[Tom21]

Hello, I'm new here.  I am a little rusty on my chemistry, and would appreciate some help with calculating a soil concentration.
I have a quantitative field instrument that will give me a concentration of "X" in water (in mg/L).
"X" is extremely soluble in water and I will assume that when I extract from soil, all of the "X" will go into solution.

If I extract 3 grams of soil into 15 ml of water, analyze the water and get a concentration of, say, 50 mg/L, what calculation will give me the concentration of "X" in the soil (in mg/kg)?

Thanks for any *delete me*

[enahs]

Since you are assuming you are getting 100% extraction, you can make the assumption that the density of the solution/water is exactly 1g/mL  (a pretty good assumption). Meaning, 1 L of water = 1000mL = 1000g = 1kg.

So in short, no calculation, just change your units.

[Tom21]

I don't know why, when I said "Thanks for any h-e-l-p" (without the dashes)  it came through as "Thanks for any *delete me*.  Strange.

As for the calculation:  

50 mg/L is the concentration for 3 grams of soil in 15 ml of water.  Shouldn't a calculation be required to take into account the 3g/15 ml?

[Tom21]

That is,

50 mg/L x 0.015 L/0.003 Kg = 250 mg/Kg "X" in soil

That right?

[enahs]

50 mg/L = 50mg/kg


In words, you have 50mg per 1 kg of sample. That is what your instrument is telling you (assuming you are reading it correct).

So, if you had half a kg of sample you would have 25 mg.
To get 250mg you would need to have 5kg of sample.


50mg/L is not for "3 grams of soil in 15mL of water". It is the amount per unite mass. It is a concentration. Concentration does not depend on the amount you have.


Are you trying to figure out the actual amount in mg of whatever you are analyzing for is actually in 3 g?

[Borek]

I don't know why, when I said "Thanks for any h-e-l-p" (without the dashes)  it came through as "Thanks for any *delete me*.  Strange.

It is an old joke built into forum many years ago - many kids come here crying "please help me", that gets translated to "please delete me".

I am not saying its a good joke, I am just explaining what is going on.

[Tom21]

It appears we are talking apples and oranges.
Just because my solution says I have 50 mg/L, it doesn't mean I have 50 mg/kg in the soil.  That is because I extracted from a certain amount of soil into a certain amount of water.  It is like a dilution.  I need to calculate the actual mg of "X" in the soil.

[fledarmus]

That is,

50 mg/L x 0.015 L/0.003 Kg = 250 mg/Kg "X" in soil

That right?

This looks right to me - your concentration of "X" in your extract is 50 mg/L, your total volume of extract is 0.015L, so the total amount of "X" in your soil sample (assuming 100% extraction) is 50mg/L x 0.015 L. Dividing this by the size of your soil sample (0.003 kg) gives you the amount of "X" as a fraction of the sample size - 250mg "X" per kg of soil.

I think enahs is assuming you diluted your sample to 1 liter before testing it.

[Tom21]

Thank you very much!  I appreciate everyone's patience in helping, and the great support!