1.- Paradichlorobenzene, contains only C, H, and Cl and has amolar mass of about

Question

1.- Paradichlorobenzene, contains only C, H, and Cl and has amolar mass of about 147 g. Given that combustion of 1.68 g of thiscompound produces 3.02 g CO2 and 0.412 g H2O,determine its empirical and molecular formulas. 2.- The % composition of cysteine is 29.74 % C, 5.82 % H,26.41 % O, 11.56 % N, and 26.47 % S. What is the molecular formulaif its molar mass is aprox. 121 g? 3.- Consider the combustion of carbon monoxide (CO) in oxygengas: 2CO(g) + O2(g) ------> 2CO2(g) Starting with 3.60 moles of CO, calculate the number of molesof CO2 produced if there is enough oxygen gas to reactwith all the CO. 1.- Paradichlorobenzene, contains only C, H, and Cl and has amolar mass of about 147 g. Given that combustion of 1.68 g of thiscompound produces 3.02 g CO2 and 0.412 g H2O,determine its empirical and molecular formulas. 2.- The % composition of cysteine is 29.74 % C, 5.82 % H,26.41 % O, 11.56 % N, and 26.47 % S. What is the molecular formulaif its molar mass is aprox. 121 g? 3.- Consider the combustion of carbon monoxide (CO) in oxygengas: 2CO(g) + O2(g) ------> 2CO2(g) Starting with 3.60 moles of CO, calculate the number of molesof CO2 produced if there is enough oxygen gas to reactwith all the CO.

Explanation / Answer

We Know that :      The combustion of P-dichlorobenzene on combustion gives CO2 and H2O       The number of moles C inthe compound is = 3.02 g / 44 g / mol                                                                         = 0.068 mol        The number of molesof H in the compound = 2 x 0.412 g / 18 g /mol                                                                             = 0.0457 mol          The weightof C = 0.068 mol x 12 g / mol                                      = 0.816 g           Theweight of H =   0.0457 g          The weightof Cl in the compound is 1.68 - ( 0.816 + 0.0457)                                                                  = 1.5527g           Thenumber of moles of Cl in the compound = 1.5527g/ 35.5 g / mol                                                                                 = 0.0437 mol          The empericalformula of the compound =  C3H2Cl                n = Molecular weight / emperical formula weight                    = 127 / 63.5                     = 2          Molecularformula of the compound = C6H4Cl2        similarly the second bitcan be solved in the above procedure.           Thegiven equation is :           2CO(g)+ O2(g) ------> 2CO2(g)         From the balancedequation it is clear that 2 moles of CO gives 2 molesCO2 so 3.60 moles of CO gives 3.6moles         of CO2 in the presence of the excess of Oxygen.                n = Molecular weight / emperical formula weight                    = 127 / 63.5                     = 2          Molecularformula of the compound = C6H4Cl2        similarly the second bitcan be solved in the above procedure.           Thegiven equation is :           2CO(g)+ O2(g) ------> 2CO2(g)         From the balancedequation it is clear that 2 moles of CO gives 2 molesCO2 so 3.60 moles of CO gives 3.6moles         of CO2 in the presence of the excess of Oxygen.          

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