1. Percent Composition, Empirical Formulas, Molecular Formulas

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

3. Percent Composition Example: What is the percent composition of Potassium Permanganate (KMnO 4 )? Molar Mass of KMnO 4 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 (KMnO 4 )? = 158 g % K Molar Mass of KMnO 4 39.1 g K 158 g x 100 =24.7 % % Mn 54.9 g Mn 158 g x 100 = 34.8 % % 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

5. Percent Composition Determine the percentage composition of sodium carbonate (Na 2 CO 3 )? Molar MassPercent 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= 106 g

6. 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%

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 79.90 g ___________ 119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.

8. Percent Composition Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C 6 H 14 N 2 O 2. 1. Molar Mass of C 6 H 14 N 2 O 2 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

9. Hydrates Hydrated salt – salt that has water molecules trapped within the crystal lattice Examples: CuSO 4 5H 2 O, CuCl 2 2H 2 O Anhydrous salt – salt without water molecules Examples: CuCl 2 Can calculate the percentage of water in a hydrated salt.

10. Percent Composition Calculate the percentage of water in sodium carbonate decahydrate, Na 2 CO 3 10H 2 O. 1. Molar Mass of Na 2 CO 3 10H 2 O 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 H 2 O = 10(18.02) = 180.2

11. Percent Composition Calculate the percentage of water in Aluminum bromide hexahydrate, AlBr 3 6H 2 O. 1. Molar Mass of AlBr 3 6H 2 O 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

12. Percent Composition If 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain? MgSO 4. 7 H 2 O 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 H 2 O = 7 x 18.02 g = 126.1 g Total MM = 120.4 g + 126.1 g = 246.5 g 2. % MgSO 4 120.4 g 246.5 g X 100 =48.84 % 3. Grams anhydrous MgSO 4 0.4884 x 125 =61.1 g

13. Percent Composition If 145 grams of copper (II) sulfate pentahydrate is completely dehydrated, how many grams of anhydrous copper sulfate will remain? CuSO 4. 5 H 2 O 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 H 2 O = 5 x 18.02 g = 90.1 g Total MM = 159.6 g + 90.1 g = 249.7 g 2. % CuSO 4 159.6 g 249.7 g X 100 =63.92 % 3. Grams anhydrous CuSO 4 0.6392 x 145 =92.7 g

14. Percent Composition A 5.0 gram sample of a hydrate of BaCl 2 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 %

15. Percent Composition A 7.5 gram sample of a hydrate of CuCl 2 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 %

16. Percent Composition A 5.0 gram sample of Cu(NO 3 ) 2 nH 2 O 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 2. Percent of water 1.1 g water 5.0 g hydrate x 100 = 22 % 3. Amount of water 0.22 x 18.02 =4.0

17. Percent Composition A 7.5 gram sample of CuSO 4 nH 2 O 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 2. Percent of water 2.1 g water 7.5 g hydrate x 100 = 28 % 3. Amount of water 0.28 x 18.02 =5.0

18. 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: C 4 H 10 - molecular C 2 H 5 - empirical C 6 H 12 O 6 - molecular CH 2 O - empirical

19. Formulas Is H 2 O 2 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. 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

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 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, Al 2 O 3

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. 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 Co0.1544 mol Cl 0.07721 = 2= 1 CoCl 2

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. 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

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. 1.3813 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

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 0.00667 mol H 0.006667 mol As 0.02667 mol O 0.006667 = 1.000 mol Pb = 1.00 mol H = 1.000 mol As = 4.000 mol O PbHAsO 4

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 C1 mol C 12.01 g C = 5.302 mol C 12.38 g N1 mol N 14.01 g N = 0.8837 mol N 9.80 g H1 mol H 1.01 g H = 9.72 mol H 14.14 g O1 mol O 16.00 g O = 0.8832 mol 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 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 C 6 NH 11 O

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 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 (P 2 O 5 ) 2 = P 4 O 10

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? C = 12.01 g H = 1.01 g 13.01 g 78 g/mol 13.01 g/mol = 6 (CH) 6 = C 6 H 6