Percent Composition Empirical Formulas Molecular Formulas Percent Composition

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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 Percent composition Mass of element in 1 mol of a compound or =__________ x 100% Mass of 1 molecule

Percent Composition Example (Step 1): What is the percent composition of Potassium Permanganate (KMn. O 4)? Molar Mass of KMn. O 4 K = 1(39. 1) = 39. 1 Mn = 1(54. 9) = 54. 9 O = 4(16. 0) = 64. 0 MM = 158. 0 g

Percent Composition Example (Step 2): What is the percent composition of Potassium Permanganate (KMn. O 4)? Molar Mass of KMn. O 4 % K 39. 1 g K 158. 0 g = 158. 0 g x 100 = 24. 7 % 54. 9 g Mn x 100 = 34. 7 % % Mn 158. 0 g K= 1(39. 1) = 39. 1 Mn = 1(54. 9) = 54. 9 O = 4(16. 0) = 64. 0 MM = 158. 0 64. 0 g O x 100 = 40. 5 % % O 158. 0 g

Percent Composition Determine the percentage composition of sodium carbonate (Na 2 CO 3)? Molar Mass Percent Composition 46. 0 g/mol x 100% =43. 4% Na = 2(23. 0) = 46. 0 % Na =106. 0 g/mol C = 1(12. 0) = 12. 0 g/mol x 100% =11. 3 % O = 3(16. 0) = 48. 0 % C = 106. 0 g/mol MM= 106. 0 g/mol 48. 0 g/mol x 100% =45. 3% % O = 106. 0 g/mol

Percent Composition Determine the percentage composition of ethanol (C 2 H 5 OH)? % C = 52. 13%, % H = 13. 15%, % O = 34. 72% ________________________ Determine the percentage composition of sodium oxalate (Na 2 C 2 O 4)? % Na = 34. 31%, % C = 17. 93%, % O = 47. 76%

Formulas Percent composition allows you to calculate 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: C 4 H 10 - molecular C 2 H 5 - empirical C 6 H 12 O 6 - molecular CH 2 O - empirical

Formulas Is H 2 O 2 an empirical or molecular formula? Molecular, it can be reduced to HO HO = empirical formula

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. 4. 151 g Al 1 mol Al = 0. 154 mol Al 27. 0 g Al 3. 692 g O 1 mol O 16. 0 g O = 0. 231 mol O

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. 154 moles Al 0. 154 = 1. 00 mol Al 0. 231 moles O 0. 154 = 1. 50 mol O 4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds) O = 1. 50 x 2 = 3 Al = 1. 00 x 2 = 2 therefore, Al 2 O 3

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. 4. 550 g Co 1 mol Co 58. 9 g Co 5. 475 g Cl 1 mol Cl 35. 5 g Cl 0. 0772 mol Co 0. 0772 =1 Co. Cl 2 = 0. 0772 mol Co = 0. 154 mol Cl 0. 0772 =2

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 2. 000 g Fe 1 mol Fe 55. 9 g Fe 0. 573 g O 1 mol O 16. 0 g = 0. 0358 mol Fe 1: 1 Fe. O

Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains 1. 3813 g lead, 0. 00672 g of hydrogen, 0. 4995 g of arsenic, and 0. 4267 g of oxygen. Calculate the empirical formula for lead arsenate. 1. 3813 g Pb 1 mol Pb 207. 2 g Pb 0. 00672 g. H 1 mol H 1. 0 g H = 0. 006667 mol Pb = 0. 0067 mol H 0. 4995 g As 1 mol As = 0. 006667 mol As 74. 92 g As 0. 4267 g Fe 1 mol O 16. 0 g O = 0. 0267 mol O

Calculating Empirical Formula A sample of lead arsenate, an insecticide used against the potato beetle, contains 1. 3813 g lead, 0. 00672 g 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. 0067 mol H 0. 0067 = 1. 00 mol H 0. 006667 mol As = 1. 000 mol As 0. 0067 0. 0267 mol O 0. 0067 = 4. 000 mol O Pb. HAs. O 4

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. 00 g 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 12. 0 g C 12. 38 g N 1 mol N 14. 0 g N = 5. 28 mol C 9. 80 g H 1 mol H 1. 0 g H = 0. 884 mol N 14. 14 g O 1 mol O 16. 0 g O = 9. 8 mol H = 0. 884 mol O

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. 28 mol C 0. 884 = 6 mol C 0. 884 mol N 0. 884 = 1 mol N 9. 8 mol H 0. 884 = 11 mol H 0. 884 mol O 0. 884 = 1 mol O 6: 1: 1 C 6 NH 11 O

Calculating Molecular Formula A white powder is analyzed and found to have an empirical formula of P 2 O 5. The compound has a molar mass of 283. 88 g. What is the compound’s molecular formula? Step 3: Multiply Step 1: Molar Mass P = 2 x 31. 0 g = 62. 0 g O = 5 x 16. 0 g = 80. 0 g 142. 0 g Step 2: Divide MM by Empirical Formula Mass 238. 88 g = 1. 682 = 2 142. 0 g (P 2 O 5)2 = Step 4: Write new formula P 4 O 10

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. 0 g H = 1. 0 g 13. 0 g 78 g/mol 13. 0 g/mol C 6 H 6 =6