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


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1 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 x 100% Mass of 1 mol
2 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
3 Percent Composition Example: What is the percent composition of Potassium Permanganate (KMnO 4 )? Molar Mass of KMnO 4 = 158 g K = 1(39.10) = 39.1 Mn = 1(54.94) = 54.9 O = 4(16.00) = 64.0 MM = 158 % K % Mn 39.1 g K 158 g 54.9 g Mn 158 g % O 64.0 g O 158 g x 100 = 24.7 % x 100 = 34.8 % x 100 = 40.5 %
4 Percent Composition Determine the percentage composition of sodium carbonate (Na 2 CO 3 )? Molar Mass Na = 2(23.00) = 46.0 C = 1(12.01) = 12.0 O = 3(16.00) = 48.0 MM= 106 g Percent Composition % Na = 46.0 g 106 g % C = 12.0 g 106 g % O = 48.0 g 106 g x 100% = 43.4 % x 100% = 11.3 % x 100% = 45.3 %
5 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%
6 Percent Composition Calculate the mass of bromine in 50.0 g of Potassium bromide. 1. Molar Mass of KBr 2. K = 1(39.10) = Br =1(79.90) =79.90 MM = g g = x 50.0g = 33.6 g Br
7 Percent Composition Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C 6 H 14 N 2 O Molar Mass of C 6 H 14 N 2 O 2 C = 6(12.01) = H =14(1.01) = N = 2(14.01) = O = 2(16.00) = MM = g g = x 85.0 mg = 16.3 mg N
8 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.
9 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) = C = 1(12.01) = H = 20(1.01) = 20.2 O = 13(16.00)= MM = Water H = 20(1.01) = 20.2 O = 10(16.00)= MM = or H = 2(1.01) = 2.02 O = 1(16.00) = MM H2O = g g So 10 H 2 O = 10(18.02) = x 100%= %
10 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) = Br = 3(79.90) = H = 12(1.01) = O = 6(16.00) = or MM = Water H = 12(1.01) = 12.1 O = 6(16.00)= MM = MM = For 6 H2O = 6(18.02) = g g x 100%= %
11 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 = g S = 1 x = g O = 4 x = g MM = g H = 2 x 1.01 = 2.02 g O = 1 x = g MM = g MM H 2 O = 7 x g = g Total MM = g g = g 2. % MgSO g g X 100 = % 3. Grams anhydrous MgSO x 125 = 61.1 g
12 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 = g S = 1 x = g O = 4 x = g MM = g H = 2 x 1.01 = 2.02 g O = 1 x = g MM = g MM H 2 O = 5 x g = 90.1 g Total MM = g g = g 2. % CuSO g g X 100 = % 3. Grams anhydrous CuSO x 145 = 92.7 g
13 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 2. Percent of water 5.0 g hydrate g anhydrous salt 0.7 g water 0.7 g water 5.0 g hydrate x 100 = 14 %
14 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 2. Percent of water 7.5 g hydrate g anhydrous salt 2.2 g water 2.2 g water 7.5 g hydrate x 100 = 29 %
15 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 g anhydrous salt 1.1 g water 3. Amount of water 0.22 x = Percent of water 1.1 g water 5.0 g hydrate x 100 = 22 %
16 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 g anhydrous salt 2.1 g water 3. Amount of water 0.28 x = Percent of water 2.1 g water 7.5 g hydrate x 100 = 28 %
17 Empirical and Molecular Formulas
18 Empirical Formula Empirical Formula A formula that gives the simplest wholenumber ratio of the atoms of each element in a compound. Molecular Formula H 2 O 2 C 6 H 12 O 6 CH 3 O CH 3 OOCH = C 2 H 4 O 2 Empirical Formula HO CH 2 O CH 3 O CH 2 O
19 EMPIRICAL FORMULA Mass % of elements Empirical Formula Assume 100g sample Calculate mole ratio Grams of each element Use Atomic Masses Moles of each element
20 What is an empirical formula? A chemical formula in which the ratio of the elements are in the lowest terms is called an empirical formula.
21 Example: The empirical formula for a glucose molecule (C 6 H 12 O 6 ) is CH 2 O. All the subscripts are divisible by six. C 6 H 12 O C H 2 O
22 Exceptions: Some formulas, such as the one for carbon dioxide, CO 2, are already empirical formulas without being reduced.
