 # Percent Composition, Empirical and Molecular Formulas Courtesy www.lab-initio.com.

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Percent Composition, Empirical and Molecular Formulas Courtesy www.lab-initio.com

Percent Composition Composition can be expressed in 3 ways 1.In terms of # of atoms (the formula) 2.In terms of mass per mole 3.In terms of mass of each element relative to total mass of compound

3 Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass.Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass. The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage.The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage. Percentage Composition

Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. Formula mass of magnesium carbonate: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

5 Percentage Composition Glucose has the molecular formula C 6 H 12 O 6. What is its empirical formula, and what is the percentage composition of glucose?Glucose has the molecular formula C 6 H 12 O 6. What is its empirical formula, and what is the percentage composition of glucose? Empirical Formula = smallest whole number ratio CH 2 O

6 Percentage Composition CH 2 O Total mass = 12.01 + 2.02 + 16.00 = 30.03 %C = 12.01/30.03 x 100% = 39.99% %H = 2.02/30.03 x 100% = 6.73% %O = 16.00/30.03 x 100% = 53.28%

7 Percentage Composition Saccharin has the molecular formula C 7 H 5 NO 3 S. What is its empirical formula, and what is the percentage composition of saccharin? Saccharin has the molecular formula C 7 H 5 NO 3 S. What is its empirical formula, and what is the percentage composition of saccharin? Empirical Formula is same as molecular formula MW = 183.19 g/mole %C = (7 x 12.011)/183.19 x 100% = 45.89% etc.

Formulas  molecular formula = (empirical formula) n  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11

11 Empirical Formula The empirical formula gives the ratio of the number of atoms of each element in a compound.The empirical formula gives the ratio of the number of atoms of each element in a compound. CompoundFormulaEmpirical Formula Hydrogen peroxideH 2 O 2 OH BenzeneC 6 H 6 CH EthyleneC 2 H 4 CH 2 PropaneC 3 H 8 C 3 H 8

The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula the actual ratio of elements in a compound The two can be the same. CH 2 empirical formula C 2 H 4 molecular formula C 3 H 6 molecular formula H 2 O both

Calculating Empirical Just find the lowest whole number ratioJust find the lowest whole number ratio C 6 H 12 O 6C 6 H 12 O 6 CH 4 NCH 4 N It is not just the ratio of atoms, it is also the ratio of moles of atomsIt is not just the ratio of atoms, it is also the ratio of moles of atoms In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygenIn 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen In one molecule of CO 2 there is 1 atom of C and 2 atoms of OIn one molecule of CO 2 there is 1 atom of C and 2 atoms of O

Calculating Empirical Pretend that you have a 100 gram sample of the compound.Pretend that you have a 100 gram sample of the compound. That is, change the % to grams if given %.That is, change the % to grams if given %. Convert the grams to moles for each element.Convert the grams to moles for each element. Write the number of moles as a subscript in a chemical formula.Write the number of moles as a subscript in a chemical formula. Divide each number by the least number.Divide each number by the least number. Multiply the result to get rid of any fractions.Multiply the result to get rid of any fractions.

Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g soAssume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 g C38.67 g C x 1mol C = 3.220 mole C 12.01 g C 16.22 g H x 1mol H = 16.09 mole H 1.01 g H16.22 g H x 1mol H = 16.09 mole H 1.01 g H 45.11 g N x 1mol N = 3.219 mole N 14.01 g N45.11 g N x 1mol N = 3.219 mole N 14.01 g N

3.220 mole C3.220 mole C 16.09 mole H16.09 mole H 3.219 mole N3.219 mole N C 3.22 H 16.09 N 3.219 If we divide all of these by the smallest one It will give us the empirical formula

Example The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol NThe ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol NThe ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N C 1 H 5 N 1 is the empirical formulaC 1 H 5 N 1 is the empirical formula

43.6 g P x 1mol P = 1.4 mole P 30.97 gP43.6 g P x 1mol P = 1.4 mole P 30.97 gP 56.36 g O x 1mol O = 3.5 mole O 16 gO56.36 g O x 1mol O = 3.5 mole O 16 gO P 1.4 O 3.5 A compound is 43.64 % P and 56.36 % O. What is the empirical formula?

Divide both by the lowest one The ratio is 3.52 mol O = 2.5 mol O 1.42 mol P 1 mol PThe ratio is 3.52 mol O = 2.5 mol O 1.42 mol P 1 mol P P 1.4 O 3.5 P 1 O 2.5 Multiply the result to get whole numbers (get rid of fractions). P 1 O 2.5 2 X = P 2 O 5

20 Empirical Formula A compound’s empirical formula can be determined from its percent composition.A compound’s empirical formula can be determined from its percent composition. A compound’s molecular formula is determined from the molar mass and empirical formula.A compound’s molecular formula is determined from the molar mass and empirical formula.

Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1.Treat % as mass, and convert grams to moles

Empirical Formula Determination 2. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

Empirical Formula Determination 3. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 352 Empirical formula: C3H5O2C3H5O2

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular 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 molecular mass by the mass given by the empirical formula.

Finding the Molecular Formula 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4