15 
SERIES NO. 1. 
Date. 
Name of person. 
Condition. 
Per 
cent of 
hemo- 
globin. 
Colorime- 
ter read- 
ings at the 
end of 24 
hours com- 
pared with 
standard 
phenol- 
phthalein 
solution 
No. l,a set 
at 5 divi- 
sions on the 
colorimeter 
scale. 
Quan- 
tity of 
phenol- 
phthal- 
ein 
formed, 
in 
milli- 
grams. & 
Quan- 
tity of 
phenol- 
phthal- 
in ox- 
idized, 
in 
milli- 
grams. 
Per 
cent 
oxi- 
dized 
Oxidiz- 
ing 
power 
of the 
blood, 
in 
terms 
of nor- 
mal 
= 100. 
Apr. 22 
Amoss 
Normal 
96 
3.4 
0. 468 
0. 471 
29.4 
100 
22 
Collier 
Pernicious anemia . . . 
20-21 
19.5 
.0816 
.082 
5.1 
17 
22 
Hvnson 
do 
27 
11.6 
.137 
.138 
8.6 
29 
SERIES NO. 2. 
Apr. 24 Amoss... 
24 Callahan 
24 Johnson 
Normal 
96 
6.6 
0. 24 
0. 242 
15.1 
100 
Fibroma, uterus 
31 
21.75 
.0731 
.074 
4.6 
30.4 
Disease unknown 
49 
12.5 
.127 
.128 
8 
53 
a Standard phenolphthalein solution No. 1 contained 0.0318 gram phenolphthalein in 100 c. c. alco- 
hol. 1 c. c. of this solution measured out by pipette in usual manner and mixed with 5 c. c. of 
distilled water and 1 c. c. N/10 sodium hydroxide furnished the standard for the color comparison. 
!>The quantity of phenolphthalein formed in milligrams and the quantity of phenolphthalin 
oxidized, also in milligrams, has been calculated in the following manner: As already pointed out, 
5 c. c. of the phenolphthalin reagent was used throughout the work. This reagent contained 0.032 
gram of phenolphthalin in 100 c. c. of the solution: 1 c. c. of the reagent contained 0.00032 gram of 
phenolphthalin, and 5 c. c. contained 0.00032x5=0.0016 gram of phenolphthalin. If this were com- 
pletely oxidized, it would yield 0.00159 grams of phenolphthalein. The standard solutions of phe- 
nolphthalein employed in making the color comparisons all contained 0.0318 gram of phenolphthalein 
in 100 c. c. of solution. Hence. 1 c. c., the amount employed in making up the standard for the color 
comparison, contained 0.000318 gram of phenolphthalein or one-fifth of the amount of phenolphthalein 
which would have been formed had all of the phenolphthalin employed in each test been oxidized. 
In making the color comparisons in the colorimeter, the colorimeter scale on the side of the standard 
was arbitrarily placed at 5 divisions and the other scale adjusted until the colors matched on the two 
sides of the instrument. If, now, in order to match the standard, which had been set at 5, the 
other scale should have had to be placed at 5, then the color of the solution originally would have 
been exactly equal to that of the standard or to the color produced by 0.000318 gram of phenol- 
phthalein. or 0.318 milligram of phenolphthalein would have resulted from the oxidation. Had it 
been necessary to set the scale of the test at 1 in order to match the standard at 5, then, obviously, 
five times the' amount of phenolphthalein present in the standard, or 0.000318x5=0.00159 gram, or 
1.59 milligrams, would have been formed. 
Hence, in order to calculate the amount of phenolphthalein formed during these experiments we 
have from the foregoing the following proportions: 
5 : 5:: .000318 : x 
and 
1:5:: .000318 : x 
from which it follows that the amount of phenolphthalein formed is equal to 0.00159 divided by the 
reading on the colorimeter scale, when the standard is set at 5 on the same scale, and since 
318 : 320 : : 1 : 1.006 
we multiply the number of milligrams of phenolphthalein formed by 1.006 in order to get the amount 
of phenolphthalin oxidized in milligrams. 
To take the first case under Series No. 1, April 22, the colorimeter reading was 3.4 in terms of the 
standard=5. Hence, to calculate the amount of phenolphthalein formed, we would have: 
0.00159 
— n-j— =.000468 or 0.468 milligrams 
and for the phenolphthalin oxidized: 
(See Series No. 1, Apr. 22.) 
0.468x1.006=0.471. 
