Calculate the number of moles of phosphorus in 15.95g of tetraphosphorus decaoxide I've got this question and I cannot figure out how to get the correct answer.


  Calculate the number of moles of Phosphorus in $15.95\:\mathrm{g}$ of tetraphosphorus decaoxide. ($\ce{P4O10}$)


I tried doing it like this:

Molar Mass of (element) = Number of atoms * Relative Atomic Mass = ### grams/mole

Mass of Compound in Grams = $15.95\:\mathrm{g}$

$M(\ce{P4}) = 4(30.97376) = 123.89504 \:\mathrm{g/mol}$
$M(\ce{O10}) = 10(15.9994) = 159.994 \:\mathrm{g/mol}$  

$M(\ce{P4O10}) = 123.89504\:\mathrm{g/mol} + 159.994\:\mathrm{g/mol} = 283.88904 \:\mathrm{g/mol}$

$\ce{P}: (123.89504 \:\mathrm{g/mol})/(283.88904 \:\mathrm{g/mol}) = 0.4364207 \cdot (100\% = 43.64\% \ce{P})$

$15.95\:\mathrm{g} \cdot 0.4364207 = 6.9565\:\mathrm{g} \ce{P}$

$\ce{P} = 6.9565\:\mathrm{g}/ (123.89504\:\mathrm{g/mol}) = 0.056148 \:\mathrm{mol}$

But the answer in the back of the book is $0.2247 \:\mathrm{mol}$
You can go about it more directly as follows.  You have $(15.95/283.889)=\pu{0.0562 moles}$ of $\ce{P4O10}$.  From the molecular formula you see that for every mole of $\ce{P4O10}$ you have 4 moles of phosphorous.  Therefor you have $4 \times 0.0562 = \pu{0.2247 moles}$ of phosphorous. 

Your error involved converting moles of $\ce{P4}$ to moles of $\ce{P}$.  If you multiply your answer by 4 you get the correct answer.