1.- Paradichlorobenzene, contains only C, H, and Cl and has amolar mass of about
ID: 76176 • Letter: 1
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.Related Questions
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