Percent Composition, Empirical Formulas, Molecular
Description: Percent Composition, Empirical Formulas, Molecular Formulas Percent Composition Percent Composition the percentage by mass of each element in a compound (note: not the ratio of elements in the compound) Law of Definite Proportions
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slide1. Percent Composition, Empirical Formulas, Molecular Formulas<br>
slide2. Percent Composition Percent Composition – the percentage by mass of each element in a compound (*note: not the ratio of elements in the compound)
Law of Definite Proportions – compounds always contain the same proportions of elements by mass. Percent = _______ Part Whole x 100% So… Percent composition
of a compound or =
molecule Mass of element in 1 mol ____________________ Mass of 1 mol of compound x 100%<br>
slide3. 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/mol<br>
slide4. Percent Composition Example: What is the percent composition of Potassium Permanganate (KMnO4)? (assume a 1.0mol sample) = 158 g/mol % K Molar Mass of KMnO4 39.1 g K 158 g x 100 = 24.7 % % Mn 54.9 g Mn 158 g x 100 = 34.7 % % O 64.0 g O 158 g x 100 = 40.5 % K = 1(39.10) = 39.1 Mn = 1(54.94) = 54.9 O = 4(16.00) = 64.0 MM = 158<br>
slide5. Percent Composition Determine the percentage composition of sodium carbonate (Na2CO3)? Molar Mass Percent Composition % Na = 46.0 g 106 g x 100% = 43.4 % % C = 12.0 g 106 g x 100% = 11.3 % % O = 48.0 g 106 g x 100% = 45.3 % Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= 106g/mol<br>
slide6. 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%<br>
slide7. 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 79.90 g ___________ 119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.<br>
slide8. 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 MM = 146.2 28.02 g ___________ 146.2 g = 0.192 3. 0.192 x 85.0 mg = 16.3 mg N 2. N = 2(14.01) = 28.02 O = 2(16.00) = 32.00<br>
slide9. Homework Read section 6.6
Page 286 #1,2
Page 287 #3
Page 288 # 1-5,7,8<br>
slide10. 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.<br>
slide11. 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 MM = 286.2 H = 20(1.01) = 20.2 O = 13(16.00)= 208.00 H = 20(1.01) = 20.2 Water O = 10(16.00)= 160.00 MM = 180.2 2. 3. 180.2 g _______ 286.2 g 67.97 % x 100%= or H = 2(1.01) = 2.02 O = 1(16.00) = 16.00 MM H2O = 18.02 So…
10 H2O = 10(18.02) = 180.2<br>
slide12. 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 MM = 374.8 H = 12(1.01) = 12.12 O = 6(16.00) = 96.00 H = 12(1.01) = 12.1 Water O = 6(16.00)= 96.00 MM = 108.1 2. 3. 108.1 g _______ 374.8 g 28.85 % x 100%= or MM = 18.02
For 6 H2O = 6(18.02) = 108.2<br>
slide13. 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 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 H = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g
MM = 18.02 g MM H2O =
7 x 18.02 g = 126.1 g Total MM =
120.4 g + 126.1 g = 246.5 g 2. % MgSO4 120.4 g 246.5 g X 100 = 48.84 % 3. Grams anhydrous MgSO4 0.4884 x 125 = 61.1 g<br>
slide14. 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 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 H = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g
MM = 18.02 g MM H2O =
5 x 18.02 g = 90.1 g Total MM =
159.6 g + 90.1 g = 249.7 g 2. % CuSO4 159.6 g 249.7 g X 100 = 63.92 % 3. Grams anhydrous CuSO4 0.6392 x 145 = 92.7 g<br>
slide15. 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? 1. Amount water lost 5.0 g hydrate
4.3 g anhydrous salt
0.7 g water 2. Percent of water 0.7 g water 5.0 g hydrate x 100 = 14 %<br>
slide16. 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? 1. Amount water lost 7.5 g hydrate
5.3 g anhydrous salt
2.2 g water 2. Percent of water 2.2 g water 7.5 g hydrate x 100 = 29 %<br>
slide17. Formulas 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 Percent composition allow you to calculate the simplest ratio among the atoms found in compound. Examples: C4H10 - molecular C2H5 - empirical C6H12O6 - molecular CH2O - empirical<br>
