Apatite Homework Help - Find P in 1150 kg of Apatite

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the amount of pure phosphorus (P) in 1150 kg of apatite (Ca5(PO4)3F), focusing on the application of chemical formulas and molar mass in stoichiometric calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants discuss the need to find the formula weight of apatite to convert to moles.
  • One participant suggests using the molar mass of apatite, calculated as 504.31 g/mol.
  • There is a query about how many moles of phosphorus are present per mole of apatite, with some participants indicating it is three moles based on the empirical formula.
  • Confusion arises regarding the relationship between moles of apatite and moles of phosphorus, with differing interpretations of the stoichiometry involved.
  • Calculations are presented for the mass of phosphorus, with varying results and corrections being offered by participants.
  • One participant expresses uncertainty about the calculations and requests confirmation of their results.
  • Another participant emphasizes the importance of considering significant figures in the calculations.

Areas of Agreement / Disagreement

There is no consensus on the calculations or the interpretation of the stoichiometric relationships, with multiple competing views and ongoing confusion among participants.

Contextual Notes

Participants have not fully resolved the steps needed to accurately calculate the mass of phosphorus, and there are indications of missing assumptions regarding the application of stoichiometric principles.

chawki
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Homework Statement


Apatite (Ca5(PO4)3F) is used to manufacture fertilizer.

Homework Equations


How many kilograms of pure phosphorus (P) is included in 1150 kg of apatite?



The Attempt at a Solution


How we solve this, tell me only first step...
we can't even use the composition percentage law
 
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Start with the formula weight of apatite. Convert to moles.

Can you take it from there?
 


can you please tell me how to find the formula weight?
i have learned chemistry ages ago..and it was in french..

maybe you meant the molar mass ? if it's that, then it is 504.31g/mol
504.31 = 1150000/n
n= 1150000/504.31
n= 2280.34mol
 
Last edited:


Right, that's how many moles of apatite. How many moles of phosphorous per mole of apatite?
 


That's where I'm stuck
 


Just look at the empirical formula. You do this part by inspection... nothing to calculate.
 


1mole
 


So if you were given 1 mole of Ca5(PO4)3F, you would get only one mole of P out of it?
 


i'm not sure, I'm confused
 
  • #10


mp/30.97 = 1150000/504.31
mp= 7062.22g
 
  • #11


AxBy gives you x moles of A and y moles of B per mole of AxBy.

How about the moles of P in one mole of Ca5(PO4)3F?
 
  • #12


Three
 
  • #13


chemisttree said:
So if you were given 1 mole of Ca5(PO4)3F, you would get only one mole of P out of it?

Not necessarly i think
 
  • #14


chawki said:
mp/30.97 = 1150000/504.31
mp= 7062.22g

70622.23
 
  • #15


Please tell me if I'm right
 
  • #16


No. You have still ignored fact that there are three moles of P per apatite molecule.
 
  • #17


mp/3*30.97 = 1150000/504.31
mp= 211866.709g
mp=211.866Kg
 
  • #18


Please reply to my posts!
 
  • #19


We are not your private tutorial service, aren't we?

Watch significant figures.
 
  • #20


No, You are not. and i want to say that i appreciate having time for posting and replying.
God bless you all.
 

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