There is here a deficiency of 5.76 per cent. due to oxygen. Nothing else could be found, and it is known that in the sulphates the metals exist as oxides. By multiplying the weight of the copper by 1.252, the weight of copper oxide (CuO) will be ascertained; in this case it equals 10.57 per cent. The ferrous iron multiplied by 1.286 will give the ferrous oxide (FeO); in this case 15.19 per cent. The ferric iron multiplied by 1.428 will give the ferric oxide (Fe_{2}O_{3}); in this case 0.54 per cent. The zinc multiplied by 1.246 will give the zinc oxide (ZnO); in this case it equals 0.35 per cent. The a.n.a.lysis will be reported as--
Water 44.51 Sulphuric oxide 28.82 Copper oxide 10.57 equal to copper 8.44% Ferrous oxide 15.19 Ferric oxide 0.54 Zinc oxide 0.35 ----- 99.98
The following (A) is an a.n.a.lysis of a sample of South American copper ore, which will serve as a further ill.u.s.tration. The a.n.a.lysis showed the presence of 6.89 per cent. of ferrous oxide, and some oxide of copper.
The a.n.a.lysis (B) is that of an ore from the same mine after an imperfect roasting. It will be seen that the carbonates have been converted into sulphates. If the total sulphur simply had been determined, and the sulphate overlooked, the "oxygen and loss" would have been 5.65 per cent., an amount which would obviously require an explanation.
A. B.
Water 0.25 0.59 Organic matter 0.54 -- Sulphur 29.50 21.33 Copper 10.92 9.80 {Copper 9.57 {Copper oxide 0.28 Iron 32.09 39.73 {Iron 34.32 {Ferric oxide 7.73 Lead 0.35 0.12 Zinc 0.86 0.69 Cobalt 0.06 0.11 Lime 5.25 7.69 Magnesia 2.33 2.55 Sulphuric oxide 1.00 5.30 Carbon dioxide 8.87 -- "Insoluble silicates" 5.12 8.38 Oxygen and loss 2.86 2.47 ----- Potash 0.15 100.00 Soda 1.09 ----- 100.00
WATER.
Water occurs in minerals in two forms, free and combined. The term "moisture" ought, strictly, to be limited to the first, although, as has already been explained, it is more convenient in a.s.saying to apply the term to all water which is driven off on drying at 100 C. The combined water is really a part of the mineral itself, although it may be driven off at a high temperature, which varies with the base. In some cases a prolonged red heat is required; whilst with crystallised salts it is sometimes given off at the ordinary temperatures. This latter phenomenon, known as efflorescence, is mostly confined to artificial salts.
The determination of the combined water may often be made by simply igniting the substance from which the moisture has been removed. The quant.i.ty of water may be determined, either indirectly by the loss, or directly by collecting it in a calcium chloride tube, and weighing. In some cases, in which the loss on ignition does not give simply the proportion of combined water, it can be seen from the a.n.a.lysis to what else the loss is due; and, after a proper deduction, the amount of water can be estimated. For example, 1 gram of crystallised iron sulphate was found to contain on a.n.a.lysis 0.2877 gram of sulphuric oxide; and on igniting another gram, 0.2877 gram of ferric oxide was left. As the salt is known to be made up of ferrous oxide, sulphuric oxide, and combined water, the combined water can be thus calculated: 0.2877 gram of ferric oxide is equal to 0.2589 gram of ferrous oxide,[98] and consequently, the loss on ignition has been diminished by 0.0288 gram, which is the weight of oxygen absorbed by the ferrous oxide during calcining. The loss on ignition was 0.7123 gram, to which must be added 0.0288 gram; hence 0.7411 gram is the weight of the combined sulphuric oxide and water present. Deducting the weight of sulphuric oxide found, 0.2877 gram, there is left for combined water 0.4534 gram. The composition of 1 gram of the dry salt is then:--
Water 0.4534 Sulphuric oxide 0.2877 Ferrous oxide 0.2589 ------ 1.0000
The following is another example:--A sample of malachite lost on ignition 28.47 per cent., leaving a residue which was found on a.n.a.lysis to be made up of oxide of copper (equal to 70.16 per cent. on the mineral), and silica and oxide of iron (equal to 1.37 per cent.). Carbon dioxide and water (but nothing else) was found to be present, and the carbon dioxide amounted to 19.64 per cent.; deducting this from the loss on ignition, we have 8.82 as the percentage of water present. The a.n.a.lysis was then reported as follows:--
Cupric oxide 70.16 equal to 56.0% copper.
Silica and ferric oxide 1.37 Carbon dioxide 19.64 Water 8.82 ----- 99.99
[Ill.u.s.tration: FIG. 63.]
