posted by Rog on .
How many phosphorus atoms are in a 0.3352 g sample of Ca3(PO4)2 ?
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.
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.