slide18. Formulas Is H2O2 an empirical or molecular formula? Molecular, it can be reduced to HO
HO = empirical formula<br>
slide19. 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 26.98 g Al = 0.1539 mol Al 3.692 g O 1 mol O 16.00 g O = 0.2308 mol O<br>
slide20. 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 0.1539 = 1.000 mol Al 0.2308 moles O 0.1539 = 1.500 mol O 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<br>
slide21. 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 58.93 g Co = 0.07721 mol Co 5.475 g Cl 1 mol Cl 35.45 g Cl = 0.1544 mol Cl 0.07721 mol Co 0.1544 mol Cl 0.07721 0.07721 = 2 = 1 CoCl2<br>
slide22. 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. 2.000 g Fe 1 mol Fe 55.85 g Fe = 0.03581 mol Fe 0.573 g O 1 mol O 16.00 g = 0.03581 mol Fe Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g 1 : 1 FeO<br>
slide23. 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. 1.3813 g Pb 1 mol Pb 207.2 g Pb = 0.006667 mol Pb 0.00672 gH 1 mol H 1.008 g H = 0.00667 mol H 0.4995 g As 1 mol As 74.92 g As = 0.006667 mol As 0.4267g Fe 1 mol O 16.00 g O = 0.02667 mol O<br>
slide24. 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 0.00667 mol H 0.006667 mol As 0.02667 mol O 0.006667 0.006667 0.006667 0.006667 = 1.000 mol Pb = 1.00 mol H = 1.000 mol As = 4.000 mol O PbHAsO4<br>
slide25. 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 12.01 g C = 5.302 mol C 12.38 g N 1 mol N 14.01 g N = 0.8837 mol N 9.80 g H 1 mol H 1.01 g H = 9.72 mol H 14.14 g O 1 mol O 16.00 g O = 0.8832 mol O<br>
slide26. 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 0.8837 = 6.000 mol C 0.8837 mol N 0.8837 = 1.000 mol N 9.72 mol H 0.8837 = 11.0 mol H 0.8837 mol O 0.8837 = 1.000 mol O 6:1:11:1 C6NH11O<br>
slide27. Homework Read section 6.7
Page 292 #1
Page 293 #2-9<br>
slide28. 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 1: Molar Mass P = 2 x 30.97 g = 61.94g
O = 5 x 16.00g = 80.00 g
141.94 g Step 2: Divide MM by
Empirical Formula Mass 238.88 g 141.94g = 2 Step 3: Multiply (P2O5)2 = P4O10<br>
slide29. Calculating Molecular Formula A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula? C = 12.01 g
H = 1.01 g
13.01 g 78 g/mol 13.01 g/mol = 6 (CH)6 = C6H6<br>
slide2. Percent Composition Percent Composition – the percentage by mass of each element in a compound (*note: not the ratio of elements in the compound)
Law of Definite Proportions – compounds always contain the same proportions of elements by mass. Percent = _______ Part Whole x 100% So… Percent composition
of a compound or =
molecule Mass of element in 1 mol ____________________ Mass of 1 mol of compound x 100%<br>
slide3. 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/mol<br>
slide4. Percent Composition Example: What is the percent composition of Potassium Permanganate (KMnO4)? (assume a 1.0mol sample) = 158 g/mol % K Molar Mass of KMnO4 39.1 g K 158 g x 100 = 24.7 % % Mn 54.9 g Mn 158 g x 100 = 34.7 % % O 64.0 g O 158 g x 100 = 40.5 % K = 1(39.10) = 39.1 Mn = 1(54.94) = 54.9 O = 4(16.00) = 64.0 MM = 158<br>
slide5. Percent Composition Determine the percentage composition of sodium carbonate (Na2CO3)? Molar Mass Percent Composition % Na = 46.0 g 106 g x 100% = 43.4 % % C = 12.0 g 106 g x 100% = 11.3 % % O = 48.0 g 106 g x 100% = 45.3 % Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= 106g/mol<br>
slide6. 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%<br>
slide7. 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 79.90 g ___________ 119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.<br>
slide8. 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 MM = 146.2 28.02 g ___________ 146.2 g = 0.192 3. 0.192 x 85.0 mg = 16.3 mg N 2. N = 2(14.01) = 28.02 O = 2(16.00) = 32.00<br>
slide9. Homework Read section 6.6
Page 286 #1,2
Page 287 #3
Page 288 # 1-5,7,8<br>
slide10. 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.<br>
