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
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
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 %
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 %
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%
Percent Composition Calculate the mass of bromine in 50.0 g of Potassium bromide. 1. Molar Mass of KBr 2. 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
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 N = 2(14.01) = 28.02 O = 2(16.00) = 32.00 MM = 146.2 2. 28.02 g 146.2 g = 0.192 3. 0.192 x 85.0 mg = 16.3 mg N
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.
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 H = 20(1.01) = 20.2 O = 13(16.00)= 208.00 2. MM = 286.2 Water H = 20(1.01) = 20.2 O = 10(16.00)= 160.00 MM = 180.2 or H = 2(1.01) = 2.02 O = 1(16.00) = 16.00 MM H2O = 18.02 3. 180.2 g 286.2 g So 10 H 2 O = 10(18.02) = 180.2 x 100%= 67.97 %
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 H = 12(1.01) = 12.12 O = 6(16.00) = 96.00 2. or MM = 374.8 Water H = 12(1.01) = 12.1 O = 6(16.00)= 96.00 MM = 108.1 3. MM = 18.02 For 6 H2O = 6(18.02) = 108.2 108.1 g 374.8 g x 100%= 28.85 %
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
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
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 - 4.3 g anhydrous salt 0.7 g water 0.7 g water 5.0 g hydrate x 100 = 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 - 5.3 g anhydrous salt 2.2 g water 2.2 g water 7.5 g hydrate x 100 = 29 %
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 3. Amount of water 0.22 x 18.02 = 4.0 2. Percent of water 1.1 g water 5.0 g hydrate x 100 = 22 %
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 3. Amount of water 0.28 x 18.02 = 5.0 2. Percent of water 2.1 g water 7.5 g hydrate x 100 = 28 %
Empirical and Molecular Formulas
Empirical Formula Empirical Formula A formula that gives the simplest whole-number 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
EMPIRICAL FORMULA Mass % of elements Empirical Formula Assume 100g sample Calculate mole ratio Grams of each element Use Atomic Masses Moles of each element
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.
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 6 6 6 6 C H 2 O
Exceptions: Some formulas, such as the one for carbon dioxide, CO 2, are already empirical formulas without being reduced.
Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. Steps 1. Find mole amounts. 2. Divide each mole by the smallest mole.
1. Find mole amounts. 2.128 g Cl x 1 mol Cl = 0.0600 mol Cl 35.45 g Cl 1.203 g Ca x 1 mol Ca = 0.0300 mol Ca 40.08 g Ca
2. Divide each mole by the smallest mole. Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300 Ca = 0.0300 mol Ca = 1.00 mol Ca 0.0300 Ratio 1 Ca: 2 Cl Empirical Formula = CaCl 2
A compound weighing 298.12 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
A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Percent to mass: Mg (72.2%/100)*298.12 g = 215.24 g N (27.8%/100)*298.12 g = 82.88 g Mass to mole: Mg 215.24 g * ( 1 mole ) = 8.86 mole 24.3 g N 82.88 g * ( 1 mole ) = 5.92 mole 14.01 g Divide by small: Mg - 8.86 mole/5.92 mole = 1.50 N - 5.92 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
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 ~124.06 and empirical formula is CH 2 O 3. 2. EFM = 62.03 g 3. 124.06/62.03 = 2 4. 2(CH 2 O 3 ) = C 2 H 4 O 6
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.
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
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
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.
Empirical formula. A. Find mole amounts. 4.90 g N x 1 mol N = 0.350 mol N 14.01 g N 11.2 g O x 1 mol O = 0.700 mol O 16.00 g O
B. Divide each mole by the smallest mole. N = 0.350 = 1.00 mol N 0.350 O = 0.700 = 2.00 mol O 0.350 Empirical Formula = NO 2 Empirical Formula Mass = 46.01 g/mol
Molecular formula Molar Mass = 92.0 g/mol = 2.00 Emp. Formula Mass 46.01 g/mol Molecular Formula = 2 x Emp. Formula = N 2 O 4
A 528.39 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 ~222.25 g/mol. What is its molecular formula?
A 528.39 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 ~222.25 g/mol. What is its molecular formula? g C (48.38/100)*528.39 g = 255.64 g g H (8.12/100)*528.39 g = 42.91 g g O (43.5/100)*528.39 g = 229.85 g mole C - 255.64 g * ( 1 mole ) = 21.29 mol 12.01 g mole H 42.91 g * ( 1 mole ) = 42.49 mol 1.01 g mole O 229.85 g * ( 1 mole ) = 14.37 mol 16.00 g
A 528.39 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 ~222.25 g/mol. What is its molecular formula? From last slide: 21.29 mol C, 42.49 mol H, 14.27 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
A 528.39 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 ~222.25 g/mol. What is its molecular formula? From last slide: Empirical formula = C 3 H 6 O 2 EFM = 74.09 Molar mass = 222.24 = ~3 EFM 74.09 3(C 3 H 6 O 2 ) = C 9 H 18 O 6
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 www.nisd.net/communicationsarts/pages/chem Ratio of eggs to cookies = 12.5 dozen cookies
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 www.nisd.net/communicationsarts/pages/chem
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 www.nisd.net/communicationsarts/pages/chem
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 www.nisd.net/communicationsarts/pages/chem
Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS Molar Mass (g/mol) MOLES 6.02 10 23 particles/mol NUMBER OF PARTICLES Molarity (mol/l) LITERS OF SOLUTION Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem