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Which of the following samples contains the largest number of atoms?

a.    2.0 moles of $\ce{H3PO4}$
b.    3.0 moles of $\ce{H2SO3}$
c.    4.0 moles of $\ce{HNO3}$
d.    6.0 moles of $\ce{HClO}$
e.    8.0 moles of $\ce{HBr}$

My logic would tell me that every mole has $6.022\times10^{23}$ atoms in it. Then you divide the amount of grams of each compound by the number of moles?

so for A I'd  use the following:

(Avg # *2) / 98 grams/per mole of $\ce{H3PO4}$  = $1.22\times10^{22}$ atoms 

and so on, but I don't get the right answer :/

$(2\times(6.022\times10^{23})$ = number of atoms = $1.20\times10^{23}$ $\ce{H3PO4}$ = 98 grams/mole

The answer is C btw, I just don't know to set up the equation right to get to this answer.

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Geeze, NaCl has 1 mole of Na and 1 mole of Cl per mole of NaCl. So that is 2 moles of atoms per mole of NaCl.

$\ce{K2SO4}$ has 2 moles of potassium, 1 mole of S and 4 moles of O per mole of $\ce{K2SO4}$. So that is 7 moles of atoms per mole $\ce{K2SO4}$.

You don't need to convert to atoms using Avogadro's number then back to moles. That complicates the problem needlessly. Just use the stoichiometry and you can do the arithmetic in your head.

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