slide11. 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 MM = 286.2 H = 20(1.01) = 20.2 O = 13(16.00)= 208.00 H = 20(1.01) = 20.2 Water O = 10(16.00)= 160.00 MM = 180.2 2. 3. 180.2 g _______ 286.2 g 67.97 % x 100%= or H = 2(1.01) = 2.02 O = 1(16.00) = 16.00 MM H2O = 18.02 So…
10 H2O = 10(18.02) = 180.2<br>
slide12. 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 MM = 374.8 H = 12(1.01) = 12.12 O = 6(16.00) = 96.00 H = 12(1.01) = 12.1 Water O = 6(16.00)= 96.00 MM = 108.1 2. 3. 108.1 g _______ 374.8 g 28.85 % x 100%= or MM = 18.02
For 6 H2O = 6(18.02) = 108.2<br>
slide13. 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 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 H = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g
MM = 18.02 g MM H2O =
7 x 18.02 g = 126.1 g Total MM =
120.4 g + 126.1 g = 246.5 g 2. % MgSO4 120.4 g 246.5 g X 100 = 48.84 % 3. Grams anhydrous MgSO4 0.4884 x 125 = 61.1 g<br>
slide14. 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 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 H = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g
MM = 18.02 g MM H2O =
5 x 18.02 g = 90.1 g Total MM =
159.6 g + 90.1 g = 249.7 g 2. % CuSO4 159.6 g 249.7 g X 100 = 63.92 % 3. Grams anhydrous CuSO4 0.6392 x 145 = 92.7 g<br>
slide15. 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? 1. Amount water lost 5.0 g hydrate
4.3 g anhydrous salt
0.7 g water 2. Percent of water 0.7 g water 5.0 g hydrate x 100 = 14 %<br>
slide16. 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? 1. Amount water lost 7.5 g hydrate
5.3 g anhydrous salt
2.2 g water 2. Percent of water 2.2 g water 7.5 g hydrate x 100 = 29 %<br>
slide17. Formulas 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 Percent composition allow you to calculate the simplest ratio among the atoms found in compound. Examples: C4H10 - molecular C2H5 - empirical C6H12O6 - molecular CH2O - empirical<br>
slide18. Formulas Is H2O2 an empirical or molecular formula? Molecular, it can be reduced to HO
HO = empirical formula<br>
slide19. 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 26.98 g Al = 0.1539 mol Al 3.692 g O 1 mol O 16.00 g O = 0.2308 mol O<br>
slide20. 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 0.1539 = 1.000 mol Al 0.2308 moles O 0.1539 = 1.500 mol O 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<br>
slide21. 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 58.93 g Co = 0.07721 mol Co 5.475 g Cl 1 mol Cl 35.45 g Cl = 0.1544 mol Cl 0.07721 mol Co 0.1544 mol Cl 0.07721 0.07721 = 2 = 1 CoCl2<br>
slide22. 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. 2.000 g Fe 1 mol Fe 55.85 g Fe = 0.03581 mol Fe 0.573 g O 1 mol O 16.00 g = 0.03581 mol Fe Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g 1 : 1 FeO<br>
slide23. 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. 1.3813 g Pb 1 mol Pb 207.2 g Pb = 0.006667 mol Pb 0.00672 gH 1 mol H 1.008 g H = 0.00667 mol H 0.4995 g As 1 mol As 74.92 g As = 0.006667 mol As 0.4267g Fe 1 mol O 16.00 g O = 0.02667 mol O<br>
slide24. 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 0.00667 mol H 0.006667 mol As 0.02667 mol O 0.006667 0.006667 0.006667 0.006667 = 1.000 mol Pb = 1.00 mol H = 1.000 mol As = 4.000 mol O PbHAsO4<br>
slide25. 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 12.01 g C = 5.302 mol C 12.38 g N 1 mol N 14.01 g N = 0.8837 mol N 9.80 g H 1 mol H 1.01 g H = 9.72 mol H 14.14 g O 1 mol O 16.00 g O = 0.8832 mol O<br>
slide26. 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 0.8837 = 6.000 mol C 0.8837 mol N 0.8837 = 1.000 mol N 9.72 mol H 0.8837 = 11.0 mol H 0.8837 mol O 0.8837 = 1.000 mol O 6:1:11:1 C6NH11O<br>
slide27. Homework Read section 6.7
Page 292 #1
Page 293 #2-9<br>
slide28. 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 1: Molar Mass P = 2 x 30.97 g = 61.94g
O = 5 x 16.00g = 80.00 g
141.94 g Step 2: Divide MM by
Empirical Formula Mass 238.88 g 141.94g = 2 Step 3: Multiply (P2O5)2 = P4O10<br>
slide29. Calculating Molecular Formula A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula? C = 12.01 g
H = 1.01 g
13.01 g 78 g/mol 13.01 g/mol = 6 (CH)6 = C6H6<br>