Analysis of Chloridized Ore Assay

Analysis of Chloridized Ore Assay

The following paper deals with a method for the “Approximate Analysis ” of Chloridized Ores, which are free from arsenic, antimony and lime. When these are present the analysis is rendered far more complicated, and many of the methods which I am about to describe could not be applied. Sufficient scope is therefore offered for discussing the analysis, when these metals are present, in a separate paper, which I hope to contribute at a future date.

I am fully aware that this method is far from being perfect. No doubt many modifications and improvements will suggest themselves to my readers, and I trust that this paper will be the means of giving a certain impetus to the study of Proximate Analysis which at the present time is practically dormant. For the benefit of my readers who are not initiated into the details of ordinary estimations, I have briefly described well known methods, which, however, will be found fully described in most text-books of analysis.

Assuming that the ore contains sulphates and chlorides of soda and magnesia, sulphates, sulphides, chlorides and oxides of copper, lead, iron, zinc, manganese and silver chloride, the analysis would be conducted as follows:—The first compounds to estimate are the chlorides and sulphates of soda and magnesia.

Take a convenient portion of the roasted ore (according to the percentage of (Na Cl) as used in chloridizing; say the ore was chloritized with 10 % sodium chloride, take about 20 gms.) and digest with cold water, filter, wash until all soluble salts are removed. Acidify the filtrate slightly with hydrochloric acid, and through it pass hydrogen sulphide to saturation. Boil to expel H2S and filter off the insoluble ppt. To the filtrate add ammonium chloride, ammonia in excess, then ammonium sulphide. Boil and filter. The filtrate now contains sodium and magnesium salts only.

These are estimated as follows:

Make the solution up to a convenient bulk, and divide into two equal portions. In one portion estimate chlorine (Cl) and sulphuric oxide (SO3) as follows—Evaporate, the solution to dryness and ignite gently to drive off ammonium salts. After cooling take up the residue with water, make up to a definite bulk and divide into two portions.

In one, estimate chlorine. The chlorine is estimated volumetrically by means of a standard solution of silver nitrate. The silver solution is prepared by dissolving 16.96 grammes of pure silver nitrate in water and diluting up to 1 litre.

According to the equation Ag NO3 + X Cl = Ag Cl + X NO3.
169.6 = 35.37.
In decinormal solution 16.96 Ag NO3 = 3.537 Cl.
∴ 1 c.c. = .003537 grams, chlorine.

To standardize the solution, dissolve 5.837 grammes of pure and dry sodium chlorine in 1 litre of water. One c.c. of this solution contains .003537 grammes chlorine.

Take 5 c.c. of the solution and add 1 c.c. of a 1 % solution of yellow potassium chromate, Make the bulk up to 20 c.c.

Titrate with silver solution until a permanent red coloration of the solution is produced.

The actual determination should be conducted under similar conditions of bulk of solution and temperature of solution.

Note the amount of chlorine found.

In the other portion of solution, estimate sulphuric oxide by the following method:—The principle of the method depends upon the precipitation of SO3 by means of a standard solution of barium chloride, and subsequently estimating the excess of barium solution volumetrically by means of a standard solution of potassium chromate.

To prepare the barium chloride solution, dissolve 60.98 grammes of the pure crystallized salt in 1 litre of water. One c. c. of this solution should correspond to 0.02 grams of H2SO4 or 0.016 SO3. The neutral chromate solution is prepared by dissolving 18.451 grammes of potassium bichromate in H2O, adding ammonia until the reddish yellow colour has changed to pale yellow and diluting to 1 litre.

To check these solutions, proceed as follows:

Take 10 c.c. of the barium solution, dilute to 50 c.c. and add 20.4 c.c. of the chromate solution. After the ppt. of barium chromate has subsided, the supernatant fluid must appear yellowish from slight excess of chromate of potassa.

Now add barium chloride drop by drop. If the two solutions are in correct relation to each other, 0.2 c. c. will be required.

In the actual determination, add barium chloride to the sulphate solution until no further precipitation takes place. Then titrate with chromate solution until the excess is marked by the yellowish coloration. When this point is reached, add barium chloride until the solution is completely decolorized.

Divide the number of c. cs. of chromate used by 2, then deduct the result from the whole of the number of c. cs. of chloride of barium used, and from the remainder calculate the SO3.

Having now determined the amount of chlorine and sulphuric oxide in portion of the solution which represents ¼ of the total amount of salts contained in the original portion of the ore operated upon, multiply by 4, and we have the total amount of chlorine and sulphur trioxide, which is combined with sodium and magnesium.

