5. Phosphorus fertilizers are derived from phosphate rocks, called fluorapatite, Ca5(PO4)3F(s) . Fluorapatite is insoluble in water, so it must first be dissolved using excess sulfuric acid to form water-soluble calcium dihydrogen phosphate Ca(H2PO4)2. If a powdered 5.00 gram sample of a rock containing fluorapatite is reacted with sulfuric acid and 6.60 x 10-3 moles of HF(g) are released during the process, what is the % mass of fluorapatite in the rock sample?

Question

5. Phosphorus fertilizers are derived from phosphate rocks, called fluorapatite, Ca5(PO4)3F(s) . Fluorapatite is insoluble in water, so it must first be dissolved using excess sulfuric acid to form water-soluble calcium dihydrogen phosphate Ca(H2PO4)2. If a powdered 5.00 gram sample of a rock containing fluorapatite is reacted with sulfuric acid and 6.60 x 10-3 moles of HF(g) are released during the process, what is the % mass of fluorapatite in the rock sample?

2 Ca5(PO4)3F(s) + 7 H2SO4(aq)  3 Ca(H2PO4)2 (aq) + 7 CaSO4(aq) + 2 HF (g)

Answer 1

mols HF produced = 0.00660

Convert mols HF to mols Ca5(PO4)3F by using the coefficients in the balanced equation.
0.00660 mols HF x (2 mol Ca3(PO4)3F/2 mols HF) = ?
Convert mols Ca5(PO4)3F to grams. g = mols x molar mass and this is the theoretical yield (TY). Assuming 100% efficiency in the process, then % fluorapitite = (mass fluorapitite/mass sample)*100= ?
Post your work if you get stuck.

Answer 2

To find the % mass of fluorapatite in the rock sample, we need to determine the moles of fluorapatite and the mass of the rock sample.

First, let's find the moles of HF released during the process. From the balanced chemical equation, we can see that 2 moles of Ca5(PO4)3F produce 2 moles of HF. Therefore, the moles of HF released are equal to the moles of fluorapatite in the rock.

Given that 6.60 x 10^(-3) moles of HF are released, we can conclude that there are also 6.60 x 10^(-3) moles of fluorapatite present in the rock sample.

Next, we need to find the molar mass of Ca5(PO4)3F. From the chemical formula, we can calculate it as follows:

Ca: 1 atom x 40.08 g/mol = 40.08 g/mol
P: 3 atoms x 30.97 g/mol = 92.91 g/mol
O: 12 atoms x 16.00 g/mol = 192.00 g/mol
F: 1 atom x 18.99 g/mol = 18.99 g/mol

Summing up the masses, we get:
40.08 g/mol + 92.91 g/mol + 192.00 g/mol + 18.99 g/mol = 343.98 g/mol

Now, to find the mass of the rock sample, we can use the molar mass of Ca5(PO4)3F. Given that the sample weighs 5.00 grams, we can divide this mass by the molar mass to find the moles of the rock sample:

5.00 g / 343.98 g/mol = 0.0145 moles

Finally, we can calculate the % mass of fluorapatite in the rock sample by dividing the moles of fluorapatite by the total moles of the rock sample (including all its components) and multiplying by 100:

(6.60 x 10^(-3) moles / 0.0145 moles) x 100 = 45.5%

Therefore, the % mass of fluorapatite in the rock sample is approximately 45.5%.