~Direct Determination of Combined Water.~--Transfer about 3 grams of the substance to a piece of combustion tube (8 or 10 inches long), attached (as in fig. 63) at one end to a ~U~-tube containing sulphuric acid, and at the other end to a calcium chloride tube. The last is weighed previous to the determination. The tube should be warmed to ensure complete dryness, and must be free from a misty appearance. Aspirate a current of air through the apparatus, heat the mineral by means of a Bunsen burner, cautiously at first, and afterwards to redness (if necessary). The water is driven off and condenses in the calcium chloride tube, which is afterwards cooled and weighed. The increase in weight is due to the water. If the substance gives off acid products on heating, it is previously mixed with some dry oxide of lead or pure calcined magnesia.
EXAMINATION OF WATERS.
The a.s.sayer is occasionally called on to test water for the purpose of ascertaining the nature and quant.i.ty of the salts contained in it, and whether it is or is not fit for technical and drinking purposes.
In mineral districts the water is generally of exceptional character, being more or less charged, not only with earthy salts, but also frequently with those of the metals. Distilled water is only used by a.s.sayers in certain exceptional cases, so that by many it would be cla.s.sed among the rarer oxides. Water of ordinary purity will do for most purposes, but the nature and quant.i.ty of the impurities must be known.
The following determinations are of chief importance:--
~Total Solids at 100 C.~--Where simply the amount is required, take 100 c.c. and evaporate on the water-bath in a weighed dish; then dry in the water-oven, and weigh.
~Total Solids Ignited.~--The above residue is very gently ignited (keeping the heat well below redness), and again weighed. A larger loss than 4 or 5 parts per 100,000 on the water requires an explanation.
~Chlorine.~--Take 100 c.c. of the water in a porcelain dish, add 2 c.c.
of a 5 per cent. solution of neutral pota.s.sic chromate, and t.i.trate with a neutral standard solution of nitrate of silver, made by dissolving 4.789 grams of crystallised silver nitrate in distilled water, and diluting to 1 litre. The addition of the nitrate of silver is continued until the yellow of the solution a.s.sumes a reddish tint. The reaction is very sharp. Each c.c. of nitrate of silver used is equal to 1 part by weight of chlorine in 100,000 of water. At inland places this rarely amounts to more than 1 in 100,000; but near the sea it may amount to 3 or 5. More than this requires explanation, and generally indicates sewage pollution.
~Nitric Pentoxide (N_{2}O_{5}).~--It is more generally reported under the heading, "nitrogen as nitrates." Take 250 c.c. of the water and evaporate to 2 or 3 c.c.; acidulate with a few drops of dilute sulphuric acid, and transfer to a nitrometer (using strong sulphuric acid to wash in the last traces). The sulphuric acid must be added to at least twice the bulk of the liquid. Shake up with mercury. The mercury rapidly flours, and nitric oxide is given off (if any nitrate is present). The volume of the nitric oxide (corrected to normal temperature and pressure), multiplied by 0.25, gives the parts of nitrogen per 100,000; or, multiplied by 0.965, will give the nitric pentoxide in parts per 100,000. In well and spring waters the nitrogen may amount to 0.3 or 0.4 parts per 100,000; or in richly cultivated districts 0.7 or 0.8 parts per 100,000. An excess of nitrates is a suspicious feature, and is generally due to previous contamination.
~Ammonia.~--Take 500 c.c. of the water and place them in a retort connected with a Liebig"s condenser. Add a drop or two of a solution of carbonate of soda and distil over 100 c.c.; collect another 50 c.c.
separately. Determine the ammonia in the distillate colorimetrically (with Nessler"s solution, as described under _Ammonia_) and compare with a standard solution of ammonic chloride containing 0.0315 gram of ammonic chloride in 1 litre of water. One c.c. contains 0.01 milligram of ammonia. The second distillate will show little, if any, ammonia in ordinary cases. The amounts found in both distillates are added together, and expressed in parts per 100,000.
Waters (other than rain and tank waters) which contain more than 0.003 per 100,000 are suspicious.
~Organic Matter.~--The organic matter cannot be determined directly; but for ordinary purposes it may be measured by the amount of permanganate of pota.s.sium which it reduces, or by the amount of ammonia which it evolves on boiling with an alkaline permanganate of pota.s.sium solution.
A. _Alb.u.minoid Ammonia._--To the residue left after distilling the ammonia add 50 c.c. of a solution made by dissolving 200 grams of potash and 8 grams of pota.s.sium permanganate in 1100 c.c. of water, and rapidly boiling till the volume is reduced to 1 litre (this should be kept in a well stoppered bottle, and be occasionally tested to see that it is free from ammonia). Continue the distillation, collecting 50 c.c. at a time, until the distillate is free from ammonia. Three or four fractions are generally sufficient. Determine the ammonia colorimetrically as before.
