Obtaining the Molecular Formula from Empirical Formula: A Mathematical Approach - Prof. Ja, Study notes of Chemistry

Download Obtaining the Molecular Formula from Empirical Formula: A Mathematical Approach - Prof. Ja and more Study notes Chemistry in PDF only on Docsity! 1 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 AaBbCc, where each atom making up the compound is identified by its chemical symbol, e.g. A, and the number of atoms of type A in the compound is identified 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 δ, which allows us to represent the set as {a b c} as {δa’ δb’ δc’} = δ{a’ b’ c’}. Now if we define the symbol AaBbCc as the molecular formula, the formula Aa’Bb’Cc’ represents a different formula, the empirical formula. In the case where δ = 1, the molecular formula and the empirical formula are identical. However in the case where δ is a whole number greater than 1 then a’, b’ and c’ are all smaller than a, b and c by the factor δ. 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, C6H12O6. This compound is characterized by the set {6 12 6}, which can be written {6.1 6.2 6.1} = 6{1 2 1}, i.e. δ = 6 and a’ = c’ = 1 with b =2. Thus we can state the empirical formula for glucose as CH2O. Associated with each molecular and empirical formula is a molecular mass (MM) and empirical mass (EM). The former can be calculated as follows: a b CA B A A B C MM a M b M c M= ⋅ + ⋅ + ⋅ (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: a ' b ' c 'A B A A B C EM a ' M b ' M c ' M= ⋅ + ⋅ + ⋅ (2) At this point we leave it to the student to show that a b c a ' b ' c 'A B C A B C MM EM = δ (3) With the above definitions 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 2 make the solution more concrete, we will assume the molecular formula above (AaBbCc). Further, we will define the amount of material in a sample of the compound, a b c sam A B Cmass as: ( ) a b c a b c a b c a b c sam sam sam A B C A B C A B C A B C A B Cmass n MM n a M b M c M= ⋅ = ⋅ + ⋅ + ⋅ (4) Where a b c sam A B Cn 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. ( ) a b c a b c ' ' ' a b ca b c sam sam sam , ' ' A B C A B C A B C A B C A B Cmass n EM n a M b M c M= ⋅δ ⋅ = ⋅δ ⋅ + ⋅ + ⋅ (5) We see that if ' ' ' a b ca b c A B C A B C1, EM MMδ = = and the results are unambiguous. However, if δ > 1, ' ' ' a b ca b c A B C A B C EM MMδ ⋅ = (6) Let us now evaluate the fractional composition of element A, fA in the sample of AaBbCc. a b c a b c a b c a b c sam A B C A A A sam A B C A B C A B C n a M a Mf n MM MM ⋅ ⋅ ⋅ = = ⋅ (7) In passing we note that the formulation of the problem as a ratio in equation 7 has eliminated one of the variables, a b c sam A B Cn . 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. a b c a b c a ' b ' c ' a ' b ' c ' sam A B C A A A sam A B C A B C A B C n a ' M a ' Mf n EM EM ⋅δ ⋅ ⋅ ⋅ = = ⋅δ ⋅ (8) We see that owing to the division operation we have removed δ 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 fB and fC: a ' b ' c ' a ' b ' c ' CB B C A B C A B C c ' Mb ' Mf ; f EM EM ⋅⋅ = = (9) Equations 8 and 9 form a system of three equations in ten variables: fA, fB, fC, a ' b ' c 'A B C EM , MA, MB, MC, a’, b’, and c’. Of these variables, six are known quantities (fA, fB, fC, MA,