Previous to calculating the sodium chloride and sulphate, magnesium sulphate and chloride must be found.

To this end proceed as follows—Evaporate the second half of the solution to dryness and expel ammonium salts by gentle ignition. Take up the residue with water. Add ammonia in excess, and some ammonium chloride to the solution. Precipitate the magnesia with a slight excess of ammonium phosphate. This yields magnesium phosphate and ammonium chloride and sulphate.

The excess of ammonium phosphate must be removed as follows—Drive off free ammonia by continuous boiling, if necessary, neutralize with hydrochloric acid. Add ferric chloride until the fluid is yellowish, then ammonium carbonate until the solution is neutral or only acid with carbon dioxide, boil and filter off the insoluble basic iron phosphate, wash, evaporate and drive off ammonium salts by gentle ignition.

Having thus removed ammonium salts take up the residue with water and estimate chlorine and sulphuric oxide, by methods already described, multiply the amounts found by 2. The total amounts will be found to have diminished.

From these data, the amount of sodium and magnesium, chloride and sulphate, is calculated as follows—Suppose the total amount of chlorine found was one gramme.

After removing magnesia, we find only .7 grams of chlorine. Then 1 gramme — .7 gramme = .3 gramme i.e. 0.7 grams of Cl is combined with sodium, and since the molecular weight of sodium chloride is 58.37 and the atomic weight of chlorine is 35.37, then 35.37 chlorine = 58.37 sodic chloride,

∴ 58.37 x 0.7/35.37 =1.15 grms. NaCl.

Also 0.3 gramme of Cl is combined with magnesium. The molecular weight of magnesium chloride is 94.74 and the chlorine 70.74

∴ 94.74 x 0.3/70.74 = 0.40 grams MgCl2

In the same way sodium and magnesium sulphate is calculated from the following:

79.94 SO3 = 141.92 sodium sulphate
79.94 SO3 = 119.92 magnesium sulphate.

Having now found the amounts of the sodium and magnesium salts, the next operation is the determination of the quantity of lead chloride or lead sulphate or both, soluble in cold water. Digest a portion of the one with cold water, filter and wash. To the filtrate add HNO3 to peroxidize the iron. Neutralize the filtrate with ammonium carbonate, then add an excess of ammonium acetate.

Estimate the lead volumetrically by titration with potassium bi-chromate.

This method depends upon the, precipitation of lead from its solution by means of a standard solution of potassium bi-chromate. The end reaction is determined by means of neutral silver nitrate solution. When the bi-chromate has been added in excess, a drop of the fluid which is being titrated is brought in contact with a drop of silver solution, when the well known red coloration of silver chromate occurs.

Note the amount of lead found and also the amount of bi-chromate required.

Treat another portion of the ore with cold water. Per-oxidize the iron with nitric acid, and add sufficient bi-chromate to precipitate lead.

Filter and neutralize nitric acid present by means of dilute hydrochloric acid, through the solution pass H2S to saturation. Filter and wash. To filtrate, add ammonium chloride, ammonia in excess, then ammonium sulphide, filter, wash, evaporate to dryness and ignite. Take up the residue with water, make up to a definite bulk and divide into two portions.

In one estimate chlorine and in the other sulphuric oxide. An increase of these will be found. This is the amount of acid combined with lead which is soluble in cold water. From these quantities we calculate lead chloride and sulphate according to 70.74 chlorine = 277.14 lead chloride, and 79.94 sulphuric oxide = 302.32 lead sulphate. The amount of lead found by this calculation should agree with the amount found by titration.

The insoluble residue which has been freed from its most soluble salts by digestion with cold water should now be treated with boiling water, and the soluble lead salts estimated in the same way.

Next in order to be estimated are the salts of manganese.

Treat as before with cold water, and remove lead by potassium bi-chromate. Filter and wash. To the filtrate, add ammonium chloride, then ammonia to remove iron. Filter rapidly and neutralize the filtrate with hydrochloric acid.

The solution is now titrated with a solution of potassium permanganate of such a strength that 30 c. c correspond to about 1 gramme of manganese.

“This method depends upon the fact that when a solution of potassium permanganate is added to a dilute neutral or barely acid solution of manganese at 8o° temperature, the whole of the manganese, both in the original solution and in the precipitant, is precipitated as MnO2HO. The end of the reaction is evidenced by the pink coloration of the fluid.

