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Percent Composition, Empirical and Molecular Formulas What Is Percent Composition? A relative measure of the mass of each different element present in a compound. Obviously, it is expressed as a percent (%). It can be determined regardless of the # of elements in the compound. The sum of all the % values MUST equal 100. Why Is % Comp Important? Sometimes it is important to know what percent of a compound’s mass is due to a certain element. A mining company may wish to know what % of 400 tons of Fe2O3 is due to iron. They can then work out if it is worth trying to recover the iron from the sample of iron oxide. STEPS TO Calculate % Comp 1. 2. 3. 4. Calculate the mass of an element Calculate the molar mass of the compound. Divide the mass of the element by the molar mass of the substance. Multiply by 100 to make it a percentage. Calculating % Composition Calculate the % composition of magnesium carbonate, MgCO3 Molar mass of magnesium carbonate: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 24.31 Mg 100 28.83% 84.32 12.01 C 10014.24% 84.32 48.00 O 100 56.93% 84.32 100.00 Percent Composition Percent = mass of element in 1 mole x 100 by mass molar mass of compound Let’s try some 1. What is the percent composition by mass of carbon in carbon dioxide (CO2)? C = 12.011 x 1 atom = 12.011 g O = 15.999 x 2 atoms = 31.998 g Total Mass = 44.009 g Percent comp of carbon = = 27.292 % of carbon 12.011 g x 100% 44.009 g 2. What is the percent composition of both hydrogen and oxygen in water? H = 1.008 x 2 atoms = 2.016 g O = 15.999 x 1 atoms = 15.999 g Total Mass = 18.015 g Percent Comp H: 2.016/18.015 = 11.19 % Percent Comp O: 15.999/18.015 = 88.81 % Your percentage for the entire compound should always add to 100 percent! Let’s practice Complete Percent Comp Worksheet Complete Percent composition POGIL Determining Formulas from Percent Composition Definitions of Formulas Molecular formula: the true number of atoms of each element in the formula of a compound. Empirical formula: the lowest whole number ratio of atoms in a compound. molecular formula = C6H6 = (CH)6 empirical formula = CH Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl Al2(SO4)3 MgCl2 K2CO3 Formulas (continued) Formulas for covalent (molecular) compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11 Let’s start with Calculating Empirical Formulas from Percent Composition! Empirical Formula Determination In the problem, you will be given the % comp of the compound. 1.Base your calculation on 100 grams of compound. 2.Determine the # of moles of each element in 100 grams of the compound. (use molar mass) 3.Divide each value of moles by the smallest of the mole values. 4.Multiply each number by an integer to obtain all whole numbers. Rhyme to help you remember 1.Percent to mass 2.Mass to mole 3.Divide by small 4.Multiple ‘til whole Problem 1: Empirical Formula Calculation 1. Cinnamon contains cinnamaldehyde. A molecule of cinnamaldehyde contains 81.79% carbon, 6.10% hydrogen, and 12.11% oxygen. Determine the molecules empirical formula. Empirical Formula Calculation Step 1: Percent to mass C: 81.79% 81.79 g H: 6.10% 6.10 g O: 12.11% 12.11 g Empirical Formula Calculation Step 2: Mass to Mole C 81.79g x 1 mole/12g = 6.61 mol H 6.10g x 1 mole/1g = 6.05 mol O 12.11g x 1 mole/16g =0.757 mol Empirical Formula Calculation Step 3: Divide by small C 6.61 mol/ 0.757 mol = 9 H 6.05 mol/ 0.757 mol = 8 O 0.757 mol/ 0.757 mol = 1 Empirical Formula Calculation Step 4: Multiple ‘til whole C9 H8 O (in this case, we didn’t have to) Problem 2: Empirical Formula Calculation 2. Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% carbon, 8.16 hydrogen, and 43.20% oxygen. Empirical Formula Calculation Step 1: Percent to mass C: 48.64% H: 8.16 O: 43.20% Empirical Formula Calculation Step 2: Mass to Mole C 48.64g x 1 mole/12g = H 8.16g x 1 mole/1g = O 43.20g x 1 mole/16g = Empirical Formula Calculation Step 3: Divide by small C 4.050 mol / 2.70 mol = 1.5 H 8.10 mol / 2.70 mol = 3 O 2.70 mol / 2.70 mol = 1 Empirical Formula Calculation Step 4: Multiple ‘til whole C 1.5 x 2 = 3 H3x2=6 O1x2=2 C3H6O2 Problem 3: Empirical Formula Determination 3. Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? Problem 3: Empirical Formula Determination 1. Percent to mass and 2. mass to mole 49.32 g carbon 1 m olcarbon 4.107m olcarbon 12.01g carbon 6.85 g hydrogen 1 m olhydrogen 6.78m olhydrogen 1.01g hydrogen 43.84 g oxygen 1 m oloxygen 2.74 m oloxygen 16.00 g oxygen Empirical Formula Determination 3. Divide by small Carbon: 4.107m olcarbon 1.50 2.74 m ol Hydrogen: 6.78m olhydrogen 2.74 m ol Oxygen: 2.47 2.74 m oloxygen 1.00 2.74 m ol Empirical Formula Determination 4. Multiple ‘til whole Carbon: 1.50 x 2 3 Hydrogen: 2.50 x 2 5 Empirical formula: C3H5O2 Oxygen: 1.00 x 2 2 STOP Work with a partner to complete worksheet: Determining Empirical Formulas Now let’s calculate Molecular Formulas from Percent Composition! Molecular Formula Determination In these problems, either you will be given the empirical formula or you will be asked to calculate the empirical formula. You will also be given the molar mass of the compound. Molecular Formula Determination 1. 2. 3. Calculate the molar mass of the empirical formula. Divide the given molar mass by the molar mass of the empirical formula. Multiply the empirical formula by this number to get the molecular formula. Given molar mass = number of times Empirical molar mass Finding the Molecular Formula 1. The empirical formula for adipic acid is C3H5O2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the molar mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 2. Divide the molar mass by the mass given by the empirical formula 146 2 73 Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 146 2 73 (C3H5O2) x 2 = C6H10O4 3. Multiply the empirical formula by this number to get the molecular formula Pulling it all together 2. Naphthalene, commonly found in mothballs, is composed of 93.7% carbon and 6.3% hydrogen. The molar mass of naphthalene is 128 g/mole. Determine the empirical and molecular formulas for naphthalene. Pulling it all together 3. Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% carbon, 5.08% hydrogen, and 54.24% oxygen and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid. STOP Work on completing worksheet: Determining Molecular Formulas (True)