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# GENERAL CHEMISTRY CHEM 142

UW

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This 4 page Class Notes was uploaded by Carmela Kilback on Wednesday September 9, 2015. The Class Notes belongs to CHEM 142 at University of Washington taught by James Callis in Fall. Since its upload, it has received 26 views. For similar materials see /class/192545/chem-142-university-of-washington in Chemistry at University of Washington.

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Date Created: 09/09/15

Obtaining the Molecular Formula from its Percent Composition an Algebraic Formulation James B Callis Professor of Chemistry University of Washington Let us begin with a compound whose chemical formula can be represented as AaBch where each atom making up the compound is identi ed by its chemical symbol e g A and the number of atoms of type A in the compound is identi ed with a subscript e g a which represents a whole number Thus the compound under discussion has a atoms of type A combined with b atoms of type B and c atoms of type C For the set of numbers a b c we consider two possibilities i the set has no common multiplier other than one or ii the set has a whole number common multiplier say 5 which allows us to represent the set as a b c as 5a 5b 5c 5a b c Now if we define the symbol AaBbCC as the molecular formula the formula Aabich represents a different formula the empirical formula In the case where 5 l the molecular formula and the empirical formula are identical However in the case where 5 is a whole number greater than 1 then a b and c are all smaller than a b and c by the factor 5 Generally we denote the case where a b and c is the set containing the smallest possible whole numbers we have the chemical empirical formula As an example consider the molecular formula for glucose C5H1205 This compound is characterized by the set 6 12 6 which can be written 6391 6392 6391 61 2 l ie 5 6 and a c l with b 2 Thus we can state the empirical formula for glucose as CHzO Associated with each molecular and empirical formula is a molecular mass MM and empirical mass EM The former can be calculated as follows MMAEBEAC aMAbMBcMC 1 Where MMAaBbcc is the molar mass in grams of the compound AaBbCC and MA is the atomic mass in grams of the element A and similarly for MB and Mc The calculation for the empirical mass is similar EMA BE A a MAb MBc MC 2 At this point we leave it to the student to show that MMAEBECE EMA Bbicci 6 3 With the above de nitions in mind we are now ready to pose the problem stated in the title Find the molecular formula of a compound given its fractional composition by mass together with the values of the atomic masses using Dalton s atomic theory To make the solution more concrete we will assume the molecular formula above AaBch Further we will de ne the amount of material in a sample of the compound massif abc as mass njf3bCEMM n2 1 3bccaMAbMBcMC 4 sa AaBbC AaBbC Where n2ch is the number of molecules in the sample The results of equation 4 can be expanded to include the possibility of an empirical formula that is different than the molecular formula sam sam sam 39 39 massAaBbcc nAaBbCE 6EMA B C nAaBbCE 5a MA b MB c MC 5 abc We see that if 5 1 EM A B C MM and the results are unambiguous However if AaBbC 5gtl 8EMA B C MMAEBECE 6 ab c Let us now evaluate the fractional composition of element A f A in the sample of AaBch f z nAEBbcc39a39MA a MA 7 A Sam nAaBbC MMAaBbC MMAaBbC In passing we note that the formulation of the problem as a ratio in equation 7 has eliminated one of the variables nifle We no longer need be concerned with this variable As was done in equation 5 we expand equation 7 to include the possibility of an empirical formula whose coefficients are smaller than those in the molecular formula Sam v f nAaBbCE SaMA a MA 8 A sam nAaBbCE 39 5 39 EMAE BE CE EMAE BE CE We see that owing to the division operation we have removed 5 from the equation This implies that we have lost the ability to determine the molecular formula and can only recover the empirical formula Similarly we may derive expressions for f3 and fc b M M fB B fC 0 0 9 EMAE BE CE EMAE BE CE Equations 8 and 9 form a system of three equations in ten variables f A fB fc EM A Bb c MA MB Mc a b and c Ofthese variables six are known quantities fA fB fc MA MB MC and four are unknown a b c EM is underdetermined Moreover we seem to have eliminated the ability to determine the molecular formula because neither it nor 5 is present in the set of equations CE As currently stated this problem A13b Nevertheless we continue Our next manipulation is to eliminate the known the mass of each element that is present in one mole of sample This is accomplished by dividing each fraction by its corresponding atomic mass Equations 8 and 9 become A a39 f B b39 f C 39 10 EM M EM M EM A 13 icci c A Bb1c51 B A Bb1cE1 At this point we have a system of three equations in 7 variables fA fB fc EM AE BE CE a b and c Of these variables three are known quantities f A fB fc and four are unknown a b c EM AE BE CE To simplify the problem further we eliminate the empirical mass by dividing each of the equations by the equation with the smallest value Suppose this is the first equation in 10 Then we have fA a MA EMA 1B a b 1 11 fA a MA EMA1Bb1 E fB b39 ME EMAEBECE z 12 fA a a39 MA EMAE BE E fC a MC 2 EMA Bb1cE Cy 13 fiA a a39 MA EMA Bb1cE We see that taking the ratio of ratios has eliminated equation 11 it is now trivial and reduced the problem to one of a system of two equations 12 and 13 in three unknowns a b c We have now arrived at the simplest possible expression ofthe problem as one that is illposed Fortunately there is a further constraint that we can require for the solution that reduces the problem to one that is wellposed This constraint may be stated as follows a b cw1mlenos 14 mln Thus the coefficients a b c must be such that they form the minimum set of whole numbers that satisfy equations 12 and 13 We can also restate equation 14 in terms of a Cartesian space with axes along the coef cient directions In this case we are asking for the point given by coordinates a b c that express that the minimum set of coefficients is the lowest possible distance from the origin in the Euclidean sense ie that 2 2 2 whole nos a39 b Cv 15 To provide a concrete illustration of the concepts in this work we consider the case of adipic acid The molecular formula for this compound is C5H1004 MM 14614 gmol The empirical formula is C3H50z EM 7307 gmol and 5 2 With this data we can calculate the fractional elemental abundances C 04931 H 06897 and O 04379 We are now ready to pose the central problem as applied to adipic acid Find the empirical formula of adipic acid given its fractional composition by mass together with the values of the atomic masses using Dalton s atomic theory Our first step is to eliminate the number of molecules in the original sample by forming the following ratios using the definitions found in equation 10 24931204106 E06897 MC 1201 f 4379 006932 0 39 02737 16 H 1007 MO 1600 Next we eliminate the empirical mass from the equations by dividing by the lowest value of fXMx In this case it is fOMo 02737 Using the definitions found in equations 11 7 13 we find the following ratios f0 fc LH MO2737L MCZIMN6L mz MH1m2Z f0 02737 fg 02737 fg 02737 17 M0 M0 M0 The above set of numbers is identical within experimental and round off error to the set of rational numbers 1 32 52 These can then be converted to the smallest set of whole numbers by multiplication by 2 least common multiplier This final set of whole numbers is identical to the empirical formula of adipic acid that we started with Our final task is to show how to go from the empirical formula to the molecular formula Equation 3 holds the key If we have the set a b c and the empirical molar mass we need only one more parameter According to equation 3 it may be either 5 or MM which then yields 5 AaBbC I Experimentally it is usually easier to obtain MM A EEC

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