Scan_Doc0098 - of molecules with the formula C1 2 C 2 H 4 ....

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94 Chapter Three Stoichiometry .;» ....• Figure 3.7 The two forms of dichloroethane. What information do we need to find the empirical formula? ./ Mass of each element in 100.00 g of compound ./ Moles of each element How do we get there? What is the mass of each element in 100.00 g of compound? CI 71.65 g C 24.27 g H 4.07 g What are the moles of each element in 100.00 g of compound? 1 mol CI 71.65 g-et X = 2.021 mol CI 35.45 g-et 1 mol C 24.27 g-C X 12.01 g-C = 2.021 mol C 1 mol H 4.07 g-H X = 4.04 mol H 1.008 g-H What is the empirical formula for the compound? Dividing each mole value by 2.021 (the smallest number of moles present), we find empirical formula CICH 2 . What is the molecular formula for the compound? Compare the empirical formula mass to the molar mass. Empirical formula mass = 49.48 glmol (Confirm this!) Molar mass is given = 98.96 glmol Molar mass 98.96 glmol --------- = = 2 Empirical formula mass 49.48 glmol Molecular formula = (CICH 2 )2 = Cl 2 C 2 H 4 This substance is composed
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Unformatted text preview: of molecules with the formula C1 2 C 2 H 4 . Note: The method we employ here allows us to determine the molecular formula of compound but not its structural formula. The compound Cl 2 C 2 H 4 is called dichloroethane: There are two forms of this compound, shown in Fig. 3.7. The form on the right was fOT-merly used as an additive in leaded gasoline. SEE EXERCISES 3.81 AND 3.82 EXAMPLE 3.11 I r. A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% 0",_'-gen by mass. The compound hasa molar mass of 283.88 g/mol. What are the compound-s empirical and molecular formulas? Solution Where are we going? To find the empirical and molecular formulas for the given compound What do we know? ./ Percent of each element ./ Molar mass of the compound is 283.88 glmol...
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