[tex]\huge \text{$\boxed{\boxed{\rm KNO_3}}$}[/tex]
Empirical formula is the chemical formula of a compound where its constituent elements are in the simplest mole ratio.
To determine a compound's empirical formula, we must first calculate the number of moles of each element.
However to do this, we require the mass in grams. From the percentage compositions, we can say, "let the mass of the compound be 100 grams."
[tex]\large \textsf{$\therefore $ There is 38.7 g of potassium, 13.9 g of nitrogen, and 47.4 g}\\ \large \textsf{\ \ \ \,of oxygen in 100 g of compound.}[/tex]
To find the number of moles of each element (with symbol n ), we can divide the mass of each element (in grams, with symbol m ), by the molar mass of each element (in g/mol, with symbol M ), which can be found on an international standard IUPAC Periodic Table.
[tex]\boxed{\begin{tabular}{c}\Large\text{$\therefore$ number of moles = $\frac{\rm mass\ present}{\rm molar\ mass}$} \\\\ \huge\textsf{$\Rightarrow n=\frac{m}{M}$ }\\\end{tabular}}[/tex]
Therefore, applying this formula to all of the elements in the compound:
[tex]\large \textsf{$n(\rm K) = \frac{38.7}{39.10}$}\\\\\large \textsf{$\phantom{n(\rm K)}=0.9898\ \rm mol$}\\\large \textsf{$n(\rm N) = \frac{13.9}{14.01}$}\\\\\large \textsf{$\phantom{n(\rm N)}=0.9921\ \rm mol$}\\\\\large \textsf{$n(\rm O) = \frac{47.4}{14.01}$}\\\\\large \textsf{$\phantom{n(\rm O)}=3.383\ \rm mol$}[/tex]
∴ The ratio of K : N : O = 0.9898 : 0.9921 : 3.383. Simplifying this ratio by dividing all parts by 0.9898, will give us:
[tex]\large \text{1 : 1.002 : 3.418}\\\\\large \text{$\implies$ 1 : 1 : 3}[/tex]
Hence, inputting these values as the subscripts of each elemental symbol in the formula, the empirical formula is thus:
[tex]\Large \text{$\boxed{\boxed{\implies \rm KNO_3}}$}[/tex]
Note: the compound found, is a common ionic compound known as potassium nitrate.
To learn more about the empirical formula:
brainly.com/question/14044066
