeine 1 mol caffeine 1 mol 28.87g N x 194.2g caffeine 56.07g N = 100g caffeine 1 mol caffeine 1 mol 16.49g O x 194.2g caffeine 32.02g O = 100g caffeine 1 mol caffeine 1 mol
38 Example then convert to moles C 8 H 10 N 4 O g C 1 mol caffeine x 1 mol C g C = 8.00 mol C 1 mol caffeine 10.0g H 1 mol caffeine x 1 mol H 1.008g H = 9.92 mol H 1 mol caffeine 56.07g N 1 mol caffeine x 1 mol N 14.01g N = 4.00 mol N 1 mol caffeine 32.02g O 1 mol caffeine x 1 mol O 16.00g O = 2.00 mol O 1 mol caffeine
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