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# How many atoms of oxygen are contained in $47.6\,g$ of $A{l_2}{\left( {C{O_3}} \right)_3}$? The molar mass of $A{l_2}{\left( {C{O_3}} \right)_3}$ is $233.99\,g\,mo{l^{ - 1}}$ .

Last updated date: 06th Sep 2024
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Hint: The total mass occupied by Avogadro’s number of particles I.e., $6.022 \times {10^{23}}$ particles of a substance is known as its molar mass. It is generally provided in the units of $gram\left( {gm} \right)$ and is numerically equal to the molar mass of the molecule. A molecule of$A{l_2}{(C{O_3})_3}$ has $3CO_3^{2 - }$ i.e., three carbonate ions each of which contains$3O$atoms thus making a total of $9$ atoms of Oxygen in the whole molecule which would be used to obtain the required answer.

We know that, one molecule of $A{l_2}{\left( {C{O_3}} \right)_3}$ contains $3 \times 3 = 9$ oxygen atoms
Now, Number of moles $\left( n \right)$ of $A{l_2}{\left( {C{O_3}} \right)_3}$ is equal to $\dfrac{w}{M}$
(Where w is the given weight of $A{l_2}{\left( {C{O_3}} \right)_3}$ and M is the molar weight of $A{l_2}{\left( {C{O_3}} \right)_3}$)
So, $n = \dfrac{{47.6g}}{{233.99g\,mol{\,^{ - 1}}}}$
$= 0.203\,mol$
Now, $1$mole of$A{l_2}{(C{O_3})_3}$ contains $9moles$ of Oxygen atom
Therefore, $0.203mole$ of $A{l_2}{(C{O_3})_3}$ contains $0.203 \times 9 = 1.827moles$ of oxygen atoms,
So, the number of oxygen atoms is equal to $1.827 \times 6.022 \times {10^{23}} = 11.022 \times {10^{23}}$ atoms.
Note: Here precaution must be taken while handling the numerical calculations as a small error in calculation could make a drastic change thus, we must try to do calculations that are precise to the very decimal points. Also, we must take note about the units of weight and molar mass in the question as this sometimes leads to grave calculative errors especially when weight is given in units other than the standard S.I. units of $grams\left( {gm} \right)$, commonly the weight is provided in units of microgram, milligram, nanogram etc. other than the standard units of gram. Another common mistake done by students is not reading what is asked in the question properly; sometimes we are tasked with finding the number of oxygen atoms in moles and not in grams. In those cases we should only provide answers to what is asked and not to what all we can.