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Percent Composition

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

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Percent Composition

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  1. Percent Composition

  2. Percent Composition • Percent Composition – the percentage by mass of each element in a compound Part _______ Percent = x 100% Whole 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. 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.

  9. 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%= 62.96 % 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

  10. 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

  11. Empirical and Molecular Formulas

  12. Formulas Percent composition allows you to calculate the simplest ratio among the atoms found in a 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

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

  14. 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

  15. 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

  16. 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

  17. 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 O 16.00 g 1 : 1 FeO

  18. 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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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