Question

The label of a fertilizer states that the fertilizer contains
18.0% P_{4}O_{10} by mass. What is the mass
percentage of phosphorous in the fertilizer?

Answer #1

100g of fertilizers contains 18.0g of P4O10. Thus, moles of P4O10 can be calculated as follows:

Moles of P4O10 = Wt. of P4O10 / molar mass of P4O10

Moles of P4O10 = (18.0 g) / (283.89g/mol)

Moles of P4O10 = 0.0634

One mole of P4O10 contains four moles of phosphorous (P). Thus, number of moles of P can be calculated as follows:

Number of moles of P = 4 x moles of P4O10

Number of moles of P = 4 X 0.0634

Number of moles of P = 0.2536

Thus, mass of P can be calculated as follows:

Mass of P = moles of P x molar mass of P

Mass of P = 0.2536 mol x (31.0 g/mol)

Mass of P = 7.9 g

Since, mass of P in 100g of fertilizer is 7.9g. Therefore, mass
percentage of phosphorous in the fertilizer is **7.9%**

The label on a bottle of "organic" liquid fertilizer concentrate
states that it contains 7.40 grams of phosphate per 100.0 mL and
that 16.0 fluid ounces should be diluted with water to make 32.0
gallons of fertilizer to be applied to growing plants. What version
of the dilution equation can be used to calculate the final
phosphate concentration? Express your answer in terms of Vi, Ci,
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Mass of small beaker
112.64 g
Mass of Fertilizer and Beaker
123.03 g
Mass of fertilizer sample
10.39 g
Mass of filter paper
0.44
g
Mass of filter paper and dry MgNH4PO4*6H2O
precipitate
0.96 g
Mass of dry MgNH4PO4*6H2O precipitate
0.52 g
Moles of MgNH4PO4*6H2O
0.0021
mol
Moles of Phosphorous
0.0021
mol
Mass of Phosphorous
0.0065 g
% P in fertilizer sample
.62
%
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