I think the term is held over from titration analysis used to measure the actual reactive concentration of alum in solution.  I.e. a measured weight of the alum sample is added to water and dissolved, then a measured (drop by drop) amount of (e.g.) potassium hydroxide of known concentration is added until the solution reaches neutral pH, indicating all of the sulfate ion has been reacted.  The result can then be expressed as percent of the sulfate in the original sample, but because the formula weight of the alum varies depending on how much water the dry sample has absorbed, it is more convenient to use "equivalent weight of Al2O3" as a fixed/known formula weight for the percentage calculation.  Hope that's all as clear as mud.

Thanks, btrueblood.  Another question is....so, how should I calculate the amount of alum from, for example, 8.5% Al2O3?   

The manufacture of alum entails just one step additional to the aluminum sulfate production process. The uses of aluminum sulfate and alums are similar and the compounds are largely interchangeable. The term "alum" is loosely applied.

A true alum is a double sulfate of aluminum or chromium and a monovalent metal (or a radical, such as ammonium). Aluminum sulfate is very important industrially and although it is not a double sulfate, it is often called alum or papermakers' alum. Alum has been know since ancient times. The writings of the Egyptians mention its use as a mordant for madder and in certain medical preparations. The Romans employed it to fireproof their siege machines and prepare it from alunite which is plentiful in Italy.

True Alum: Ammonium aluminum sulfate, or ammonia alum, NH4Al(SO4)2·12H2O, is used in tanning, in dyeing and fireproofing textiles, in vegetable glues and porcelain cements, and in water purification. Chromium potassium sulfate, or chrome alum, KCr(SO4)2·12H2O, is used as a mordant in dyeing, in tanning, and in photographic fixing baths to harden gelatin films and plates.

Alum has been replaced to a large extent by aluminum sulfate, which has a greater aluminum equivalent per unit weight. Aluminum sulfate is Al2(SO4)3·18H2O. The most important single application of it is in clarifying water, more than half of its total amount being manufactured being so consumed. In water treatment aluminum sulfate is used to produce an aluminum hydroxide floc.

Aluminum sulfate, also called alum, became an industrial product in the 19th century. It was made by treating either bauxite or china clay with sulfuric acid. Unlike true alum, aluminum sulfate can not be conveniently purified through recrystallization because of its greater solubility in water. This is one of the reasons why it often contains varying proportions of silica, iron and free sulfuric acid.

Because of its greater concentration of alumina (Al2O3) and cheaper production procedures, aluminum sulfate saved costs in papermaking and therefore replaced aluminum potassium sulfate for most purposes in papermaking and particularly in rosin sizing towards the mid-19th century. This earned it the name "papermaker's alum."

Al2O3 is the active ingredient and that is what you are paying for when you buy aluminum sulfate. 8.5% Al2O3 is sold because that concentration has the highest freezing temperature which is practical consideration for shipping and handling.

Percent Al2O3 is a common method of quoting the strength of aluminium-based coagulants on a w/w basis. Another method of referring to concentration is to state the aluminium content, which is roughly half the Al2O3 content. For example, liquid alum of 7.5% w/w Al2O3, which is the same as 4.0% w/w aluminium (Al).

The term percent weight/weight or % w/w is similar to the previous term and is the number of kilograms of active chemical per 100 kilograms of liquid chemical. For example, liquid alum at 7.5% w/w Al2O3 has 7.5 kg of Al2O3 in solution for every 100 kg of liquid chemical delivered.

When calculating coagulant doses, it is important to first state on what basis the dose is to be expressed. The most commonly used unit in water treatment is mg/L, which is a weight per volume unit. This is also the same as parts per million (ppm), but only when quoted as ppm on a weight/weight basis.

If you state a chemical dose as ppm using the volume of chemical dosed to the volume of raw water, then what you have calculated is ppm v/v and this is not the same as
mg/L.

To calculate the dose of a coagulant or other chemical in mg/L, you will need to know its % w/w strength and specific gravity.

The formula is:
mg/L = 10,000 X % w/w X SG
For example, consider alum at 49 % w/w strength aluminium sulphate (as Al2(SO4)3.18H2O) and with a SG of 1.29. Using the formula above, the concentration of the alum in mg/L, will be: 10,000 X 49 X 1.29 = 632,000 mg/L or 632 g/L. If you are dosing alum at a rate of 150 mL/min as found from a drop test, and the raw water flow is 50 L/s, then the alum dose will be: (150 X 632,000/(1000 X 60))/50 = 31.6 mg/L or ppm (w/w) or: 1,000,000 X (150/(60 X 1000))/50 = 50 ppm (v/v).

In customary US units:
Chemical Feed Setting (mL/min) = [(Flow, MGD)(Alum Dose, mg/L)(3.785L/gal)(1,000,000 gal/MG)]/[(Liquid Alum, mg/mL)(24 hr/day)(60 min/hr)]

The only element which requires explanation is the liquid concentration.  The liquid concentration, given in mg/mL, is the amount of dry chemical (in mg) mixed with the amount of water (in mL):

So, if you added 230 mg of alum to 100 mL of water, you would have a liquid alum solution with a concentration of 2.3 mg/mL.

