A Straightforward Derivation of Stoichiometric Mass Relationships The most fundamental type of problem in chemistry is that of calculating the explicit mass relationships that are implied by any given chemical equation. For example, ifwe take 10.0g of glucose, what mass of oxygen will beconsumed in the following reaction?
In contemporary texthooks, it has become fashionable to develop a solution based on the analysis of the units involved, in the fallowing general way: (10.0 g glucose)
(6 mo10,) (32.0 g 0,) X (1 mol glucose) (1.00 mol0,)
(1.00 mol glucose) - 10.7 g O? (180 g glucose)
Intheaboveexpresn~on,allthr tlnmcsnccl crrept "g 0.",and o n ~ r rthus led tothrnnsncr "10.7 K O ".Thicanlurr iacorrwt; but the method utrd to ohrlin it ohscures the logic on uhlrh the ~ u l c u l n t ~13 mb s 4 In order to make that log^ clear, let u s hegin by uriting a pcnerul~,drquntwn thr follcw~ngua).
The arrow in eq 3 implies that there is no appreciable reverse reaction, so that with appropriate amounts of reagents there will be complete conversion into products. Dividing the mass of each substance, mt, by its molar mass, M1, gwss us the quantity nt:
(4)
n, = mtlMI
this quantity measures the relative numbers of particles in comparison witha stipulated standard.' When there is complete conversion of the reactants into products, the respective values of (ndvd must all be equal:
Substitution of eq 4 into 5 and rearrangement gives the following relationship between mAand any other mass ml:
Application of eq 6 to the example under discussion gives us
This calculation is done in strict conformance w t h the rulrs $of the cumpurdricmnl method ultrn rnlled "fa.t
