A fertilizer sample, as received, contains 24.0% P2O5 and 5.00% H2O. The sample is oven dried and %H2O becomes 0.50%. Calculate the %P2O5 in the oven dried sample.
1) consider the % as a grams, then your original sample will be 24.0g of P2O5, 5.00 g H2O and the rest 71.0 g of something else that we don't know what it is and we assume that is not volatile and will evaporate at the temperature used to drie the H2o from sample. 2) after evaporation your water is going to be 0.05 g in place of the original 5.00g. Your total sample is not going to be 100g but 100g -( 5.00g - 0.05g) = 100-4.95= 95.05g 3) Since the mass of P2O5 has not changed, now you can calculate the % of P2O5 in the dried sample considering the original mass of the P2O5 (24.0g) and the new mass of the dried sample (95.05g) 100 x 24.0g /95.05 = 25.25%
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