36 replies to this topic

[SunsetSatine]

I'm using the following recipe, does this math/reasoning sound right?

For my enrichment, I currently have 40 mL in a serum bottle with all the components, except for my carbohydrate, NaHCO3, and CyS.

Per every 100 mL, the recipe says, I need 0.4g of NaHCO3 & 0.6g Carbohydrate

I take 0.4g NaHCO3 in 10 mL of DI Water (Sparge & Autoclave)
  0.6g Carbohydrate in 10 mL of DI Water (Sparge & Autoclave)

Since, I only have 40 mL in my serium bottle, I would need to add 0.16 mL of NaHCO3, and 0.24 mL of Carbohydrate in my serium bottle....correct?

[pito]

for 100ml you need 0,4 grams,

so for 40 ml you need: 0,16 grams (2.5 times less)


If you hav 10ml DI water with 0,4 grams and you only need 0,16 grams, then how much DI water do you need to take?
10ml = 0,4 grams
xml = 0,16 grams.

X= ?

if you cant figure it out, make easier for yourself.

10 ml = 0,4 grams ==> 0,1 grams is 10/4ml or 2,5 ml
so for 0,16 grams, you mulitply 2,5 with 1,6, wich makes: 4ml.

I dont understand why you are making it yourself so difficult,

why do you make that 10ml solution?
WHy not just add the amount you need to the serum boittle at once?

[SunsetSatine]

Hi, thanks!

I have to set up multiple enrichments with multiple carbohydate sources, so I prepared a large amount of the media except for those three components and divided it into multiple serum bottles.

[SunsetSatine]

So, I think I should create a stock solution of my carbohydrate in case I need to prepare this enrichment again.

The original recipe calls for 0.6g of carbohydate for every 100 ml of enrichment prepared. I can prepare a stock of 0.3g in 50 mL so that my stock solution is 0.6%. If I prepare only 40 mL of enrichment, then I will take 0.24 mL of my 0.6% carbohydrate stock solution correct?

In retrospect, I'm thinking now that I should have not made my enrichment the even 40, but accounted for adding the carbohydrate, reducing agent, and Na2CO3......

[pito]

SunsetSatine, on 06 December 2012 - 08:13 AM, said:

So, I think I should create a stock solution of my carbohydrate in case I need to prepare this enrichment again.

The original recipe calls for 0.6g of carbohydate for every 100 ml of enrichment prepared. I can prepare a stock of 0.3g in 50 mL so that my stock solution is 0.6%. If I prepare only 40 mL of enrichment, then I will take 0.24 mL of my 0.6% carbohydrate stock solution correct?

In retrospect, I'm thinking now that I should have not made my enrichment the even 40, but accounted for adding the carbohydrate, reducing agent, and Na2CO3......

I do not understand what you mean.

If you need 0,6grams for every 100ml and you prepare 0,3grams in 50ml... then you have 0,6grams per 100ml ... (or 0,6%)

So if you take 0,24ml of this 0,6% and add this in 40ml you dont have 0,6% in the end..

[SunsetSatine]

Hi,

I'm decreasing the quantity because I want to use a small serium bottle for my stock solution.

Hmmm, I see your point. I can't use m1v1=m1v2 to solve for how much of the stock solution I need to use since I want to keep the concentration the same right? Maybe there is a more efficient way to do this? Now, I think I'm just confusing myself.

Maybe I should be preparing a greater concentration for my stock solution?

[pito]

SunsetSatine, on 06 December 2012 - 03:52 PM, said:

Hi,

I'm decreasing the quantity because I want to use a small serium bottle for my stock solution.

Hmmm, I see your point. I can't use m1v1=m1v2 to solve for how much of the stock solution I need to use since I want to keep the concentration the same right? Maybe there is a more efficient way to do this? Now, I think I'm just confusing myself.

Maybe I should be preparing a greater concentration for my stock solution?

you can only dilute (stock) solutions. So if you allready have (stock) solution of 0,6gram per liter then you cant "dilute" this anymore without changing the "dilution".

If C1 is allready C2 then there is nothing to change/dilute...

[SunsetSatine]

How does this sound?

For my NaHCO3, I prepare a 20% stock solution (10g in 50 mL DI Water)
M1V1=M2V2
(20%)(x)=(50mL)(0.4%)
X=1 mL

Reasoning, I have a 20% stock solution. I want to calculate how much of the 20% stock solution I have to add to my 50 mL of media, so that the final concentration of the NaHCO3 in the media is 0.4%.

Therefore, I add 1 mL of my 20% stock solution to 49 mL of media so that the final concentration of the NaHCO3 in the media is 0.4%.

[pito]

SunsetSatine, on 12 December 2012 - 07:36 AM, said:

How does this sound?

For my NaHCO3, I prepare a 20% stock solution (10g in 50 mL DI Water)
M1V1=M2V2
(20%)(x)=(50mL)(0.4%)
X=1 mL

Reasoning, I have a 20% stock solution. I want to calculate how much of the 20% stock solution I have to add to my 50 mL of media, so that the final concentration of the NaHCO3 in the media is 0.4%.

Therefore, I add 1 mL of my 20% stock solution to 49 mL of media so that the final concentration of the NaHCO3 in the media is 0.4%.

1ml of a 20% stock solution means that of that 1ml only 20% is your "substance X", meaning 0,2ml
so 0,2ml in 50 ml total is indeed 0,4%
(0,2ml in 50 means, 0,4 in 100ml, so yeah...)

[SunsetSatine]

That makes so much more sense. I will prepare my carbohydrate the same way now. I never thought to look at it that way before.

Thank you so much! That really helps!

[pito]

SunsetSatine, on 12 December 2012 - 10:06 AM, said:

That makes so much more sense. I will prepare my carbohydrate the same way now. I never thought to look at it that way before.

Thank you so much! That really helps!

Its the same as what you did... I just "checked" whether you were right...

Its important to understand the logic behind it and not just use formulas you learned by heart.

[SunsetSatine]

That makes sense. If I understand the logic behind it, then I can apply it to other situations in the future.

[SunsetSatine]

I'm having trouble getting my sodium bicarbonate to dissolved. I measured out 10g into 50 mL of DI water and it is just a cloudy solution. Since I can't heat the solution up, is there another way to dissolve it?

[hobglobin]

SunsetSatine, on 12 December 2012 - 12:58 PM, said:

I'm having trouble getting my sodium bicarbonate to dissolved. I measured out 10g into 50 mL of DI water and it is just a cloudy solution. Since I can't heat the solution up, is there another way to dissolve it?

9 g in 100 mL seems the maximum amount at room temperature.

[pito]

SunsetSatine, on 12 December 2012 - 12:58 PM, said:

I'm having trouble getting my sodium bicarbonate to dissolved. I measured out 10g into 50 mL of DI water and it is just a cloudy solution. Since I can't heat the solution up, is there another way to dissolve it?

You should have checked the solubility.
From wiki: Solubility in water 9 g/100 mL

So bad luck...

Heating could help a bit, but in this case... that wont do much (and since you cant even do it...)
Make a new batch or dilute it more.