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Sorry if my question is a bit obvious, but for some reason I'm getting confused with the percentages in this question:

Concentration of $\ce{A}$ is $\pu{80 at\%}$ and concentration of $\ce{B}$ is $\pu{19.8 at\%}$ for $\ce{AB2}.$ What would be the mass for $\ce{A}$ and $\ce{B}$ individually, if the atomic weight of $\ce{A}$ is $\pu{5 g/mol}$ and $\ce{B}$ is $\pu{10 g/mol}?$

I understand that the molecular weight $M$ for my compound would be:

$$M(\ce{AB2}) = (5 + 2\cdot 10)~\pu{g/mol} = \pu{25 g/mol}$$

How do I include the atomic percentage to get the mass of $\ce{A}$ and $\ce{B}?$

I am just not sure how to use $\pu{at\%}.$ My understanding is that if I assume $\pu{1 mol}$ of $\ce{AB2},$ then I can calculate the mass of $\ce{A}$ by assuming that $\pu{80 at\%}$ is $\pu{0.80 mol}?$

I have another problem where they specify $\pu{mol\%}$ (mole percentage) instead. So I am getting confused as to how to use $\pu{at\%}$ and $\pu{mol\%}.$

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    $\begingroup$ Atom percent is used in X-ray photoelectron spectroscopy or energy dispersive spectrometry. Could you provide some context? Is this an experimental data or just an textbook exercise? $\endgroup$ – M. Farooq Jun 11 at 3:39
  • $\begingroup$ Any kernel with nucleon number 5 decays extremely fast, so we can definitely exclude option of experimental data. $\endgroup$ – Poutnik Jun 11 at 4:26
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    $\begingroup$ at% = mol% in case matter is atomised. $\endgroup$ – Poutnik Jun 11 at 4:42

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