Calculate the number of moles the Phosphorus in $15.95:mathrmg$ of tetraphosphorus decaoxide. ($ceP4O10$)

I tried act it favor this:

Molar fixed of (element) = variety of atoms * family member Atomic mass = ### grams/mole

Mass of link in Grams = $15.95:mathrmg$

$M(ceP4) = 4(30.97376) = 123.89504 :mathrmg/mol$$M(ceO10) = 10(15.9994) = 159.994 :mathrmg/mol$

$M(ceP4O10) = 123.89504:mathrmg/mol + 159.994:mathrmg/mol = 283.88904 :mathrmg/mol$

$ceP: (123.89504 :mathrmg/mol)/(283.88904 :mathrmg/mol) = 0.4364207 cdot (100\% = 43.64\% ceP)$

$15.95:mathrmg cdot 0.4364207 = 6.9565:mathrmg ceP$

$ceP = 6.9565:mathrmg/ (123.89504:mathrmg/mol) = 0.056148 :mathrmmol$

But the price in the ago of the publication is $0.2247 :mathrmmol$

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edited Mar 4 "18 in ~ 2:50

Gaurang Tandon
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You have the right to go about it an ext directly together follows. You have $(15.95/283.889)=pu0.0562 moles$ of $ceP4O10$. Native the molecular formula you check out that for every mole that $ceP4O10$ you have 4 mole of phosphorous. Therefor you have $4 imes 0.0562 = pu0.2247 moles$ of phosphorous.

Your error connected converting moles of $ceP4$ to mole of $ceP$. If friend multiply her answer by 4 you get the exactly answer.

You are watching: How many atoms of phosphorus are in tetraphosphorus decoxide?

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edited Mar 4 "18 in ~ 2:51

Gaurang Tandon
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answered Jun 11 "14 at 23:28

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