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6 1 mol C2H6S 62.1 gC2H6S 6.02 23 moleculesC2H6S 10 1 molC2H6S = 8.1 18 molecules C2H6S 10
Given that a normal drop of liquid is about0.05mL a volume of 1.0 L is only 0.02 of a drop Mole of an Element  A second look We took one mole of an element to be 6.022141023 element .This is the only definition possible for elements such as iron,sodium,and copper.But the atoms of some elements are clustered together to form molecules.Bulk samples of these elements are composed of collections of molecules. Molecular forms of elemental oxygen (O2) 33 Composition of Chemical Compounds halothane This picture tells us that per mole of halothane there are two moles of C atoms,one mole each of H,Br,and Cl atoms,and three moles of F atoms. Example Using Relationships Derived from a Chemical Formula. How many moles of F atoms are in a 75.0mL sample of halothane (d= 1.871 g/mL) ? Solution 1.convert the volume of the sample to mass; 2.convert the mass of halothane to its amount in moles; 3.the final convert factor is based on the formula of halothane. 1.871 g C2HBrClF3 ? mol F = 75.0 mL C2HBrClF3 1 mLC2HBrClF3 1 molC2HBrClF3 3 mol F 197.4 gC2HBrClF3 1 molC2HBrClF3 = 2.13 mol F Calculating Percent Composition from a Chemical Formula Mass % element = number of atoms of element per formul a unit molar mass of element 100 % molar mass of compd Example Calculation the Mass Percent of a Compound.What is the mass percent composition of halothane,C2HBrClF3 ? Solution First, determine the molar mass of C2HBrClF3.this was shown on page 73 to be 197.38 g/mol. Then,for one mole of compound, formulate mass ratios and percents .
%C = %H = %Br = %Cl = %F = (2 12.01) g 197.38 g 1.01 g 197.38 g 79.90 g 197.38 g 35.45 g 197.38 g (3 19.00)g 197.38 g
100% = 12.17% 100% = 0.51% 100% = 40.48% 100% = 17.96% 100% = 28.88% Establishing Formulas from the Experimentally Determined Percent Composition of compounds Mass % composition of compound Assume 100 g Empirical formula
Grams of each element 1/molar mass
Calculate mole ratios Moles of each element Example Determining the Empirical and Molecular Formulas of a compound from its Mass Percent Composition. Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58%C,9.63% H,and 27.79% O.Its experimentally determined molecular mass in 230u .What are the empirical and molecular formulas of dibutyl succinate? Solution
Step 1. Determine the mass of each element in a 100.0g sample 62.58 g C, 9.63 g H, 27.79 g O Convert each of these masses to an amount in moles
? mol C = 62.58 g C 1 mol C = 5.210 mol C 12.011g C 1 mol H ? mol H= 9.63 g H = 9.55 mol H 1.008g H 1 mol O ? mol C = 62.58 g O = 1.737 mol O 15.999g O Step 2. Step 3. Write a tentative formula based on these number of moles C5.21H9.55O1.74 Step 4. Divide each of the subscripts of the tentative formula by the smallest (1.74) C 5.21 H 9.55 O 1.74 = C2.99H5.49O
1.74 1.74 1.74 Step 5. Round off any subscript from step 4 that differ only slightly from w...
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This note was uploaded on 05/07/2010 for the course FIN 401 taught by Professor Chemistry during the Spring '10 term at Uni Potsdam.
 Spring '10
 Chemistry

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