Help me solve this problem with details please. Familiarize yourself with calcul
ID: 1061731 • Letter: H
Question
Help me solve this problem with details please.
Familiarize yourself with calculations that will be performed in the lab report: e.g. A sample of a commercial fertilizer was analyzed for phosphorous. A 1.7532 g sample was dissolved in water and diluted to 1-L in a volumetric flask to give solution A. A 5-mL aliquot of solution A was diluted to 100-mL in a volumetric flask to give solution B. Solution B was analyzed by molecular absorption spectroscopy and determined to have a concentration of 8.015 mu g P/mL. a. Calculate the % P_2O_5 in the fertilizer. Express the results to 3 significant figures. (aw P = 31; fw P_2O_5 = 142) b. If the standard deviation in the concentration of the diluted solution B was 0.08314 mu g P/mL. Calculate the standard error in the % P_2O_5 in the commercial fertilizer.Explanation / Answer
4a) Start with the absorption data and work backwards to arrive at the answer.
Solution B has 8.015 µg P/ mL. Volume of solution B = 100 mL. Therefore, mass of P in solution B = (8.015 µg/1 mL)*(100 mL) = 801.5 µg P.
Now, solution B was prepared by taking 5 mL aliquot from solution A and diluting to 100 mL. Since solution A was the sole supplier of P, hence, we can assume that the mass of P in A was much higher. Infact, the mass of P in solution A can be obtained as
(801.5 µg P)*(1000 mL/5 mL) = 160300 µg = (160300 µg)*(1 g/106 g) [1 g = 106 µg] = 0.1603 g (remember we prepared 1000 mL solution A and took 5 mL from the solution for our test).
Molar mass of P = 31; moles of P = (0.1603 g)*(1 mole/31 g) = 5.171*10-3 mol.
The fertilizer had P2O5 which decomposes to give P as
2 P2O5 ------> 4 P + 5 O2
Moles P2O5: moles P = 2:4
Therefore, 5.171*10-3 mol P = (5.171*10-3 mol P)*(2 mole P2O5/ 4 mole P) = 2.5855*10-3 mol P2O5.
Molar mass of P2O5 = 142 g/mol.
Therefore, mass of P2O5 in the fertilizer = (2.5855*10-3 mol)*(142 g/1 mol) = 0.3671 g.
We took 1.7532 g of the original sample.
Therefore, percentage of P2O5 in the fertilizer sample = (0.3671 g/1.7532 g)*100 = 20.9388 20.939% (ans).
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