If the total alb.u.minoid ammonia does not exceed 0.005 in 100,000, the water may be regarded as clean as regards organic matter; if it amounts to more than 0.015, it is dirty.
B. _Oxygen Consumed._--A standard solution of permanganate of potash is made by dissolving 0.395 gram of the salt in water and diluting to 1 litre. Each c.c. equals 0.1 milligram of available oxygen. The following are also required:--1. A solution of sodium hyposulphite containing 1 gram of the salt (Na_{2}S_{2}O_{3}.5H_{2}O) in 1 litre of water. 2.
Dilute sulphuric acid, made by adding one part of the acid to three of water, and t.i.trating with the permanganate solution till a faint pink persists after warming for several hours. 3. Starch paste. 4. Pota.s.sium iodide solution.
Take 250 c.c. of the water in a stoppered bottle, add 10 c.c. of sulphuric acid and 10 c.c. of the permanganate, and allow to stand in a warm place for four hours. Then add a few drops of the solution of pota.s.sium iodide, and t.i.trate the liberated iodine with "hypo," using starch paste towards the end as an indicator. To standardise the hyposulphite, take 250 c.c. of water and 10 c.c. of sulphuric acid, and a few drops of pota.s.sium iodide; then run in 10 c.c. of the "permanganate" solution, and again t.i.trate; about 30 c.c. of the "hypo"
will be used. The difference in the two t.i.trations, divided by the last and multiplied by 10, will give the c.c. of permanganate solution used in oxidising the organic matter in the 250 c.c. of water. Each c.c.
represents 0.04 parts of oxygen in 100,000.
~Metals.~--These may for the most part be estimated colorimetrically.
~Lead.~--Take 100 c.c. of the water in a Nessler tube, and add 10 c.c.
of sulphuretted hydrogen water, and compare the tint, if any, against a standard lead solution, as described under _Colorimetric Lead_. Report in parts per 100,000.
~Copper.~--Proceed as with the last-mentioned metal; but, if lead is also present, boil down 500 c.c. to about 50 c.c., then add ammonia, filter, and estimate the copper in the blue solution, as described under _Colorimetric Copper_.
~Iron.~--Take 50 c.c., or a smaller quant.i.ty (if necessary), dilute up to the mark with distilled water, and determine with pota.s.sium sulphocyanate, as described under _Colorimetric Iron_.
~Zinc.~--Zinc is the only other metal likely to be present; and, since it cannot be determined colorimetrically, it must be separately estimated during the examination of the "total solids."
~Examination of "Total Solids."~--Evaporate 500 c.c. to dryness with a drop or two of hydrochloric acid. Take up with hydrochloric acid, filter, ignite, and weigh the residue as "silica." To the filtrate add a little ammonic chloride and ammonia, boil and filter, ignite, and weigh the precipitate as "oxide of iron and alumina." Collect the filtrate in a small flask, add a few drops of ammonium sulphide or pa.s.s sulphuretted hydrogen, cork the flask, and allow to stand overnight; filter, wash, and determine the zinc gravimetrically as oxide of zinc. If copper or lead were present, they should have been previously removed with sulphuretted hydrogen in the acid solution. To the filtrate add ammonic oxalate and ammonia, boil for some time, allow to stand, filter, wash, ignite, and weigh as "lime." Evaporate the filtrate with nitric acid, and ignite. Take up with a few drops of dilute hydrochloric acid, add baric hydrate in excess, evaporate, and extract with water. The residue contains the magnesia; boil with dilute sulphuric acid, filter, precipitate it with phosphate of soda and ammonia, and weigh as pyrophosphate. The aqueous extract contains the alkalies with the excess of barium. Add sulphuric acid in slight excess, filter, evaporate, and ignite strongly. The residue consists of the sulphates of the alkalies (which are separately determined, as described under _Potash_).
~Sulphuric Oxide (SO_{3}).~--Take 200 c.c. and boil to a small bulk with a little hydrochloric acid, filter (if necessary), add baric chloride solution in slight excess to the hot solution, filter, ignite, and weigh as baric sulphate.
