Phosphorus fertilizers are derived from phosphate rocks, called florapatite, Ca5(PO4)3F. 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 posdered 5 gram sample of a rock containing fluorapatite is reacted with sulfuric acid and 6.6*10^-3 moles of HF are released during the process, then what is the mass % of of fluorapatite in the rock sample?
2Ca5(PO4)3F+7H2SO4=3Ca(H2PO4)2+7CaSO4+2HF
posdered was supposed to be powdered
Thank you for the equation. See your post above and change the equation I proposed to the proper one. Post your work there if you run into trouble and I can help you through it. For this problem,
mols Ca5(PO4)3F = grams/molar mass = 5/504.3 = approx 0.006 but you need a more accurate answer for this and all of the other calculations I've made. I've rounded here and there also.
Then convert mols Ca5(PO4)3F to mols HF. Since 1 mol apatite produces 1 mol HF. mols fluoroapatite = mols HF.
Then grams HF = mols HF x molar mass HF = ?. This is the theoretical yield (TY). The actual yield (AY) from the problem is 0.0066 g.
Then % yield = (AY/TY)*100 = ?
To find the mass % of fluorapatite in the rock sample, we need to determine the moles of fluorapatite and calculate the mass % based on that.
First, let's calculate the moles of HF released during the reaction. We have the balanced equation:
2Ca5(PO4)3F + 7H2SO4 = 3Ca(H2PO4)2 + 7CaSO4 + 2HF
From the equation, we can see that 2 moles of fluorapatite produce 2 moles of HF. Therefore, the moles of HF released would be equal to the moles of fluorapatite.
Given that 6.6*10^-3 moles of HF are released, we can conclude that the moles of fluorapatite in the rock sample are also 6.6*10^-3 moles.
Now, let's calculate the molar mass of fluorapatite (Ca5(PO4)3F):
Ca: 1 * 5 = 5 (since there are 5 calcium atoms)
P: 1 * 3 = 3 (since there are 3 phosphorus atoms)
O: 16 * 12 = 192 (since there are 12 oxygen atoms)
F: 19 (since there is 1 fluorine atom)
Adding these up, the molar mass of fluorapatite is:
5 * 40.1 + 3 * 31 + 12 * 16 + 19 = 307 g/mol
Now, let's calculate the mass of fluorapatite in the rock sample:
Mass of fluorapatite = moles of fluorapatite * molar mass of fluorapatite
Mass of fluorapatite = 6.6*10^-3 moles * 307 g/mol
Finally, to find the mass % of fluorapatite, we divide the mass of fluorapatite by the original mass of the rock sample (5 grams) and multiply by 100:
Mass % of fluorapatite = (mass of fluorapatite / mass of rock sample) * 100
Substituting the values, we get:
Mass % of fluorapatite = (6.6*10^-3 moles * 307 g/mol / 5 g) * 100
Calculating this expression will give you the mass % of fluorapatite in the rock sample.