<p>In order to convert an area to a mass of soil you need the <strong>depth</strong> and <strong>bulk density</strong> of the section of the soil profile you are interested in.</p>

<p>The term "furrow-slice" is a bit archaic, but common in agriculture, and refers to some area sliced to plow-depth (approx 15 cm). For example, if someone said they applied 50 kg-P/ha, they meant 50 kg-P per ha furrow slice.</p>

<p><strong>A common short-cut:</strong></p>

<p>$1\text{ ha furrow slice}\approx2.2\times10^6$ kg, or $1\text{ acre furrow slice}\approx2\times10^6 \text{ lbs}$</p>

<p>From there its just unit conversion, something similar to the follow should suffice:
$$
\frac{50\text{ kg-N}}{2.2\times10^6\text{ kg-soil}}=\frac{\text{X g-N}}{20\text{ g-soil}}\longrightarrow \frac{\text{X g-N}}{\text{%N of fertilizer}}\longrightarrow\text{X g-fertilizer}
$$</p>

CC BY-SA 3.0
2017-06-13T09:32:07.960Z
2017-06-13T09:32:07.960Z
geosol
<p>You probably mean 50kgN/ha, which sounds like a plausible number for fertilizer application.</p>

<p>I don't know what you are doing exactly, but you need to know how much surface area your soil sample represents. If you do planting tests, it will be simply the surface of your bed.</p>

<p>Then it's just mulitplying the surface area by 50kgN/ha (mind the units!). Now you know how much N you need.</p>

<p>If you buy fertilizer, the fertilizer value in kg/t, kg/m³ or some similar measure will be given for the important macronutrients NPK. Else you need to know how much N is in Ammonia per weight but I think that is the easy part.</p>

<p>Disclaimer: I've never done any soil science, but I worked with farmers for a while and was in a field where fertilizer etc. matters.</p>

CC BY-SA 3.0
2016-04-13T09:09:10.957Z
2016-04-13T09:09:10.957Z
mart
<p>I want to add nitrogen to my soil samples in a lab as fertilizer (ammonium nitrate). I need to add it equivalent of 50 kgN/ha. I have a clump of soil, that weighs 20g. What is the best way to convert it to concentration, molarity or any other measure of how to prepare my fertilizer solution without knowing density, depth or any similar physical properties of the soil.</p>

<p>Basically, I would like to know if there is a way to convert nitrogen load from kg/ha to 20 gr. Thanks everyone in advance!</p>

CC BY-SA 3.0
2016-04-12T09:12:33.243Z
Interesting question. Could you describe for which purpose do you want to add the fertilizer to the soil? What is the target of your experiment (plant growth, bacterial conversion of the ammonium nitrate, transport of fertilizer through the soil)? Did you dig out the soil sample?
daniel.heydebreck
CC BY-SA 3.0
2016-04-12T13:08:45.277Z
Without knowing much, I think that you will have to assume density and depth. Otherwise it's not possible. I might be wrong.
Gimelist
CC BY-SA 3.0
2016-04-30T22:07:45.493Z
I vote to close the question because the questioner did show no motivation to provide more information on the problem. The currently presented information is not sufficient to give a "correct" answer.
daniel.heydebreck
<soil><agriculture><nitrogen><laboratory>
How to find molarity of a fertilizer?
iirisa
# How to find molarity of a fertilizer?

(Asked by: username_0 on 2016-04-11)

I want to add nitrogen to my soil samples in a lab as fertilizer (ammonium nitrate). I need to add it equivalent of 50 kgN/ha. I have a clump of soil, that weighs 20g. What is the best way to convert it to concentration, molarity or any other measure of how to prepare my fertilizer solution without knowing density, depth or any similar physical properties of the soil.

Basically, I would like to know if there is a way to convert nitrogen load from kg/ha to 20 gr. Thanks everyone in advance!

**Tags:** soil, agriculture, nitrogen, laboratory

> **Comments:**
> **username_1**: Interesting question. Could you describe for which purpose do you want to add the fertilizer to the soil? What is the target of your experiment (plant growth, bacterial conversion of the ammonium nitrate, transport of fertilizer through the soil)? Did you dig out the soil sample?  
> **username_2**: Without knowing much, I think that you will have to assume density and depth. Otherwise it's not possible. I might be wrong.  
> **username_1**: I vote to close the question because the questioner did show no motivation to provide more information on the problem. The currently presented information is not sufficient to give a "correct" answer.  

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## 2 answers

### Answer by username_4 on 2016-04-13. Score: 3 | Accepted answer

You probably mean 50kgN/ha, which sounds like a plausible number for fertilizer application.

I don't know what you are doing exactly, but you need to know how much surface area your soil sample represents. If you do planting tests, it will be simply the surface of your bed.

Then it's just mulitplying the surface area by 50kgN/ha (mind the units!). Now you know how much N you need.

If you buy fertilizer, the fertilizer value in kg/t, kg/m³ or some similar measure will be given for the important macronutrients NPK. Else you need to know how much N is in Ammonia per weight but I think that is the easy part.

Disclaimer: I've never done any soil science, but I worked with farmers for a while and was in a field where fertilizer etc. matters.


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### Answer by username_3 on 2017-06-13. Score: 0

In order to convert an area to a mass of soil you need the **depth** and **bulk density** of the section of the soil profile you are interested in.

The term "furrow-slice" is a bit archaic, but common in agriculture, and refers to some area sliced to plow-depth (approx 15 cm). For example, if someone said they applied 50 kg-P/ha, they meant 50 kg-P per ha furrow slice.

**A common short-cut:**

$1\text{ ha furrow slice}\approx2.2\times10^6$ kg, or $1\text{ acre furrow slice}\approx2\times10^6 \text{ lbs}$

From there its just unit conversion, something similar to the follow should suffice:
$$
\frac{50\text{ kg-N}}{2.2\times10^6\text{ kg-soil}}=\frac{\text{X g-N}}{20\text{ g-soil}}\longrightarrow \frac{\text{X g-N}}{\text{%N of fertilizer}}\longrightarrow\text{X g-fertilizer}
$$


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