The following table was crossed out on the original. A note on the previous page beside the table was:

“All these calculations are wrong.  J.R.”

I ·1078 gr salt gave ·0304 gr BaSO4 = 15·73% Ba.
II ·1641” ”   ”·0457” ” ” = 15·53” ”
III ·2425” ”   ”·0680” ” ” = 15·65” ”
IV ·2860” ”   ”·0798” ” ” = 15·54” ”
V ·1843” ”   ”·0498” ” ” = 15·08” ”
VI ·2620” ”   ”·0708” ” ” = 15·08” ”
VII ·3230” ”   ”·0906” ” ” = 15·65” ”
VIII ·2875” ”   ”·0807” ” ” = 15·66” ”

  Calculated for C19H13O7SBa
= 15·10% Ba.

In the above determinations the salt analysed was taken from specimens made at three different times and purified in slightly different ways, Nos 1, 2, & 3 being washed with absolute alcohol. Nos V and V were made by precipitating the Ba with H2SO4 from a solution of the salt.

The water was determined by heating at 110° till constant weight was reached. Part only of the weight lost was regained on standing in the air.

·3943 gr salt lost at 110° ·0286 gr = 7.25%
Water calculated for C19H13O7SBa+2H2O = 7.35%

Although these analyses show a per cent. of Ba somewhat above that required by a compound having the formula C19H13O7SBa still this appears to be the most probable formula which can be assigned to the substance. If this is the true composition of the salt, then in sulphonfluoresceïn the anhydride condition must be broken up by boiling with BaCO3 forming the salt thus.

⎧ OH ⎧ OH
│ ╱ │ ╱
│ C6H3 │ C6H3
│ ╲ │ ╲
│ O │ O
│ ╱ │ ╱
C ⎨ C6H3    + 2H2O =  C ⎨ C6H3
│ ╲ │ ╲
│ OH │ OH
│ C6H4SO2 │ C6H4SO2OH
│  ╱ │
⎩ O ⎩ OH














⎧ OH ⎧ OH
│ ╱ │ ╱
│ C6H3 │ C6H3
│ ╲ │ ╲
│ O │ O
│ ╱ │ ╱
2  C ⎨ C6H3  + BaCO3 = C ⎨ C6H3  + Ba
│ ╲ │ ╲
│ OH │ OH
│ C6H4SO2OH │ C6H4SO2OH
│ │
⎩ OH ⎩ OH  2