I would do this.
%C in H2C2O4 is approx 27% but you should get a more accurate answer.
%C = (2*atomic mass C/molar mass)*1000.
1005g of 27% C gave a reading of 102540.
So the % C in the soil is 27% times two correction factors. One factor converts for the difference in IR reading, the other for the difference in sample size.
27% x (14515/102540) x (0.1005/0.1115)= ?
You must remember that this gives somewhat limited results; i.e., the soil sample carbon but decompose the same way oxalic acid does.
The amount of organic carbon in a sample can be obtained by IR analysis of the carbon dioxide (CO2) released following complete combustion of the sample in oxygen. A reference sample of an organic compound is usually used to calibrate the instrument response.
In an analysis of the carbon content of a soil sample using this method, calibration of the instrument, after zeroing with a blank sample, was carried out by combusting a 0.1005 g sample of oxalic acid (COOH)2 which gave an instrument reading of 102540. Subsequent analysis of a 0.1115 g soil sample gave an instrument reading of 14515. Calculate the percentage of organic carbon in the soil sample.
2 answers
Can you go through this question in more detail?