Percent Composition and Molecular Vs. Empirical Formulas Videodisk Unit 4

Percent Composition (always by mass) Divide mass of each element by total compound mass and multiply by 100 Can be found experimentally or from the formula itself Can be used to find the empirical formula of a compound

Sample Problem C: 0.09480 g/0.2370 g x 100 = 40.00 % A 0.2370 g sample of an unknown compound is extracted from the roots of a plant. Decomposition of the sample produces 0.09480 g of carbon, 0.1264 g of oxygen, and 0.0158 g of hydrogen. What is the percent composition of the compound? C: 0.09480 g/0.2370 g x 100 = 40.00 % O: 0.1264 g/0.2370 g x 100 = 53.33 % H: 0.0158/0.2370 g x 100 = 6.67 %

Empirical & Molecular Formulas Empirical formula is the simplest whole number ratio of a compound. Molecular formula shows the actual number of atoms in the molecule. The molecular formula is some multiple of the empirical formula. Ex. C6H12O6 is glucose’s molecular formula CH2O is glucose’s empirical formula

Empirical & Molecular Formulas C6H12O6 is glucose’s molecular formula It means in one mole of glucose you have 6 moles of Carbon, 12 moles of hydrogen, and 6 moles of oxygen CH2O is glucose’s empirical formula It means glucose has a ratio of 1 mole C: 2 moles H: 1 mole O

What is the empirical formula of: Molecular formula : C25H10O5 Empirical Formula: C5H2O Molecular Formula: N2O6 Empirical Formula: NO3

Steps for Determining Empirical Formulas from Percents or Masses Example: a compound is 27.3% C and 72.7% O 1. Turn percent to mass (if needed). Assume 100 g if a mass isn’t given Assuming 100g, 27.3g C and 72.7g O 2. Turn mass into moles 27.3g C (1mole/12.01g) = 2.27 moles C 72.7g O (1 mole/16.00g) = 4.54 moles O

Steps for Determining Empirical Formulas from Percents or Masses 3. Write a formula using the number of moles as subscripts. C2.27O4.54 4. Divide both by the smallest number of moles (remember the empirical formula is a ratio) C2.27/2.27O4.54/2.27 C1O2 or CO2

Steps for Determining Empirical Formulas from Percents or Masses 5. If needed, multiply until all values are whole numbers Often you will be close to whole numbers, rarely will it be exact Example: If after dividing you had something like X1Z1.47 you would multiply everything by 2 , ending up with X2Z2.94, meaning your empirical formula would be X2Z3

Sample Problem A compound with a mass of 24.975g was analyzed and found to contain (by mass): 54.1% Ca, 43.2% O, and 2.7% H. What is the empirical formula for the compound? Ca: 54.1 g Ca x 1 mol Ca = 1.35 mol Ca 40.08 g Ca O: 43.2 g O x 1 mol O = 2.70 mol O 16.00 g O H: 2.7 g H x 1 mol H = 2.7 mol H 1.008 g H

Divide each mole value by the smallest number of moles and write formula. Ca: 1.35 mol Ca / 1.35 =1 O: 2.70 mol O / 1.35 = 2 H: 2.7 mol H / 1.35 = 2 CaO2H2

Determine the empirical formula for a compound containing 2.128 g Cl Sample Problem Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca. 1.203 g Ca x 1 mol/40.08 g = 0.03001 mol Ca 2.128 g Cl x 1 mol/35.45 g = 0.06003 mol Cl Ca 0.03001/0.03001Cl 0.06003/0.03001 CaCl2

What is the empirical formula if a compound consists of: 29.1% Na 40.5% S 30.4% O Assuming a 100 g sample: 29.1 g Na x 1 mol/22.99 g = 1.27 mol Na 40.5 g S x 1 mol/32.07 g = 1.26 mol S 30.4 g O x 1 mol/16g = 1.90 mol O Na 1.27/1.26 S 1.26/1.26 O 1.90/1.26 = NaSO1.5 (NaSO1.5) x 2 = Na2S2O3

Determining Molecular Formulas Find the empirical formula Find molar mass of empirical formula Divide molecular mass by empirical molar mass Multiply empirical formula by answer. You must be given molecular mass in the problem.

Sample Problem CH2O = 30.03 g 150.0 g / 30.03 g = 4.99500= 4.995 Ribose is an important sugar that is found in RNA. Ribose has a molecular mass of 150.0 g and an empirical formula of CH2O. CH2O = 30.03 g 150.0 g / 30.03 g = 4.99500= 4.995 5 (CH2O) = C5H10O5

Sample Problem Empirical to Molecular Find the molecular formula for a compound that has a mass of 16.1g and contains 30.4% Nitrogen and 69.6% Oxygen. The molecular mass of the compound is 92.0 g/mol. N: 30.4 g N x 1 mol N = 2.17 mol N 14.01 g N O: 69.4 g O x 1 mol O = 4.34 mol O 16.00 g O N 2.17/2.17 O 4.35/2.17 = NO2 (empirical formula)

Sample Problem Empirical to Molecular NO2 = 14.01 + 2(16) = 46.01 g/mol 92.0 g/mol/46.01 g/mol = 2 2 x (NO2) = N2O4 (molecular formula)