Experiment to Determine Soil Texture by Touch and Physical Analysis Essay Sample
- Pages: 3
- Word count: 792
- Rewriting Possibility: 99% (excellent)
- Category: soil
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Introduction of TOPIC
The most important measure of soil is texture and size distribution of mineral particles. This is because the divided soil particles have a greater surface are per unit mass than the combine particles (Coarse particles). Therefore it can be said that a large mass of gravel or sand, will be less relevant in relation to chemical reactions than a minor amount of fine silt and clay. This is because it allows for better exposure to nutrients and the retention of moisture.
Soil texture is the categorization of soil particles into three groups. Firstly, clay which is less than 0.002mm. Secondly, silt which is between 0.002 and 0.005, and third, sand which is between 0.005 and 2mm. With the use of Stoke’s law that is the governing rate of sedimentation of particles that are suspended in water. The amount of water in the soil sample will determine the percentage of moisture. With the moisture percentage, we can calculate the bulk density. In general, we can then determine the soil texture with both the feel method and the hydrometer method.
Both the method and material can be found in the Sample Lab Manual on pages 15-21.
The soil texture allowed us to discover the true elements in the soil between our hands. The gritty feel reinforced the existence of sand. While the stickiness of the sand referred to the amount of clay, and the sliminess related to the amount of silt in the soil.
Mass of the can and lid = 59.02 grams
Mass of the can, lid, and soil = 139.45 grams
Mass of soil = 139.45 – 59.02 =
80.43 grams The initial mass and temperature of the hydrometer =
13 – 10.5 = 2.5
Where 2.5 is the silt, and 10.5 is the clay.
This resulted in the sand being 80.43 – 13 = 67.43 grams.
Therefore we can say that the:
– Percentage of silt = (10.5/80.43)x100 = 13.07%
– Percentage of clay = (2.5/80.43)x100 = 3.10%
– Percentage of sand (loamy sand) = (67.43/80.43)x100 = 83.83%
Additional results include:
Air dry weight which equals: 137.55 grams
Oven dry weight which equals: 79.53 grams
Moisture which equals: (139.45 – 137.55) = 1.9 grams
Therefore: (1.9/78.53) = 0.002
And so (0.002 x 100) = 2
Where moisture = 2%
Using the group soil sample, we added water which enables us to squeeze the soil paste in-between our fingers to discover the soil texture. This resulted in a gritty feel which would translate to a large percentage of sand. Furthermore, the use of the hydrometer reinforced the percentage of sand, silt and clay, as well as the soil texture. The soil was concluded to be largely made of sand due to the sand being the greater percentage at 83.83%m followed by the silt at 13.07%, and finally the clay at 3.10%.
In addition, another weighed sample of soil was taken and left to dry in the oven for a period of two days. After the oven dry procedure ended, we could then analyse the amount of moisture that was eliminated being at 1.9 grams, where is can be rounded up to 2% moisture, and then used to calculate the bulk density. However, our results could have been victim of human error or random error due to the sample soil being dark brown which meant that it should have originally been a silt clay loam. This could have resulted from the lack of precision and accuracy while carrying out the experiment in relation to accurate measurements and procedures.
On the whole, the overall aim of the experiment was reached. We were able to determine the soil texture through the method of feeling, which allowed us to discover that the soil sample was mostly sand. The hydrometer allowed us to analyse and get more accurate results of how much clay, silt, and sand was actually in the sample. The moisture rate was then successfully calculated which enabled us to proceed further and calculate the bulk density of the soil.