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How many phosphorus atoms are in a 0.3352 g sample of Ca3(PO4)2 ?

  • chemistry - ,

    well since nobody helped you yet I can probably get you started.

    avogadro's number =6.022 × 10^23
    this is the number of atoms in 1mol

    you'd have to first use the grams and find the molecular weight of the compound. Ca3(PO4)2 = ?g

    0.3352g ( 1mol Ca3(PO4)2 / mol weight of Ca3(PO4)2 ) = mol Ca3(PO4)2

    then find out how many moles of P you have by multiplying the

    (mol Ca3(PO4)2) (# mol P / 1mol Ca3(PO4)2) = # mol P

    then use that and multiply that by avogadro's number to find the atoms of P you have..

    #mol P (6.022*10^23 atom/ 1mol)= atoms of P

    If I remember general chem correctly this should be it.

  • chemistry - ,

    That's it, thanks.

    Remember Rog, that the number of P per mole of Calcium phosphate is 2 mole P per mole of calcium phosphate.

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