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Thread: A basic Chemistry question about Apatite

  1. #1 A basic Chemistry question about Apatite 
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    Hi there! I'm going over some past papers for a university and found this question...unfortunately they don't give the ANSWERS

    The problem is: Apatite (Ca5(PO4)3F) is used to manufacture fertilizer. How many kilograms of pure phosphorus (P) is included in 1150 kg of apatite?

    I found the molar masses to be as follows:
    Ca: 40.08
    P: 30.974
    O: 15.999
    F: 18.998

    What I did was added them up as shown in the formula (i.e. 40.08 x 5 + (30.974 + 15.999 x 4) x 3 + 18.998 = 504.308) then I divided the amount of phosphorous (3 x 30.974) by the total amount (504.308) and got approx. 0.184 ...I'm not sure if that step was right but I kept going! Since the amount of phosphorous makes up 0.184 of the total amount I went 1150(kg) x 0.184 and got approx. 211.9 kilograms.

    SO basically my answer is 211.9 kg but I have no idea if it is right or completely wrong! Can someone please help me?

    Thank you all!


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  3. #2  
    exchemist
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    Quote Originally Posted by aic18 View Post
    Hi there! I'm going over some past papers for a university and found this question...unfortunately they don't give the ANSWERS

    The problem is: Apatite (Ca5(PO4)3F) is used to manufacture fertilizer. How many kilograms of pure phosphorus (P) is included in 1150 kg of apatite?

    I found the molar masses to be as follows:
    Ca: 40.08
    P: 30.974
    O: 15.999
    F: 18.998

    What I did was added them up as shown in the formula (i.e. 40.08 x 5 + (30.974 + 15.999 x 4) x 3 + 18.998 = 504.308) then I divided the amount of phosphorous (3 x 30.974) by the total amount (504.308) and got approx. 0.184 ...I'm not sure if that step was right but I kept going! Since the amount of phosphorous makes up 0.184 of the total amount I went 1150(kg) x 0.184 and got approx. 211.9 kilograms.

    SO basically my answer is 211.9 kg but I have no idea if it is right or completely wrong! Can someone please help me?

    Thank you all!
    This must be a pretty feeble university. Sounds more like homework. Check in your textbook and see what your teacher has to say.


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  4. #3  
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    I don't have a professor, these are just practice questions -.-
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  5. #4  
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    I am surprised that you would not find yourself an online calculator to help you with this sort of question. This one should help you to check your answers.

    https://www.caymanchem.com/app/templ...vm/itemid/4002
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