# Determine the empirical formula from the percent

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Determine the empirical formula from the percent composition (assuming 100 g of compound). The molar mass may be determined from the density of the gas. The empirical formula and the molar mass may then be used to determine the molecular formula. Count the valence electrons in the empirical formula and then construct the Lewis structure.
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Moles of H = 1 mol 2.08 g H 1.008 g H = 2.063 mol H Moles of Cl = 1 mol 73.1 g Cl 35.45 g Cl = 2.062 mol Cl The preliminary formula is C 2.065 H 2.063 Cl 2.062 . Dividing all subscripts by the smallest subscript to obtain integer subscripts: 2.065 2.063 2.062 2.062 2.062 2.062 C H Cl = CHCl Thus, the empirical formula is CHCl, and its molar mass is 48.47 g/mol. The density is 4.3 g/L at STP: P d = RT M Rearranging to solve for molar mass: L•atm 4.3 g/L 0.0821 273 K mol•K = = 1.00 atm dRT P M = 96.37719 = 96 g/mol The molar mass (96 g/mol) is double the empirical formula mass (48.47 g/mol), so the empirical formula must be doubled to get the molecular formula: C 2 H 2 Cl 2 . The formula contains twenty-four valence electrons. A variety of structures may be drawn: C C Cl H H Cl C C H Cl H Cl C C H H Cl Cl or or 10.79 There are thirty-two valence electrons present. Begin four Lewis structures by placing a Cl in the center and four O atoms around it. Connect all the O atoms to the central Cl with single bonds. In the second structure, convert one of the single bonds to a double bond. In the third structure, two of the bonds are double, and in the last, three of the bonds are double. (It does not matter which bonds are chosen to become double bonds.) = 0 1.75
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10.80 Plan: Write the balanced chemical equations for the reactions and draw the Lewis structures. To find the heat of reaction, add the energy required to break all the bonds in the reactants to the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. The bond energy values are found in Table 9.2. Divide the heat of reaction by the number of moles of oxygen gas appearing in each reaction to get the heat of reaction per mole of oxygen. ) 2 ) 2
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10.81 F S F F S F F F S F