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At the end of this procedure we have a series of values, thus:

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86. Transformation of Relative Values into Absolute Values. From the preceding relative values we can get the absolute values just as soon as we know the absolute value of one of the heavier weights in the set. The 10' gram weight is usually selected for this purpose, and its value determined by comparison with a known standard weight by the method of substitution.29 The absolute values of the other weights can then be evaluated by proportion or by the method of approximation hereinafter described.

Suppose that the absolute weight of the 10′ gram weight was found to be 10,000.20 mg. by comparison with a known standard; then we have that 9920.95 units 10,000.20 mg. or for the accuracy in hand, that 1 unit = 1.01 mg.

=

We now notice that the relative value of the 50 gram weight is 1.27 units heavier than five times the relative value of the 10' gram weight; therefore, the absolute value of the 50 gram weight is 1.28 mg. heavier than five times the absolute value of the 10' gram weight, or 50,001.00 mg. + 1.28 mg. = 50,002.28 mg.

In a similar manner the relative value of the 20 gram weight is 0.45 units heavier than twice the relative value of the 10′ gram weight; therefore, the absolute value of the 20 gram weight is 0.45 mg. heavier than twice the absolute value of the 10' gram weight, or 20,000.40 mg. +0.45 mg. 20,000.85 mg.

=

29 Every laboratory should have a known standard weight in its possession. The Bureau of Standards, Washington, tests and certifies the accuracy of standards submitted, but it does not manufacture or sell such standards. For regulations governing tests, see their Circular No. 3, mentioned in § 59.

We proceed in this way to get the absolute value of all the other weights in the set carrying our calculations to the hundredth of a milligram and then round off our figures to the tenth of a milligram. Thus for the set of weights under consideration, we have:

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87. Table of Corrections.

Since it is very seldom that a set of weights is free from corrections, and it entails considerable labor to add up the values of the weights each time a weighing is made, the following scheme should be employed in order to minimize the likelihood of error and to save time.

In making a weighing always use the least number of weights possible and always use a single-primed weight in preference to a double-primed weight. This procedure then makes possible the use of a table which shows at a glance how much must be added or subtracted in any given case, according to the weight on the pan.

Thus the following table shows the corrections to be applied in connection with the use of the set of weights whose calibration we have just been discussing.

Table of Corrections

Milligrams to be added to (+) or subtracted from (—) given weight on pan

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1. From the following data obtained in the calibration of a set of analytical weights by the method of substitution, determine the true values of the respective weights.

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The value of the 10′ g. weight in comparison with a standard 10 g. weight by the same method was found to be 10.0002 g.

2. In weighing a brass object by the method of substitution and with the weights mentioned in § 86, the following weights were used: 20 g., 10′ g., 5 g., 0.5 g., 0.05 g. What was the weight of the object?

Ans. 35.5517 g.

When a 10 g. weight

3. Determine the ratio R: L from the following data. was placed in the right pan and a counterpoise weight in the left, 0.1 mg. had to be added to the left-hand pan. When the 10 g. weight was placed in the

left-hand pan and the counterpoise weight in the right, 0.2 mg. had to be added to the left-hand pan.

Ans. = 1.000015

L

4. The weight in air at a temperature of 20° of a Dumas bulb filled with benzene vapor is 24.9302 g., using brass weights. The weight of the bulb itself in air at 20° is 24.7238 g., and its capacity 143.01 c.c. What is the true weight of the benzene vapor? The density of brass may be taken as 8.5, that of glass as 2.7, the weight of 1 c.c. of air at 20° C. as 1.2 mg.

Ans. 0.3780 g.

5. A small cylinder of white metal weighed 58.1612 g. in air. When suspended by a platinum wire in water at 19° C. the weight was 50.4995 g. The wire suspended in water to the same depth weighed 0.3300 g. Brass weights were used. What was the weight "in vacuo" and what was its volume? Ans. 58.1626 g. 8.013 c.c.

6. Using brass weights of density 8.0 and assuming that 1 c.c. of air weighs 0.0012 g., calculate the difference between the weight in air of 1 g. of substance and the weight "in vacuo" when the density of the substance is 0.5, 1, 2, 4, 8 and 20. Plot the results with the densities as abscissæ and the differences as ordinates.

89.

CHAPTER VI

VOLUMETRIC MEASUREMENTS

Because of the fact that so many methods of analysis are based upon the operation known as titration, the general problem of measuring the volumes of solutions is one with which the analyst is concerned almost as frequently as he is with that of weighing. In carrying out such measurements the use of certain graduated glass vessels of special design, particularly flasks, burettes, pipettes, and cylinders, plays an important part. Before discussing these, however, it will be advantageous to present the theory which underlies titration so that we may more fully appreciate the problem which confronts us in measuring the volumes of solutions.

The operation of titration consists in having in solution the constituent which is to be determined and then adding to the solution the requisite amount of a standard solution i.e., one which contains a definite weight of reagent per unit volume, the reagent being of such a nature that under the given conditions it will react only with the constituent being determined until the exact number of chemical equivalents have been added, whereupon the next drop of reagent will be recognizable as being in excess by virtue of its producing in conjunction with an auxiliary reagent, known as an indicator, some physical change in the solution either through the appearance or disappearance of a color or the formation of a precipitate.

The scheme is: that upon adding the requisite volume of standard solution and noting the volume so added, the weight of reagent taking part in the reaction can be calculated, and from this weight, by virtue of the stoichiometrical equation involved, the weight of constituent sought can be ascertained.

Example. 1.000 g. of a mixture of sodium carbonate and potassium chloride were dissolved in 125 c.c. of water, two drops of methyl orange solution added as indicator, and the titration conducted by adding a solution of hydrochloric

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