Chemical Quantities Empirical and Molecular Formulas Chemistry Printable Version

Learning Objective TLW calculate empirical and molecular formulas (TEKS 8.C) TLW distinguish between empirical and molecular formulas TLW calculate percent error between empirical and molecular formulas (TEKS 2.G)

Formulas Example: molecular formula for benzene is C 6 H 6 (note that everything is divisible by 6) Therefore, the empirical formula = CH (the lowest whole number ratio) 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 (the lowest whole number ratio = cannot be reduced). Examples: NaClMgCl 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 (Correct formula) (Lowest whole number ratio)

Calculating Empirical  Just find the lowest whole number ratio C 6 H 12 O 6 CH 4 N  A formula is not just the ratio of atoms, it is also the ratio of moles.  In 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 O. = CH 2 O = this is already the lowest ratio.

Calculating Empirical  We can get a ratio from the percent composition. 1)Assume you have a 100 g sample - the percentage become grams (75.1% = 75.1 grams) 2)Convert grams to moles. 3)Find lowest whole number ratio by dividing each number of moles by the smallest value.

Example  Calculate the empirical formula of a compound composed of % C, % H, and %N.  Assume 100 g sample, so  g C x 1mol C = 3.22 mole C 12.0 g C  g H x 1mol H = mole H 1.0 g H  g N x 1mol N = 3.22 mole N 14.0 g N Now divide each value by the smallest value

Example  The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1 mol N  The ratio is mol H = 5 mol H 3.22 mol N 1 mol N = C 1 H 5 N 1 which is = CH 5 N  A compound is % P and % O. What is the empirical formula?  Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? = P 2 O 5 = C 4 H 5 N 2 O

Empirical to molecular  Since the empirical formula is the lowest ratio, the actual molecule would weigh more.  By a whole number multiple.  Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula  Caffeine has a molar mass of 194 g. what is its molecular formula? = C 8 H 10 N 4 O 2

Example Caffeine has an actual molar mass of 194 g. What is its molecular formula? From previous problem empirical formula is C 4 H 5 N 2 O Molar mass using empirical formula C 12 x 4 = 48 g H1 x 5 = 5 g N14 x 2 = 28 g O16 x 1 =16 g 97 g Divide actual molar mass by mass calculated using empirical formula 194 g / 97 g = 2 Multiply subscripts in empirical formula by 2, so molecular formula will be C 8 H 10 N 4 O 2

Independent Practice Empirical and Molecular Formulas – Practice Set 1Practice Set 1 – Practice Set 2Practice Set 2 – Practice Set 3Practice Set 3 – Practice Set 4 (aren’t we lucky)Practice Set 4

Lab Pre-work – Complete the Table Top Lab on Empirical Formulas of CompoundsEmpirical Formulas of Compounds – Read Empirical Formula Determination Experiment (see handout or Flinn Scientific lab manual p. 13 – 17) Complete the Pre-Lab Questions (page 14) – Prepare a Safe Lab Analysis Card to identify Potential Hazards, Precautions to Take, and PPE to UseSafe Lab Analysis Card – Conduct Lab in small groups – Complete Post-Lab Calculations and Analysis (p. 17) Show your work. Use correct units. Round to appropriate number of significant figures Use proper spelling and grammar in write up

n Stephen L. Cotton n Charles Page High School n lient&aq=0&oq=powerpoint+percent+compos &ie=UTF- 8&rlz=1T4ADBF_enUS321US349&q=powerpoin t+percent+composition Acknowledgements