A text-book of assaying : $b for the use of those connected with mines.Beringer, C. (Cornelius)
Science
A text-book of assaying : $b for the use of those connected with mines.
Beringer, C. (Cornelius)
Assaying
Very similar to this last problem is the question suggested under the
head "Indirect Titration" (p. 43). "If 100 c.c. of the standard
permanganate solution are equivalent to 1 gram of iron, how much
peroxide of manganese will they be equivalent to?" The equation for
dissolving the iron is already given; the second equation is
2FeSO_{4} + MnO_{2} + 2H_{2}SO_{4}
| = Fe_{2}(SO_{4})_{2} + MnSO_{4} + 2H_{2}O
|
V
Mn = 55
O_{2} = 32
--
87
It will be seen that 87 grams of peroxide of manganese are equivalent to
112 grams of iron. How much then is equivalent to 1 gram of iron?
112 : 87 :: 1 gram : _x_
_x_ = 0.7767 gram.
It is sometimes convenient to calculate the formula of a substance from
its analysis. The method of calculating is shown by the following
example. Required the formula of a mineral which gave the following
figures on analysis:--
Cupric oxide (CuO) 10.58
Ferrous oxide (FeO) 15.69
Zinc oxide (ZnO) 0.35
Sulphuric oxide (SO_{2}) 28.82
Water (H_{2}O) 44.71
------
100.15
First find the molecular weights of CuO, FeO, &c., and divide the
corresponding percentages by these figures. Thus, CuO = 63.3+16 = 79.3
and 10.58 divided by 79.3 gives 0.1334. Similarly FeO = 56+16 = 72 and
15.69 divided by 72 gives 0.2179. Treated in the same way the oxide of
zinc, sulphuric oxide and water give as results 0.0043, 0.3602 and
2.484.
Classify the results as follows:--
Bases. Acids. Water.
CuO 0.1334 SO_{3} 0.3602 H_{2}O 2.484
FeO 0.2179
ZnO 0.0043
---------- ------------- ------------
RO 0.3556 RO_{3} 0.3602 R_{2}O 2.484
The figures 0.3556, 0.3602 and 2.484 should be then divided by the
lowest of them--_i.e._, 0.3556; or where, as in this case, two of the
figures are very near each other the mean of these may be taken--_i.e._,
0.3579. Whichever is taken the figures got will be approximately 1, 1
and 7. The formula is then RO.SO_{3}.7H_{2}O in which R is nearly 2/5ths
copper, 3/5ths iron and a little zinc.
This formula requires the following percentage composition, which for
the sake of comparison is placed side by side with the actual results.
Calculated. Found.
Cupric oxide 11.29 10.58
Ferrous oxide 15.37 15.69
Zinc oxide nil 0.35
Sulphuric oxide 28.47 28.82
Water 44.84 44.71
----- ------
99.97 100.15
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