23 Determine the empirical formula for a compound containing g Cl and g Ca. Steps 1. Find mole amounts. 2. Divide each mole by the smallest mole.
24 1. Find mole amounts g Cl x 1 mol Cl = mol Cl g Cl g Ca x 1 mol Ca = mol Ca g Ca
25 2. Divide each mole by the smallest mole. Cl = mol Cl = 2.00 mol Cl Ca = mol Ca = 1.00 mol Ca Ratio 1 Ca: 2 Cl Empirical Formula = CaCl 2
26 A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Hint Percent to mass Mass to mole Divide by small Multiply til whole
27 A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Percent to mass: Mg (72.2%/100)* g = g N (27.8%/100)* g = g Mass to mole: Mg g * ( 1 mole ) = 8.86 mole 24.3 g N g * ( 1 mole ) = 5.92 mole g Divide by small: Mg mole/5.92 mole = 1.50 N mole/5.92 mole = 1.00 mole Multiply til whole: Mg 1.50 x 2 = 3.00 N 1.00 x 2 = 2.00 Mg 3 N 2
28 Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Mass. 3. Divide the molar mass by the EFM. 4. Multiply empirical formula by factor. Find the molecular formula for a compound whose molar mass is ~ and empirical formula is CH 2 O EFM = g /62.03 = (CH 2 O 3 ) = C 2 H 4 O 6
29 What is a molecular formula? A molecular formula is the true formula of a compound. The chemical formula for a molecular compound shows the actual number of atoms present in a molecule.
30 To find the molecular formula from the empirical formula: Find the empirical formula. Determine the empirical formula mass. Divide the molecular mass by the empirical formula mass to determine the multiple. Multiply the empirical formula by the multiple to find the molecular formula. MF mass EF mass = n (EF)n = molecular formula
31 EXAMPLE: The empirical formula for ethylene is CH 2. molecular formula if the molecular mass is 28.1g/mol. C = 1 x 12 = 12 H = 2 x 1 = +2 14g/mol = empirical formula mass Find the 28.1 g/mol = 2 14 g/mol (CH 2 ) 2 C 2 H 4
32 Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Mass. 3. Divide the molar mass by the EFM. 4. Multiply empirical formula by factor.
33 Empirical formula. A. Find mole amounts g N x 1 mol N = mol N g N 11.2 g O x 1 mol O = mol O g O
34 B. Divide each mole by the smallest mole. N = = 1.00 mol N O = = 2.00 mol O Empirical Formula = NO 2 Empirical Formula Mass = g/mol
35 Molecular formula Molar Mass = 92.0 g/mol = 2.00 Emp. Formula Mass g/mol Molecular Formula = 2 x Emp. Formula = N 2 O 4
36 A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula?
37 A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? g C (48.38/100)* g = g g H (8.12/100)* g = g g O (43.5/100)* g = g mole C g * ( 1 mole ) = mol g mole H g * ( 1 mole ) = mol 1.01 g mole O g * ( 1 mole ) = mol g
38 A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? From last slide: mol C, mol H, mol O C 21.29/14.27 = 1.49 H 42.49/14.27 = 2.98 (esentially 3) O 14.27/14.27 = 1.00 C 1.49 x 2 = 3 H 3 x 2 = 6 C 3 H 6 O 2 O 1 x 2 = 2
39 A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? From last slide: Empirical formula = C 3 H 6 O 2 EFM = Molar mass = = ~3 EFM (C 3 H 6 O 2 ) = C 9 H 18 O 6
40 Proportional Relationships 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. I have 5 eggs. How many cookies can I make? 5 eggs 5 doz. 2 eggs Courtesy Christy Johannesson Ratio of eggs to cookies = 12.5 dozen cookies
41 Proportional Relationships Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2 2 MgO Courtesy Christy Johannesson
42 Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio  moles moles Mole Molar ratio mass   moles moles moles grams Molarity  moles liters soln Molar volume  moles liters gas Core step in all stoichiometry problems!! 4. Check answer. Courtesy Christy Johannesson
43 Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0 C and 1 atm Courtesy Christy Johannesson
44 Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS Molar Mass (g/mol) MOLES particles/mol NUMBER OF PARTICLES Molarity (mol/l) LITERS OF SOLUTION Courtesy Christy Johannesson
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