Consider a situation in which the flow of the plant is 5 MGD and the alum dosage is 10 mg/L.  The liquid product is shipped to the plant as 40.6% alum (6.9% Al2O3).The liquid alum has a concentration of 516 mg/mL.  The chemical feeder setting would be determined as follows:

Chemical Feed Setting (mL/min) = [(5 MGD)(10 mg/L)(3.785L/gal)(1,000,000 gal/MG)]/[(516 mg/mL)(24 hr/day)(60 min/hr)]

The setting on the liquid alum feeder should be 254 mL/min.

As an another example, the liquid product is shipped to the plant as 48.5% alum (8.3% Al2O3). The Alum solution that you use has approximately 642 mg dry Alum per mL of liquid.  

If the plant is treating 830 gpm water with 394 mL/min alum, the calculation is as follows:

394 mL/min X 1 L / 1,000 mL = 0.394 L per min

0.394 L/min X 1440 min/Day X 642 mg/mL Alum X 1 lb/453.6 grams = 803 lb/Day.

http://www.generalchemical.com/assets/pdf/Liquid_Alum_PDS.pdf
http://www.generalchemical.com/assets/pdf/Liquid_Alum_Storage_and_Handling_TSR_32_78.pdf
http://www.generalchemical.com/assets/pdf/Densities_and_Wts_of_Std_Alum_Solutions.pdf
http://www.wvdhhr.org/oehs/eed/swap/training&certification/drinking_water/documents/classiimanual2008.pdf

Thanks a lot, bimr, but what do you mean by "active ingredient"?  Do you know the mechanism of conversion (or, extraction) of Al2O3 from Al2(SO4)3.18H2O?

I thought that, as btrueblood said, the expression with Al2O3 was for the convenience.  Is it not true?

How did you calculate 6.9% Al2O3 from 40.6% alum and 8.3% Al2O3 from 48.5% alum?

Thank you.

Well, I thought I knew what I was talking about, but I'm confused too, ddad.  I thought the "active ingredient" in alum was Aluminum sulfate, not aluminum oxide.

But, in any case, the calculation/conversion between the two quoted figures would be to first find the formula weight (grams/mole) for each (note that you use the theoretical, but practically impossible dehydrated formula for alum):

Al203 = 2*27+3*16 = 102
Al2(SO4)3 = 2*27+12*16+3*32 = 342

The weight ratio 342/102 can then be used to convert weight percentages quoted in alumina to percentage of alum:

1% equivalent Al2O3 = 3.4% Aluminum Sulfate.

But - here is where it gets really tricky.  If you look at a data sheet for "Dry Alum", such as:

http://www.generalchemical.com/assets/pdf/Dry_Alum_PDS.pdf

The formula weight for so-called "dry alum" includes 14 moles of absorbed water in the hydrated salt...so the conversion to a percentage of "dry alum" would use a different calculation.  Again, calculate the formula weight for "dry alum":

Al2(SO4)3+14(H2O) = 2*27+12*16+3*32+28*1+14*16 = 594

Now use the ratio 594/102 to convert from weight percent alumina to weight % of "dry alum":

1% alumina = 5.94% dry alum.

This checks with the above manufacturer's chart, see the third link in bimr's post:

http://www.generalchemical.com/assets/pdf/Densities_and_Wts_of_Std_Alum_Solutions.pdf

Ain't chemistry fun?
 

Al2(SO4)3 18H2O dissociates in water to yield A+++ and SO4.  Aluminum sulfate --  Al2(SO4)3 18H2O — contains 14.5 to 17.5 per cent Al2O3. (The theoretical content of Al2O3 in pure Al2(SO4)3 18H2O is 15.3%.) The water content of the commercial product also is closer to 14 than 18 moles.

Active ingredient simply means what is available for reaction. You are also buying the water, sulfate ion, and other minor constituents when you buy alum. These other inactive ingredients end up in the treated water.

Liquid alum is a clear, amber colored liquid sometimes called 50% alum. This is because a gallon of liquid alum weighs 11.2 lb (0.5 g) and contains 5.4 lb (0.2 g) of dry aluminum sulfate. Actually, it usually contains 8.5% or more of available water-soluble alumina (Al2O3) as compared with the 17% Al2O3 available in dry alum.

Aluminum reaction with bicarbonate in water:

Al2(SO4)3 18H2O + 6NaHCO3 ( water)  >> 2Al(OH)3 (precipitant) + 3Na2SO4  + 6CO2  + 18H2O

If you calculate the amonunt of aluminum that is needed by using the molecular weight of aluminum sulfate according to the reaction, it would be incorrect. You need to know how much aluminum that the aluminum sulfate contains.


 

Ok, I think we agree then bimr - you are saying the "active ingredient" in the alum solution is the Al+++ ion, and the Al203 is just a convenient way of tracking that ion, correct?

Right, that is my understanding.