111 CH3 LCT 7 emp form

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Published on January 16, 2008

Author: Dolorada

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Chapter 3:  Chapter 3 3.6 Determining Compound Formulas Slide2:  A Simple Rhyme for a Simple Formula by Joel S. Thompson Journal of Chemical Education Vol. 65, No. 8; August 1988, p. 704 When teaching the method for converting percentage composition to an empirical formula, I have devised the following rhyme: Percent to mass Mass to mole Divide by small Multiply 'til whole Here's an example of how it works. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? (1) Percent to mass: Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g nitrogen. (2) Mass to moles: for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mg for N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N (3) Divide by small: for Mg: 2.97 mol / l.99 mol = 1.49 for N: 1.99 mol / l.99 mol = 1.00 (4) Multiply 'til whole: for Mg: 2 x 1.49 = 2.98 (i.e., 3) for N: 2 x 1.00 = 2.00 and the formula of the compound is Mg3N2. Students enjoy this device and have discovered that they have both the rhyme and reason for working chemistry problems of this type. The above article is copyright © 1988 by the Division of Chemical Education of the American Chemical Society, Inc. The assumption of 100 grams in part 1 is purely for convenience sake. This means that the percentages transfer directly into grams. If you assumed that 36.7 grams were present, you would have to multiply 36.7 by each percentage. Assuming 100 grams makes it much easier. In part 3, make sure you divide ALL answers from #2 by the smallest value. In part 4, multiply ALL values from #3 by the same factor. This factor is selected so as to produce ALL whole numbers as answers. Often this factor is chosen by trial-and-error. Empirical Formula - Part Two Empirical Formula - Part Two Return to Empirical Formula article Here's the rhyme from the article: Percent to mass Mass to mole Divide by small Multiply 'til whole Here's the example problem: A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular weight of this compound is known to be approximately 140 g/mol. What is the empirical formula? What is the molecular formula? 1) Percent to mass. Assume 100 grams of the substance is present, therefore its composition is: carbon: 68.54 grams hydrogen: 8.63 grams oxygen: 22.83 grams (2) Mass to moles. Divide each mass by the proper atomic weight. carbon: 68.54 / 12.011 = 5.71 mol hydrogen: 8.63 / 1.008 = 8.56 mol oxygen: 22.83 / 16.00 = 1.43 mol (3) Divide by small: carbon: 5.71 ÷ 1.43 = 3.99 hydrogen: 8.56 ÷ 1.43 = 5.99 oxygen: 1.43 ÷ 1.43 = 1.00 (4) Multiply 'til whole. Not needed since all values came out whole. The empirical formula of the compound is C4H6O. Next we need to determine the molecular formula, knowing the empirical formula and the molecular weight. Here's how: 1) Calculate the "empirical formula weight." This is not a standard chemical term, but I believe it is understandable. C4H6O gives an "EFW" of 70.092. 2) Divide the molecular weight by the "EFW." 140 ÷ 70 = 2 3) Multiply the subscripts of the empirical formula by the factor just computed. C4H6O times 2 gives a formula of C8H12O2. This is the molecular formula. Ch.3 Molecules and Compounds:  Ch.3 Molecules and Compounds Types of Chemical Formulas Percent Composition Determining Formulas Formulas:  Formulas Formula - shorthand way of expressing the number of atoms of each type in a molecule Sucrose - C12H22O11 Ethanol - C2H6O Dimethyl ether - C2H6O Formulas:  Formulas Empirical formula - each element is written once Structural formula - provides information on how the atoms are connected Determining Empirical Formulas:  Determining Empirical Formulas Strategy %composition A ----> g of A -----> moles of A %composition B ----> g of B -----> moles of B find the mole ratio = A/B Ratio gives the formula What is the empirical formula for hydrazine?:  What is the empirical formula for hydrazine? Hydrazine contains 87.42% N and 12.58% H Strategy % composition N ----> g of N in a 100g sample 87.42g of N and 12.58 g H g of N -----> moles of N use the molar mass Empirical Formula:  Empirical Formula 12.58% H % composition H ----> g of H 12.58g of H g of H-----> moles of H Empirical Formula:  Empirical Formula 6.241 mol N 12.48 mol H find the mole ratio = A/B = 12.48mol H/6.241 mol N mole ratio = 2 mol H/ mol N Ratio gives the formula NH2 Molecular formula:  Molecular formula Indicates the number of atoms in each molecule, not just the ratio You need the molar mass as well as the percent composition to determine molecular formula What if NH2 really represented the hydrazine molecule? Molar mass = 14.007 + 1.0079 +1.0079 = 16g/mole Molecular formula:  Molecular formula Experimental data shows that the molar mass for hydrazine is 32.0g/mole Molecular formula = N2H4 For fractional mole ratios:  For fractional mole ratios Isooctane % carbon = 84.12% % hydrogen = 15.88% Fractional mole ratios:  Fractional mole ratios Mole ratio = 15.76 mol H / 7.004 mol C = 2.25 mole ratio = 2 1/4 = 9/4 Nine H atoms for every four C atoms Empirical formula is C4H9 If C4H9 then molar mass = 9x1 + 4x12 = 57g/mole but molar mass is actually 114g/mole Molecular formula is C8H18 Word Problems:  Word Problems Sn metal + I2 ------> SnxIy Mass of tin (Sn) in the crucible originally = 1.056g Mass of iodine reacted = 1.947g Mass of tin (Sn) left over after the reaction= 0.601g Mass of tin (Sn) in the original mixture = 1.056g Mass of tin (Sn) left over after the reaction= -0.601g Mass of tin (Sn) consumed in the reaction = .455g Convert grams to moles:  Convert grams to moles Mole Ratio and Formulas:  Mole Ratio and Formulas Mole ratio = 15.34x10-3 mol I / 3.83 x 10-3 mol Sn = 4 Mole ratio = 4 mol I / mol Sn Empirical Formula is SnI4 If SnI4 then molar mass = 118.71 + 4x126.9 = 626.31g/mole and the molar mass is actually 626.31g /mole Molecular formula is SnI4 The molecular formula can sometimes be the same as the empirical formula

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