Similar question has been earlier asked on same platform...In order to calculate the organic acrbon into kg/ha , you need two important considerations viz., soil depth and bulk density. For example...The amount of organic carbon to 10cm depth in soil with a carbon value of 1.5% and bulk density of 1.3g/cm3 is:
15 (g C/kg soil) x 1 300 000 (kg soil/ha) = 19.5t C/ha
or 1.5 x 1.3 x 10 = 19.5t C/ha
Please find enclosed an excellent PDF for further reading...
Soil organic carbon (kg C/ha) to be calculated according to following formula from the available soil data:
SOC(kg/ha) = SOC (%)× BD (g/cm3)× SD (cm) x 1000
where, SOC - Concentration of soil organic carbon (%); BD - Bulk density (g/cm3); SD- soil sampling depth (cm)
What Medha Bulusu says is true. Another thing to remember is that SOC calculated from soil sampling normally only includes the C within the sampled profile (SD), and there is no standard SD. Anything from 20 - 30 cm is commonly used. This affects results.
Another thing to remember is that plant input of C that contributes to SOC does not only take place within the sampling depth. There are methods to estimate the total C input, but they are not perfect.
In section 2.7 of this publication: doi:10.1016/j.agee.2011.02.029 you can find one approach to estimate an approximate value of the input of a plant throughout the entire soil profile, based on measured C stock changes within the sampling depth (SD)
a) 1% SOC = 1 dag / kg = 10 g of SOC
SOC mg / kg => 1 dag / kg * 1000 = 10000 mg
b) For example, 1% SOC:
1% of SOC = 1 dag / kg. 1 dag / kg = 10 g / kg.
dag = 10 g. '. 10 g x 1000 = 10,000 mg / kg
mg / kg * 2 = kg / ha => 10000 mg / kg * 2 = 20000 kg / ha
c) Example 2:
2.5 dag / kg SOC or 2.5% SOC x 10,000 = 25,000 mg / kg
25,000 mg / kg x 2 = 50,000 kg / ha = 50 t / ha
One can also express 2,5 dag / kg x 20 = 50 t / ha
Good day, moving from an expression of mass (e.g. % or mg/kg) to kg/ha or tons/ha will be affected by the depth of sampling and the bulk density of the soil.
Soil Scientists usually use a conversion factor of ~20,000 to move from % to kg/ha. For example, if a sample contains 1% organic carbon (OC), then this would be ~20,000 kg/ha. It is important to note, however, that this is based on the assumption that you are sampling at a depth of 15cm from a soil that has a bulk density of around 1.35 g/cm3. For a more accurate conversion, please refer to the example below:
Let's take a sampling depth of 20cm, a soil bulk density of 1.5 g/cm3, and a SOC percentage of 1. You can convert the OC content from a % to kg/ha as follows
1) You will need to calculate the mass of soil that is present at this depth over 1 ha (10,000m2) of land. To do this, you will need to use the formula Mass of soil (M) = Soil bulk density (BD) x Volume of soil (V).
2) To calculate the volume (in units of m3), multiple the depth (in units of m) by area of land (i.e. 1ha or 10,000m2). Therefore V would be 2000m3.
3) The Mass of soil (in units of kg) in a hectare of land at a depth of 20cm can now be calculated by multiplying V by BD (unit units of Kg/m3): 2000 m x 1500 Kg/m3 = 3000000 Kg of soil.
4) OC can now be converted from % to kg/ha:
(1 kg OC/100 kg soil) x 3000000 kg soil/ha = 30,000 kg/ha
SOC = (d * BD * (C tot – C min) * CF st)
where: SOCS is soil organic carbon [kg/m2],
C tot and C min are total and mineral carbon [g g-1],
d is depth of horizon/depth class [m],
BD is bulk density [kg/m3],
CF st is correction factor for coarse fragments content (1-(%gravel + %stones)/100).
Then you have to conert from the value of m2 to hectar (10000 m2)
The soil carbon mass per area (ton/ha) is computed by the formula;
Soil C ton ha-1= BD (g cm-3) x C% x d
where: d = soil depth (cm), BD = bulk density in g cm-3, and C % = percentage carbon content of the sample.
Can be written in different ways but for the units indicated A.K. Singh is correct and follow his way.
You can convert by using this % × 10000 = ppm × 2.24 = kg/ha
Melville Tricia
Take the following example:
Lets say your sample has 1% organic carbon in it. This is the same as saying you have 1 g of organic carbon per 100 g of soil (i.e. 1 g OC/100 g soil).
Now that your carbon concentration is unit less, you can go head to covert it to g/kg . The idea is to manipulate the units but ensure that you are always multiplying by one. See below.
1g OC/ 100 g soil x 1000 g soil/ 1 kg soil = 10 g OC /kg soil
Note: 1000g soil/ 1 kg soil can be simplified as 1 kg/ 1kg, so even though you are manipulating the units, you nevertheless multiply by 1.
In summary, to move from a % to g/kg, you multiply by 10.
De Shorn Edmund Bramble thank you. Is there also an option to do the reverse? Converting g/kg to a percentage?
Melville Tricia ,
10 g/ kg x 1 kg / 1000 g = 0.01. Then you multiply by 100 to covert it to a %
All together:
10 g/ kg x 1 kg / 1000 g *100 = 1%
E.g.,
171.28 mg/kg SOM
=(171.28/10000)*100
= 1.7128 %
171.28 g/kg SOM
=(171.28/1000)*100
= 17.128%
the mass in g/kg is the same as the mass in ‰. So just divide it by 10 to get %.
One thumb rule that has to be kept in mind is that soil of one hectare land of upto 15 cm depth weigh about 2.24×1000000 kg...like wise one can calculate weight of furrow slice of different areas of the agricultural soils... For example: If soil organic carbon is 0.75 per cent it means in 100 g of soil, 0.75 g organic carbon is present means in 1000 g or 1 kg, 7.5 g is present...it means 1 kg soil contain 7.5 g OC
Then 2.24×1000000 kg soil will contain =
7.5×2.24×1000000= 16800 kg OC/ha or 16.8 tonnes organic carbon per hectare...
Dear Satendra,
You can convert SOC (%) to kg/ha by using the factor as:
% × 10000 = ppm × 2.24 = kg/ha
Now for kg/ha to mg/ha, another factor may be used as:
mg/ha = kg/ha × 106
Final conversion will be
SOC%= % × 22400000000 mg/ha
Anoop Kumar Srivastava you are right sir.
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