1048 
THE VOYAGE OF H.M.S. CHALLENGER. 
Analysis of a Manganese Nodule } 
By Professor A. Benard. 
Station 276 ; 2350 fathoms, South Pacific. 
I. 0’827l grm. of substance dried at 100° gave 0’0787 grm. water (H 2 0), 0‘1600 
grm. silica (Si0 2 ), 0'0264 grm. of lime (CaO), 0‘0526 grm. alumina (A1 2 0 3 ), 0‘2208 
grm. peroxide of iron (Fe 2 0 3 ), 0'0148 grm. magnesia (MgO), 0’2354 grm. manganic 
oxide (Mn 2 0 3 ) corresponding to 0‘2189 grm. of manganous oxide (MnO), 0‘0119 grm. 
nickel (Ni) corresponding to 0'0151 grm. oxide of nickel. 
II. 0‘1425 grm. of substance dried at 100° treated with hydrochloric acid and the 
resulting gas conducted into a solution of potassium iodide liberated iodine; 12 c.c. of 
potassium thiosulphate (1 c.c. = 0'937 c.c. of the standard solution) ; 1 c.c. of the standard 
solution = ^ or whence 1 c.c. = 3'55 grms. of chlorine or 0'8 grm. of oxygen — 
1000 : 0-8 = 12 x0-9377 : a*. 
.-. 1000 : 0-8 = 11-24 : x. 
£c = 0 , 008992 grm. of oxygen capable of liberating chlorine from hydrochloric acid, 
i.e., 6'31 per cent, oxygen. 
The atomic ratio of 0 '3 84 O is required if Mn be present as Mn0 2 and Ni as Ni 2 0 3 , 
but 0‘394 0 was the ratio observed — 
Manganous oxide, 
Nickel, 
Oxygen, 
a 
26-46 
1-82 
6-31 
MnO = 71 0-372 
Ni =74-8 0-024 
O =16 0-394 
0 - 372 +^ = 0-384 
The formula Mn0 2 + ^H 2 0 requires 9T8 per cent, water. Consequently 26‘4 6 per- 
cent. manganous oxide, which corresponds to 3 2 '42 per cent, manganese binoxide, is 
equivalent to 3'28 per cent, water. 
267 per cent, ferric oxide require as limonite 4‘50 per cent, water. 
Water (H 2 0), . 
Silica (Si0 2 ), 
Lime (CaO), 
Alumina (A1 2 0 3 ), 
Ferric oxide (Fe 2 0 3 ), 
Magnesia (MgO), 
Manganous oxide (MnO), 
Nickel oxide (NiO), 
Oxygen (0 2 ), . 
I. IT 
9-51 
9-51 
19-34 
19-34 
3-19 
3-19 
6-36 
6-36 
26-70 
26-70 
1 79 
1-79 
26-46 
26-46 
1-82 
1-82 
6-31 
6-31 
101-48 
1 In the remainder of this Appendix the symbols are used with their ordinary value, H = 1 and O = 16. — J. M. 
