For my thesis, I am required to convert the amount of copper found in the soil at a particular site from mg/kg to mg/L to place in growth media solution to test my experiment. I have been trying to convert it, however I don't think I am doing it right. I have three concentrations to convert:

1. 65 mg/kg to mg/L
2. 1440 mg/kg to mg/L
3. 5600 mg/kg to mg/L

I have tried multiplying the concentration by the density of copper and tried multiplying the concentration by the density of the soil. However in both cases, I would be putting approximately 300 mL of copper solution into 50 mL of growth solution, which I know is definitely not right. I would ask my supervisor, however she is currently on vacation and I really need to get this part done so I can move on!

Any help is appreciated!

Justine

I don't have much time to write at the moment, but I can come back to post later if this doesn't make sense.

Basically, you'll want to figure out how much volume 1 kg of soil occupies. Then, for the 1st case, you'll want to put 65 mg of copper into the volume that you calculated (and then take a ratio to figure out how much copper to put in 1 liter). It sounds like you have a value for density of soil, so if you post that here, I can look further into it.

Secondly, I assume your solution is copper sulfate (or a similar ionic solution)? And is the copper you're measuring in soil elemental or ionic? If you are measuring elemental copper, you will most likely want to use an equal molarity of copper sulfate in your growth media (not knowing the details of your project, I can't say for sure. And approximating elemental copper by adding ionic copper isn't really fair either, but I'm jumping to conclusions here because I don't have the details). What concentration is your copper solution at? In the past, I've made 1M CuSO4 solution, and added it to growth media in the range of 10-100 µM (micromolar, in case the font doesn't work). But this was for experiments with baker's yeast, so it could be completely different for what you're doing.

If you provide more details, I may be able to help further.

Thanks for your reply Mendel, I think I understand!

The density of the soil was 1.32 g/cm3 however, I am using a soils data report (I attached a picture of the table, just in case), and I am assuming that they measured the concentration of elemental copper in the soil, but I am not 100% sure. My supervisor isn't overly concerned with accuracy in terms of the concentration in growth media vs. concentration in the soil, she just wants me to have a realistic approximate concentration (for instance, I was going to put 60, 120, and 200 microliters of 0.1M CuSO4 in 50 mL of growth media as per methods found in another experiment, however she wanted a concentration close to that where the plant I am studying grows, since I am looking at the effectiveness of this plant to take up and store heavy metals since heavy metal toxicity is high in the soil in this particular area.)

So, if I understand correctly I take 65 mg/kg multiply it by 1.32 g/cm3 which would give me 85.8 mg/L. If I take that 0.0858 g/L and divide it by the molar mass (249.7 g/mol), I would get 0.000344 mol/L as my molar concentration. So, if I make a solution of 0.1M of CuSO4 (which I calculated as 12.485 g CuSO4 into 500 mL of dH2O), I multiply 0.000344 mol/L by 0.05L which gives me 0.0000172 mol of CuSO4 in 50 mL of growth media, therefore I would need to put 0.000172 L or 1.72 mL of 0.1M CuSO4 into 50 mL of growth media to get an equivalent concentration to that found in the soil?

Thanks again Mendel!

Thanks for the additional info.

Your approach is correct, however you need to be more careful with the units in the first multiplication that you've done. In this case you got lucky, because the conversion factors canceled out (or maybe you already knew they would, so didn't show the work here). But just to be fully clear, the density of soil =1.32 g/ cm^3 should be converted to mg/L before multiplying by the amount of copper per kg of soil. I.e.:

(65 mg copper/ kg soil) * (1.32g soil/ cm^3) * (1kg/ 1000g) * (1000cm^3/L)

So the two conversion factors (g to kg, and cm^3 to L) cancel out.

At the end, you're off by a factor of 10.

.000172 L * 1000ml/ L= 0.172 ml = 172 µl (microliters)

This is not too far out of the range you were originally planning to add in the first place!

Actually, I just looked at the question again and your answer is not quite right--

If the 65 mg/ kg of copper is elemental copper, you'll need to use a different molar mass to calculate its molarity in soil.

Then you can add copper sulfate to the same molarity as what you calculated for elemental copper.

Let me know if this is unclear. Sorry I didn't catch that earlier.

I attached a copy of my calculation sheet so it's easier to understand. So rather than dividing the 85.8 g/L by the molecular weight of CuSO4, I divided it by the molecular weight of elemental copper. Now, when adding 0.1M of CuSO4 to 50 mL of growth media, I determine how many moles needed in 50 mL, the divide it by the molar concentration of the solution?

Thanks again for all your help!
Justine

Yes, that looks correct now! If I think of something else I'll post again (it's late here, and I'm almost asleep!), but I don't anticipate any problems with the calculation you've done.

Have you made the CuSO4 solution yet? I think it's a very lovely shade of blue! (Although I think your solution is 1/10 the concentration of what I used to make, so your blue color might be more dilute than mine was. I'd have to look up my notes to see whether my solution was 1M or 0.1M, I'm not entirely sure on that).