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Chemical Quantities Empirical and Molecular Formulas Chemistry Printable Version
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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)
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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.
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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
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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)
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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.
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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.
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Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g sample, so 38.67 g C x 1mol C = 3.22 mole C 12.0 g C 16.22 g H x 1mol H = 16.22 mole H 1.0 g H 45.11 g N x 1mol N = 3.22 mole N 14.0 g N Now divide each value by the smallest value
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Example The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1 mol N The ratio is 16.22 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 43.64 % P and 56.36 % 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
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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
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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
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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
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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
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n Stephen L. Cotton n Charles Page High School n http://www.google.com/search?sourceid=navc lient&aq=0&oq=powerpoint+percent+compos &ie=UTF- 8&rlz=1T4ADBF_enUS321US349&q=powerpoin t+percent+composition Acknowledgements
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