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Unformatted text preview: hole numbers. The result:C3H5.49O.Then multiply all the subscripts by a small whole number chosen to make them integral. Here, we multiply by 2: 2 5.49 =10.98 11 C 2 3 H 2 5.49O 2 1 = C6H10.98O2 Empirical formula :C6H11O2 result The empirical formula mass is (612)+(111.0)+(216.0)= 115u Because the experimentally determined molecular mass (230u) is twice the empirical formula mass ,we conclude Molecular formula: C12H22O4 Combustion Analysis In Combustion Analysis ,a weighed sample of a compound is burned in a stream of oxygen gas.The water vapor and carbon dioxide gas produced in the combustion are absorbed by appropriate substances. Before combustion CxHyOz and O2 After combustion xCO2 and y/2H2O After combustion, all the carbon atoms in the sample are found in the CO2.All the H atoms are in the H2O.Moreover, the only source of the carbon and hydrogen atoms was the sample being analyzed.Oxygen atoms in the CO2 and H2O could have come partly from the sample and partly from the oxygen gas consumed in the combustion.Thus ,the quantity of oxygen in the sample has to be determined indirectly. Example Determining an Empirical Formula from Combustion Analysis Data. Vitamin C is essential for the prevention of scurvy(and large doses may be effective in preventing colds).Combustion of a 0.200-g sample of this Carbon-hydrogen compound yields 0.2998 g CO2and 0.0819 g H2O.What is the empirical formula of Vitamin C ?
We can base our calculation directly on the 0.2000-g sample. All we need to do is determine the number of moles of C,H,O present. Once we have done this, we can write the empirical formula by method present earlier .These is a complication ,however.We can easily find the numbers Of moles C,H from the quantities of CO2and H2O produced.To find the number of moles of O in the sample,however, we must know the mass of O present. We can find mass of O only by difference. This means that the mass of C and H must be determined,as well as their amounts in moles. Solution 1 molCO2 1 mol C ? mol C = 0.2998 gCO2 44.010gCO2 1 mol CO = 0.006812 mol C 2 ? g C = 0.006812 molC ? mol H = 0.0819gH2O 12.011 gC = 0.08182 g C 1 mol C 2 mol H = 0.00909 mol H 18.02 gH2O 1 mol H2O 1.008g H 1 mol H = 0.00916 g H 1mol H2O ? g H = 0.00909 molH Now we can find the mass of O in the vitamin C sample by difference, ? g O = 0.2000g sample- 0.08182gC - 0.00916g H =0.1090g O and the number of moles O 1 mol O ? mol O = 0.1091gO = 0.006813 mol O 15.999 gO As trial subscripts for the empirical formila ,we can write C0.006812H0.00909O0.006813 Next, divide each subscript by the smallest-0.006812-to obtain CH1.33O Finally, multiply each subscript by 3(because 3 1.33 4) Empirical formula of vitamin C:C3H4O3 3-4 Oxidation States : A Useful Tool in Describing Chemical Compounds Oxidation state is related to the number of electrons
that an atom loses,gains,or otherwise appears to use in joining with other atoms in compound. NaCl, in this compound a Na atom,a metal,loses one electron to a Cl atom, a nonmetal.The compound consists of the ions Na+ and Cl- , Na+ is in a +1oxidation state...
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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