The permanganate is standardized by iron or by manganese. In the actual determination, the solution should be diluted to 1 litre and a temperature of 80° maintained.”

After titration, the insoluble manganese is filtered off, the solution made slightly acid with hydrochloric acid, and then saturated with H2S. Boil and filter. To filtrate add NH4HO, then NH4HS. Filter, evaporate the filtrate and expel ammonium salts by ignition. Estimate chlorine and sulphuric oxide in the residue.

Again these acids will be found to have increased.
Manganese chloride and sulphate is then calculated from the following:

70.74 chlorine = 125.54 MnCl2, and
79.94 sulphuric oxide = 150.72 MnSO4.

Take another portion of the ore, and after treating with cold water, remove lead, iron, and manganese as already described, when estimating manganese.

Acidify with HCl and saturate with H2S. Copper is precipitated as copper sulphide and free hydrochloric acid; sulphuric acid, zinc chloride- and zinc sulphate remain in solution. Filter off the copper sulphide, and after perfect removal of H2S by boiling, to the solution add silver nitrate in slight excess.

This removes free HCl, and the zinc and sodium and potassium chlorides are converted to nitrates. Filter. To the filtrate, add an excess of lead carbonate. Filter off the insoluble lead sulphate and divide the solution into two portions. In one portion, estimate sulphuric oxide. Again the amount of SO3 will be found to have increased.

This is the sulphuric oxide combined as zinc sulphate, and according to 79.94 SO3 = Z160.82 n SO4, the amount of that salt is calculated. In the other portion, estimate zinc and calculate the total.

Deduct the amount found as zinc sulphate from the total. The remainder is zinc as zinc chloride, and since the atomic weight of zinc is 64.9, therefore 64.9 Zn = 135.64 ZnCl2. There now remains to be found iron and copper chlorides and sulphates.

Take another portion of the ore and digest with boiling water. Divide the solution into three equal portions. Estimate total chlorine from one portion and total sulphuric oxide from another. In the third portion remove lead and copper by H2S. In this precipitate, estimate copper volumetrically by the potassic iodide method. By this method very small quantities of copper can be conveniently estimated. The method depends upon the fact that when potassic iodide is added in excess to a solution of copper acetate, made acid by acetic acid, cupric iodide is precipitated and an equivalent of iodine liberated. By measuring the iodine thus set free by means of a standard sodium hyposulphite solution, the amount of copper is determined. The ppt. of copper and lead should be dissolved in nitric acid, boiled to dryness and taken up with water. A slight excess of ammonia is then added and the insoluble lead hydroxide filtered. The filtrate is then made acid with glacial acetic acid, when it is ready for the addition of potassium iodide crystals and subsequent titration with hyposulphite.

The standard “ hypo ” solution is prepared by dissolving 38 grammes of the pure and dried salt in 1 litre of water. One c.c. of this solution will equal about .01 grammes of Cu. To the filtrate add ammonium chloride and ammonia in excess. The iron is precipitated as hydrate, whilst corresponding ammonium salts are formed. Filter. From the hydrate estimate iron.

Copper Subchloride Method

“The method depends upon the reduction of ferric salts to ferrous salts by means of copper subchloride. The completion of the reduction is ascertained by means of potassium sulphocyanide, on the addition of this salt to a per-salt a deep red coloration is produced, on adding the solution of sub-chloride the red coloration is bleached. The addition of sub-chloride after this point is reached produces a precipitate of copper sub-sulphocyanide, we therefore have a double indication to the end of the reduction.

The solution of subchloride is prepared by dissolving sheet copper in nitric acid, evaporate to drive off the excess of nitric, and dissolve the residue in water containing hydrochloric acid.

The solution is put into a flask and common salt, equal in weight to the dry residue of cupric nitrate, is added. Add pieces of sheet copper to the solution and boil until the solution is nearly colorless, and all the cupric chloride has been converted to subchloride. The flask is then corked and allowed to cool, the solution is then diluted with water containing hydrochloric acid until 1 c.c = 6 milligrammes of iron.

In order to prevent the copper solution from oxidation, it should be kept in a bottle having an air tight stopper to which is attached a thick spiral of copper wire.

In the actual determination few rules are to be observed. The solution should be decidedly acid and also very dilute and only a very few drops of sulphocyanide added.”

