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narkosis
12/07/2006, 08:54 PM
Doc,

I thought I had the 2-part Ca/Alk formula sussed but something just stumped me.

How is it that 2.25 cups of NaHCO3 can weigh more (600g) than 2.5 cups of CaCl2.2H2O (500g) even though the molecular weight of NaHCO3 (84) is much lower compared to that of CaCl2 (147)? Or is there a typo there?

Please, I'm not trying to be a PITA here, reason for my asking is I have 7 drinking cups with me which ALL give me different volume readings with a variance of up to 30%. So how do I know which cup size is right?

I have a scale that I can use to measure out the 500g and 600g of CaCl2 and NaHCO3 as in the formula, but there's none given for MgSO4.7H2O and MgCl2.6H2O .... Could you let me know what the measurements should be, preferably in grams cos with volume, sometimes there's compacting issues.

Actually I tried to be smart and tried to determine the grams of the 2 Mg salts to use but failed miserably cos of the inconsistency of the 2 compounds mentioned above :confused:

narkosis
12/07/2006, 11:38 PM
up

Boomer
12/08/2006, 12:08 AM
I dont' have time to go to the article now and go through it all but here is some help.

Go back and read it. IIRC that is 600 gr of baking soda, which you then cook to make sodium carbonate.

How is it that 2.25 cups of NaHCO3 can weigh more (600g) than 2.5 cups of CaCl2.2H2O (500g)

When you go by cup it is bulk density. The particles are not the same size. What weighs more a cup of sand or a cup of pebbles ? CaCl2 is a larger grain size and BS is a smaller grain size, therefore less air space in the BS, thus higher bulk density and weighs more / cup.

CaCl2 (147)

No, 40 + 35.5 +35.2 = 111 and not 147 but when you add in the water from the CaCl2.2H2O you need to add the 2 water and 2 H20 =36 and 36 + 111 = 147. So 147 is correct for CaCl2.2H2O.


none given for MgSO4.7H2O and MgCl2.6H2O .

MgSO4.7H2O = 246 and MgCl2.6H2O = 202

Mg = 24.3, S =32 , O = 16, H=1, Cl =35.5



I have a scale that I can use to measure out the 500g and 600g of CaCl2 and NaHCO3 as in the formula

That is the way to go :D

narkosis
12/08/2006, 12:21 AM
heya boomer

Great!! Thanks a lot, I didn't think about airspaces in the dry form. That makes perfect sense then.

In that case, would you know how much of 3 cups of MgSO4.7H2O and 5 cups of MgCl2.6H2O would weigh? I'd feel a lot safer using a scale than the highly regarded cup :mixed:

Also one more thing. In Recipe #1, it says to add 610ml (2.5cups) of Part 3A stock solution after adding a gallon of both parts 1 and 2. Does this mean
1G Part1 + 1G Part2 + 610ml Part 3A or
0.5G Part1 + 0.5G Part2 + 610ml Part #3A?

Sorry if my understanding of the Eng.language syntax is not good enuff :lol:

<a href=showthread.php?s=&postid=8705062#post8705062 target=_blank>Originally posted</a> by Boomer
...therefore less air space in the BS, thus higher bulk density and weighs more / cup.
I have a scale that I can use to measure out the 500g and 600g of CaCl2 and NaHCO3 as in the formula

That is the way to go :D

Randy Holmes-Farley
12/08/2006, 07:33 AM
The magnesium recipes that I made are based on the known bulk density of the products involved, so cups will work fine, especially since there are certain assumptions in determining exactly how much magnesium to dose that are not necessarily accurate for every aquarium (as I detail here: http://reefkeeping.com/issues/2006-02/rhf/index.php#21 )

The correct interpretation is

1G Part1 + 1G Part2 + 610ml Part 3A or

1 gallon of each one plus the 610 mL. :)

narkosis
12/08/2006, 08:44 PM
Randy,

Many thanks for the clarification on the dosages - sure wouldn't to double the Mg part even though its comparatively small.

As for the cup - bulk density thing, I'm still not very comfortable with this, largely due to clumping up issues. Can I assume that it would be more accurate to "de-clump" the compounds first?

And if I were to be a stickler for grams, could I use a simple math formula for MgCl2 vs CaCl2 like

203 / 147 = 1.38 (mw factor?)
5 / 2.5 = 2 (cup factor)
Then do 1.38 * 2 * 500gCaCl2 = 1380g MgCl2

Would this make any sense at all or am i shooting in the dark?

Very sorry to trouble you on something as trivial as this, but thanks anyway on any suggestions you could make.

Cheers

Randy Holmes-Farley
12/10/2006, 08:39 AM
I don't still have the exact number values handy, but I'll show you how to determine them

from the article:

"The magnesium parts of the recipe are designed to add enough magnesium so that it is not depleted by either of the two means described above. Because the magnesium supplement (either version) is 47,000 mg/L in magnesium, we need to add (9 +19.5) grams/47 g/L = 610 ml of the magnesium solution for each gallon of the other parts of Recipe #1."

So what you need to do is add the equivalent of 9 + 19.5 = 28.5 grams of magnesium after each gallon.

Magnesium chloride hexahydrate is 12% magnesium by weight, so you need to add 238 grams of the solid magnesium chloride hexahydrate after each gallon of both of the other two parts, if you used it alone without Epsom salts.

A 64 ounce container of Epsom salts (magnesium sulfate heptahydrate) measured out at about 5 cups. So that's how I came up with the 5 cups for it.

narkosis
12/11/2006, 06:21 AM
Randy,

Thanks so much for the detailed workings. I'm so much more comfortable with 9g and 19.5g :)

So 9g of Mg from Epsom which is ~10% = 90g
19.5g of Mg from MgCl2 which is ~12% = 162g

Hope I got this right this time :)

Thanks a whole bunch

Randy Holmes-Farley
12/11/2006, 11:28 AM
No, that isn't what I meant. The 9 and 19.5 are parts of the calculation in the article on how much total magnesium is needed, not what the source of it should be.

I don’t have the actual numbers any longer, and don't have the time to go back and recalculate them. If you want to, take 5/8 of that (9 + 19.5) = 17.8 grams of magnesium coming from MAG flake and 3/8 x (9 + 19.5) = 10.7 grams of magnesium from Epsom salts and you'll be set.

FWIW, I think you are making this far more complicated than is useful. The magnesium part is an estimate, and making a precise solution to fit an estimate is not useful. :)

narkosis
12/11/2006, 07:43 PM
ahhh ok Randy, thanks for the patience and explanation. Got it this time. Alright, will try to relax, and take a few deep breaths :)

Should stay to my computers hahahah

cheers

Randy Holmes-Farley
12/12/2006, 05:44 AM
:thumbsup:

Good luck. :)