An Introductory Course of Quantitative Chemical Analysis

The problems with which the analytical chemist has to deal are not, as a matter of actual fact, difficult either to solve or to understand. That they appear difficult to many students is due to the fact that, instead of understanding the principles which underlie each of the small number of types into which these problems may be grouped, each problem is approached as an individual puzzle, unrelated to others already solved or explained. This attitude of mind should be carefully avoided.

It is obvious that ability to make the calculations necessary for the interpretation of analytical data is no less important than the manipulative skill required to obtain them, and that a moderate time spent in the careful study of the solutions of the typical problems which follow may save much later embarrassment.

1. It is often necessary to calculate what is known as a "chemical factor," or its equivalent logarithmic value called a "log factor," for the conversion of the weight of a given chemical substance into an equivalent weight of another substance. This is, in reality, a very simple problem in proportion, making use of the atomic or molecular weights of the substances in question which are chemically equivalent to each other. One of the simplest cases of this sort is the following: What is the factor for the conversion of a given weight of barium sulphate (BaSO₄) into an equivalent weight of sulphur (S)? The molecular weight of BaSO₄ is 233.5. There is one atom of S in the molecule and the atomic weight of S is 32.1. The chemical factor is, therefore, 32.1/233.5, or 0.1375 and the weight of S corresponding to a given weight of BaSO₄ is found by multiplying the weight of BaSO₄ by this factor. If the problem takes the form, "What is the factor for the conversion of a given weight of ferric oxide (Fe₂O₃) into ferrous oxide (FeO), or of a given weight of mangano-manganic oxide (Mn₃O₄) into manganese (Mn)?" the principle involved is the same, but it must then be noted that, in the first instance, each molecule of Fe₂O₃ will be equivalent to two molecules of FeO, and in the second instance that each molecule of Mn₃O₄ is equivalent to three atoms of Mn. The respective factors then become

(2FeO/Fe₂O₃) or (143.6/159.6) and (3Mn/Mn₃O₄) or (164.7/228.7).