Calculate Mass of P in Apatite - 211.866kg

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SUMMARY

The calculation of the mass of pure phosphorus (P) in 1150 kg of apatite (Ca5(PO4)3F) results in 211.866 kg. This is derived using the molar mass of apatite, which is 504.31 g/mol, and the molar mass of phosphorus, which is 30.97 g/mol. The formula used is mp/(3*30.97) = 1150000/504.31, leading to the conclusion that 211.866 kg of phosphorus is present in the given amount of apatite.

PREREQUISITES
  • Understanding of molar mass calculations
  • Familiarity with chemical formulas, specifically apatite (Ca5(PO4)3F)
  • Basic knowledge of stoichiometry
  • Ability to perform unit conversions (kg to g)
NEXT STEPS
  • Study the concept of molar mass in detail
  • Learn about stoichiometric calculations in chemical reactions
  • Research the applications of apatite in fertilizer production
  • Explore the significance of phosphorus in agriculture
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Chemistry students, educators, and professionals in agriculture or fertilizer manufacturing will benefit from this discussion, particularly those focused on nutrient calculations and chemical compositions.

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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


mp/(3*30.97) = 1150000/504.31
mp= 211866.709g =211.866Kg
 
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Where is the apatite's molar mass involved ?

I saw you had already opened a thread on the same subject. Well, don't ever do that, even if you think the received answers are not satisfactory.

Thanks.
 
Molar mass is 504.31g/mol
 

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