The filtrate is evaporated and ignited to drive off salts of ammonium. The residue is then examined for chlorine and sulphuric oxide. Comparing these results with the total amount of chlorine and sulphuric oxide in the ore, a loss is found. From this chloride and sulphate of iron is found from the following:

79.94 SO3 = 151.92 Fe SO4
239.82 SO3 = 398.76 Fe2 (SO4)3
70.74 Cl = 126.74 Fe Cl2
212.22 Cl = 324.22 Fe2 Cl6

In this case we have to determine iron in both its forms. The amount of each form can be deduced from the molecular weights, but should be checked by determining the amount of each form of soluble iron on a separate portion of the ore. Calculate according to the formula.

(151.92 + 398.76) X SO3 found/(79.94 + 239.82) = amount 0f ferrous and ferric sulphate, then by proportion find the amount of ferrous sulphate, and by difference the amount of ferric sulphate. In the same way, ferrous and ferric chloride are calculated.

The copper salts are now to be dealt with.

The amount of soluble copper has already been found. Add together the amounts of sulphuric oxide and chlorine already found and then deduct these from the total amounts in the ore, which occur in a soluble form. The difference is the amount of each of these acids which is combined with copper.

As with the salts of iron, the cupric and cuprous salts have to be calculated according to the following:

79.94 SO3 = 159.32 CuSO4.
35.37 Cl = 98.77 CuCl.
70.74 Cl = 134.14 CuCl2.

This concludes the estimation of salts soluble in water, and we now have to consider the insoluble lead sulphate and oxide.

To this end proceed as follows:

Digest a portion of the ore with cold water, then with boiling water, and lastly with ammonium acetate.

Divide the ammonium acetate solution into two equal portions. In one portion estimate lead and in the other sulphuric oxide. From the molecular weight of lead sulphate and sulphuric oxide, lead sulphate is calculated. Deduct this from the total lead found in the ammonium acetate solution and the remainder is the amount of lead as lead oxide.

We next have to estimate copper and zinc oxide. After having removed soluble salts from a portion of ore, digest with a mixture of ammonia and ammonium sulphate. The solution is divided and copper and zinc estimated, and from these the amount of copper oxide and zinc oxide determined.

(The ammoniacal solution will also contain silver chloride. This had better be estimated by difference on a separate portion of the ore. For this purpose the chloridized ore is assayed for its silver contents. Portion of the ore is then washed with cold water, then with hot, and then digested with sodium hypo-sulphite. Assay the residue. The difference is silver chloride.)

We have next to estimate other insoluble oxides and also sulphides. After leaching out salts which are soluble in water, ammonium acetate, ammonia and ammonium sulphate, there remains oxides of iron, manganese, and sulphides of zinc, lead, and manganese. The first operation is the estimation of the total of these metals.

From the amount of zinc and lead found, the sulphides of zinc and lead are calculated according the following:

64.9 Zn = 96.9 ZnS
and 206.4 Pb = 238.4 PbS.

Next estimate manganese sulphide as follows:

After separation of soluble salts the ore is digested with dilute nitric acid. Iron must be separated.

“To separate the two metals, iron and manganese, they must be in the form of chloride (ferric chloride and manganese chloride). The solution is diluted considerably, then neutralized with sodium carbonate. To the iron solution add sodium acetate in excess, so as to change all the iron and manganese to neutral acetates. Add a few drops of acetic acid, then boil for a short time. The iron is precipitated as basic acetate, whilst the manganese is in solution. To ensure complete separation, the basic acetate of iron must be re-dissolved and the operation repeated.”

The manganese is estimated in the ordinary way.
From the manganese found, manganese sulphide is calculated according to

54.8 Mn = 86.8 MnS.

Deduct the manganese found as MnS from the total. The remainder is manganese oxide. The total sulphur combined to form sulphides must now be estimated.

The amount of sulphur- can conveniently be estimated by oxidation to sulphur tri-oxide, in the wet way, and then titration with barium chloride. Deduct the total sulphur combined, with zinc, lead and manganese as sulphides, from the total sulphur just found. The difference is sulphur combined with iron and copper.

Estimate copper, and since 63.4 Cu = 94.4 CuS, it is easy to calculate the amount of sulphur combined with copper. Deduct this amount from the total sulphur combined with iron and copper. The remainder is that combined with iron, from which iron sulphide is calculated according to

56 Fe = 120 FeS2 and
392 Fe = 616 Fe6S7

The total iron is now estimated and deducting the iron which occurs as sulphide from the total, the remainder is iron oxide. Here again, as in the determination of soluble salt of iron, that metal will have to be determined in both its forms. Knowing this, we are able to calculate ferrous and ferric oxide.