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Percent Composition, Empirical Formulas, Molecular Formulas

Percent Composition, Empirical Formulas, Molecular Formulas. Percent Composition. Percent Composition – the percentage by mass of each element in a compound. Part. _______. Percent =. x 100%. Whole. So…. Percent composition of a compound or = molecule. Mass of element in 1 mol.

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Percent Composition, Empirical Formulas, Molecular Formulas

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  1. Percent Composition, Empirical Formulas, Molecular Formulas

  2. Percent Composition • Percent Composition – the percentage by mass of each element in a compound Part _______ Percent = x 100% Whole So… Percent composition of a compound or = molecule Mass of element in 1 mol ____________________ x 100% Mass of 1 mol

  3. Percent Composition Example: What is the percent composition of Potassium Permanganate (KMnO4)? Molar Mass of KMnO4 K = 1(39.1) = 39.1 Mn = 1(54.9) = 54.9 O = 4(16.0) = 64.0 MM = 158 g

  4. Percent Composition Example: What is the percent composition of Potassium Permanganate (KMnO4)? Molar Mass of KMnO4 = 158 g 39.1 g K % K x 100 = 24.7 % 158 g 54.9 g Mn 34.8 % x 100 = % Mn 158 g K = 1(39.10) = 39.1 64.0 g O x 100 = 40.5 % % O Mn = 1(54.94) = 54.9 158 g O = 4(16.00) = 64.0 MM = 158

  5. Percent Composition Determine the percentage composition of sodium carbonate (Na2CO3)? Molar Mass Percent Composition 46.0 g x 100% = 43.4 % Na = 2(23.00) = 46.0 C = 1(12.01) = 12.0 O = 3(16.00) = 48.0 MM= 106 g % Na = 106 g 12.0 g x 100% = 11.3 % % C = 106 g 48.0 g x 100% = 45.3 % % O = 106 g

  6. Percent Composition Determine the percentage composition of ethanol (C2H5OH)? % C = 52.13%, % H = 13.15%, % O = 34.72% _______________________________________________ Determine the percentage composition of sodium oxalate (Na2C2O4)? % Na = 34.31%, % C = 17.93%, % O = 47.76%

  7. Percent Composition Calculate the mass of bromine in 50.0 g of Potassium bromide. 1. Molar Mass of KBr K = 1(39.10) = 39.10 Br =1(79.90) =79.90 MM = 119.0 2. 79.90 g ___________ = 0.6714 119.0 g 3. 0.6714 x 50.0g = 33.6 g Br

  8. Percent Composition Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C6H14N2O2. 1. Molar Mass of C6H14N2O2 C = 6(12.01) = 72.06 H =14(1.01) = 14.14 N = 2(14.01) = 28.02 O = 2(16.00) = 32.00 MM = 146.2 2. 28.02 g ___________ = 0.192 146.2 g 3. 0.192 x 85.0 mg = 16.3 mg N

  9. Hydrates Hydrated salt – salt that has water molecules trapped within the crystal lattice Examples: CuSO4•5H2O , CuCl2•2H2O Anhydrous salt – salt without water molecules Examples: CuCl2 Can calculate the percentage of water in a hydrated salt.

  10. Percent Composition Calculate the percentage of water in sodium carbonate decahydrate, Na2CO3•10H2O. 1. Molar Mass of Na2CO3•10H2O Na = 2(22.99) = 45.98 C = 1(12.01) = 12.01 H = 20(1.01) = 20.2 3. O = 13(16.00)= 208.00 180.2 g MM = 286.2 _______ x 100%= 67.97 % 286.2 g 2. Water H = 20(1.01) = 20.2 O = 10(16.00)= 160.00 MM = 180.2 H = 2(1.01) = 2.02 or So… 10 H2O = 10(18.02) = 180.2 O = 1(16.00) = 16.00 MM H2O = 18.02

  11. Percent Composition Calculate the percentage of water in Aluminum bromide hexahydrate, AlBr3•6H2O. 1. Molar Mass of AlBr3•6H2O Al = 1(26.98) = 26.98 Br = 3(79.90) = 239.7 H = 12(1.01) = 12.12 O = 6(16.00) = 96.00 MM = 374.8 3. 108.1 g _______ 2. Water x 100%= 28.85 % 374.8 g H = 12(1.01) = 12.1 O = 6(16.00)= 96.00 MM = 108.1 MM = 18.02 For 6 H2O = 6(18.02) = 108.2 or

  12. Percent Composition If 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain? MgSO4.7 H2O 1. Molar Mass 2. % MgSO4 Mg = 1 x 24.31 = 24.31 g S = 1 x 32.06 = 32.06 g O = 4 x 16.00 = 64.00 g MM = 120.37 g 120.4 g X 100 = 48.84 % 246.5 g 3. Grams anhydrous MgSO4 H = 2 x 1.01 = 2.02 g O = 1 x 16.00 = 16.00 g MM = 18.02 g 0.4884 x 125 = 61.1 g MM H2O = 7 x 18.02 g = 126.1 g Total MM = 120.4 g + 126.1 g = 246.5 g

  13. Percent Composition If 145 grams of copper (II) sulfate pentahydrate is completely dehydrated, how many grams of anhydrous copper sulfate will remain? CuSO4.5 H2O 1. Molar Mass 2. % CuSO4 Cu = 1 x 63.55 = 63.55 g S = 1 x 32.06 = 32.06 g O = 4 x 16.00 = 64.00 g MM = 159.61 g 159.6 g X 100 = 63.92 % 249.7 g 3. Grams anhydrous CuSO4 H = 2 x 1.01 = 2.02 g O = 1 x 16.00 = 16.00 g MM = 18.02 g 0.6392 x 145 = 92.7 g MM H2O = 5 x 18.02 g = 90.1 g Total MM = 159.6 g + 90.1 g = 249.7 g

