How many total atoms are in .660 g of $P_2O_5$ ?

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Answer 1 · 2016-05-18T05:28:04

$14 xx 10 ^21$ atoms

Let us first calculate the number of moles of Phosphorous ( $V$ ) Pentoxide i.e. $P_2O_5$

mass of $P_2$ $O_5$ = $0.660$ $g$

molar mass of $P_2$ $O_5$ = $142$ $g$ $mol^-1$

$"number of moles of"$ $P_2$ $O_5$ = $"mass" /"molar mass"$

$n = 0.660 / 142$ $g$ $mol^-1$

$n = 0.0046$ $mol$ of $P_2$ $O_5$

one mole of $P_2$ $O_5$ has 5 moles of Oxygen atoms.

using this we can write two conversion factors

1 mol $P_2$ $O_5$ / 5 mol Oxygen or 5 mol Oxygen / 1 mol $P_2$ $O_5$

using conversion factor

$0.0046$ $mol$ of $P_2$ $O_5$ x $[ (5 " mol " O_2 ) /( 1 " mol " P_2O_5)]$

= $0.023$ $mol$ of Oxygen

$0.023$ $mol xx 6.02 xx 10 ^23$ $"atom"$ / $mol$

= $14 xx 10 ^21$ $"atoms"$

How many total atoms are in .660 g of P_2O_5P_2O_5?