~Carbon Dioxide (free).~--Carbon dioxide exists in waters in two forms, free and combined. The latter generally occurs as bicarbonate, although on a.n.a.lysis it is more convenient to consider it as carbonate, and to count the excess of carbon dioxide with the free. The method is as follows:--To determine the free carbon dioxide, take 100 c.c. of the water, place them in a flask with 3 c.c. of a strong solution of calcium chloride and 2 c.c. of a solution of ammonic chloride, next add 50 c.c.
of lime-water. The strength of the lime-water must be known. Make up to 200 c.c. with distilled water, stop the flask, and allow the precipitate to settle. Take out 100 c.c. of the clear solution with a pipette, and t.i.trate with the standard solution of acid.[99] The number of c.c.
required, multiplied by two, and deducted from that required for the 50 c.c. of lime-water, and then multiplied by 0.0045, will give the carbon dioxide present other than as normal carbonates.
~Carbon Dioxide combined~ as normal carbonate.--100 c.c. of the water are tinted with phenacetolin or lacmoid; then heated to near boiling, and t.i.trated with standard acid. The number of c.c. used, multiplied by 0.0045, will give the weight in grams of the combined carbon dioxide.
~Free Acid.~--In some waters (especially those from mining districts) there will be no carbonates. On the contrary, there may be free mineral acid or acid salts. In these cases it is necessary to determine the amount of acid (other than carbon dioxide) present in excess of that required to form normal salts. This is done in the following way:--Make an ammoniacal copper solution by taking 13 grams of copper sulphate (CuSO_{4}.5H_{2}O), dissolving in water, adding solution of ammonia until the precipitate first formed has nearly dissolved, and diluting to 1 litre. Allow to settle, and decant off the clear liquid. The strength of this solution is determined by t.i.trating against 10 or 20 c.c. of the standard solution of sulphuric acid (100 c.c. = 1 gram H_{2}SO_{4}). The finishing point is reached as soon as the solution becomes turbid from precipitated cupric hydrate. At first, as each drop falls into the acid solution, the ammonia and cupric hydrate combine with the free acid to form ammonic and cupric sulphates; but as soon as the free acid is used up, the ammonia in the next drop not only precipitates an equivalent of cupric hydrate from the solution, but also throws down that carried by itself. This method is applicable in the presence of metallic sulphates _other than ferric_. The standardising and t.i.tration should be made under the same conditions. Since sulphuric acid and sulphates are predominant in waters of this kind, it is most convenient to report the acidity of the water as equivalent to so much sulphuric acid.
~Dissolved Oxygen.~--For the gasometric method of a.n.a.lysing for dissolved oxygen, and for the Schutzenberger"s volumetric method, the student is referred to Sutton"s "Volumetric a.n.a.lysis." The following is an easy method of estimating the free oxygen in a water:--Take 20 c.c.
of a stannous chloride solution (about 20 grams of the salt with 10 c.c.
of hydrochloric acid to the litre); add 10 c.c. of hydrochloric acid, and t.i.trate in an atmosphere of carbon dioxide with standard permanganate of pota.s.sium solution (made by dissolving 1.975 gram of the salt in 1 litre of water: 1 c.c. equals 0.5 milligram of oxygen). A similar t.i.tration is made with the addition of 100 c.c. of the water to be tested. Less permanganate will be required in the second t.i.tration, according to the amount of oxygen in the water; and the difference, multiplied by 0.5, will give the weight of the oxygen in milligrams.
Small quant.i.ties of nitrates do not interfere.
In REPORTING the results of the a.n.a.lysis, it is customary to combine the acids and bases found on some such principle as the following:--The sulphuric oxide is calculated as combined with the potash, and reported as pota.s.sic sulphate (K_{2}SO_{4}); the balance of the sulphuric oxide is then apportioned to the soda, and reported as sulphate of soda (Na_{2}SO_{4}); if any is still left, it is reported as calcium sulphate (CaSO_{4}), and after that as magnesic sulphate (MgSO_{4}). When the sulphuric oxide has been satisfied, the chlorine is distributed, taking the bases in the same order, then the nitric pentoxide, and lastly the carbon dioxide. But any method for thus combining the bases and acids must be arbitrary and inaccurate. It is extremely improbable that any simple statement can represent the manner in which the bases and acids are distributed whilst in solution; and, since different chemists are not agreed as to any one system, it is better to give up the attempt, and simply state the results of the a.n.a.lysis. This has only one inconvenience. The bases are represented as oxides; and, since some of them are present as chlorides, the sum total of the a.n.a.lysis will be in excess of the actual amount present by the weight of the oxygen equivalent to the chlorine present as chloride. The following is an example of such a statement:--
Parts per 100,000.
Total solids, dried at 100 C. 28.73 Chlorine 1.70 Nitrogen as nitrate 0.03 Ammonia 0.001 Alb.u.minoid ammonia 0.004 "Oxygen consumed" in 4 hours 0.01
The solids were made up as under:--
Per 100,000 of the Water.