  14. Percent Composition A 5.0 gram sample of a hydrate of BaCl2 was heated, and only 4.3 grams of the anhydrous salt remained. What percentage of water was in the hydrate? 2. Percent of water 1. Amount water lost 0.7 g water • 5.0 g hydrate • 4.3 g anhydrous salt • 0.7 g water x 100 = 14 % 5.0 g hydrate

  15. Percent Composition A 7.5 gram sample of a hydrate of CuCl2 was heated, and only 5.3 grams of the anhydrous salt remained. What percentage of water was in the hydrate? 2. Percent of water 1. Amount water lost 2.2 g water • 7.5 g hydrate • 5.3 g anhydrous salt • 2.2 g water x 100 = 29 % 7.5 g hydrate

  16. Percent Composition A 5.0 gram sample of Cu(NO3)2•nH2O is heated, and 3.9 g of the anhydrous salt remains. What is the value of n? 1. Amount water lost • 5.0 g hydrate • 3.9 g anhydrous salt • 1.1 g water 3. Amount of water 0.22 x 18.02 = 4.0 2. Percent of water 1.1 g water x 100 = 22 % 5.0 g hydrate

  17. Percent Composition A 7.5 gram sample of CuSO4•nH2O is heated, and 5.4 g of the anhydrous salt remains. What is the value of n? 1. Amount water lost • 7.5 g hydrate • 5.4 g anhydrous salt • 2.1 g water 3. Amount of water 0.28 x 18.02 = 5.0 2. Percent of water 2.1 g water x 100 = 28 % 7.5 g hydrate

  18. Formulas Percent composition allow you to calculate the simplest ratio among the atoms found in compound. Empirical Formula – formula of a compound that expresses lowest whole number ratio of atoms. Molecular Formula – actual formula of a compound showing the number of atoms present Examples: C6H12O6 - molecular C4H10 - molecular - empirical - empirical C2H5 CH2O

  19. Formulas Is H2O2 an empirical or molecular formula? Molecular, it can be reduced to HO HO = empirical formula

  20. Calculating Empirical Formula An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula. 1. Determine the number of grams of each element in the compound. 4.151 g Al and 3.692 g O 2. Convert masses to moles. 1 mol Al 4.151 g Al = 0.1539 mol Al 26.98 g Al 1 mol O 3.692 g O = 0.2308 mol O 16.00 g O

  21. Calculating Empirical Formula An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula. 3. Find ratio by dividing each element by smallest amount of moles. 0.1539 moles Al = 1.000 mol Al 0.1539 0.2308 moles O = 1.500 mol O 0.1539 4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds) O = 1.500 x 2 = 3 Al = 1.000 x 2 = 2 therefore, Al2O3

  22. Calculating Empirical Formula A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the empirical formula for this compound. 1 mol Co 4.550 g Co = 0.07721 mol Co 58.93 g Co 1 mol Cl 5.475 g Cl = 0.1544 mol Cl 35.45 g Cl 0.07721 mol Co 0.1544 mol Cl = 1 = 2 0.07721 0.07721 CoCl2

  23. Calculating Empirical Formula When a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573 g. Determine the empirical formula. Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g 1 mol Fe 2.000 g Fe = 0.03581 mol Fe 55.85 g Fe 1 mol O 0.573 g O = 0.03581 mol Fe 16.00 g 1 : 1 FeO

  24. Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate. 1 mol Pb 1.3813 g Pb = 0.006667 mol Pb 207.2 g Pb 0.00672 gH 1 mol H = 0.00667 mol H 1.008 g H 1 mol As 0.4995 g As = 0.006667 mol As 74.92 g As 1 mol O 0.4267g Fe = 0.02667 mol O 16.00 g O

  25. Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate. 0.006667 mol Pb = 1.000 mol Pb 0.006667 0.00667 mol H = 1.00 mol H PbHAsO4 0.006667 0.006667 mol As = 1.000 mol As 0.006667 0.02667 mol O = 4.000 mol O 0.006667

  26. Calculating Empirical Formula The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6. Step 1: In 100.00g of Nylon-6 the masses of elements present are 63.38 g C, 12.38 g n, 9.80 g H, and 14.14 g O. Step 2: 63.38 g C 1 mol C 9.80 g H 1 mol H = 5.302 mol C = 9.72 mol H 12.01 g C 1.01 g H 12.38 g N 1 mol N = 0.8837 mol N 14.01 g N 14.14 g O 1 mol O = 0.8832 mol O 16.00 g O

  27. Calculating Empirical Formula The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6. Step 3: 5.302 mol C = 6.000 mol C 6:1:11:1 0.8837 0.8837 mol N = 1.000 mol N 0.8837 C6NH11O 9.72 mol H = 11.0 mol H 0.8837 0.8837 mol O = 1.000 mol O 0.8837

  28. Calculating Molecular Formula A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88g. What is the compound’s molecular formula? Step 3: Multiply Step 1: Molar Mass P = 2 x 30.97 g = 61.94g O = 5 x 16.00g = 80.00 g 141.94 g (P2O5)2 = P4O10 Step 2: Divide MM by Empirical Formula Mass 238.88 g = 2 141.94g

  29. Calculating Molecular Formula A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula? (CH)6 = C = 12.01 g H = 1.01 g 13.01 g C6H6 78 g/mol = 6 13.01 g/mol

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