Answer:
[tex]\Large \textsf{$\boxed{\boxed{\rm (NH_4)_2CO_3}}$}[/tex]
Explanation:
When working with percentage compositions, we can say, "let the mass of the compound be 100 grams"
[tex]\large \textsf{$\therefore$ there is 29.15 g of nitrogen, 8.41 g of hydrogen, 12.50 g of carbon, }\\ \large \textsf{\ \ \ and 49.9 g of oxygen in 100 g of compound.}[/tex]
Empirical Formula
The empirical formula of a compound is it's formula in which the constituent elements are in the simplest mole ratio
To find the number of moles of each element (denoted by symbol [tex]\textsf{$n$}[/tex]), we can divide the mass of each element (in grams, denoted by symbol [tex]\large \textsf{$m$}[/tex]), by the molar mass of each element (in g/mol, denoted by symbol [tex]\textsf{$M$}[/tex]), which can be found on an international standard IUPAC Periodic Table.
[tex]\Large \textsf{$\therefore \rm number\ of\ moles=\frac{mass\ present}{molar\ mass}$}[/tex]
[tex]\Large \textsf{$\implies \boxed{n= \frac{m}{M}}$}[/tex]
Now we can apply this to the above masses of each element:
[tex]\large \textsf{$n(\rm N) = \frac{29.15}{14.01}$}\\\\\large \textsf{$\phantom{n(\rm N)}=2.0807\ \rm mol$}\\\large \textsf{$n(\rm H) = \frac{8.41}{1.008}$}\\\\\large \textsf{$\phantom{n(\rm H)}=8.3433\ \rm mol$}\\\\\large \textsf{$n(\rm C) = \frac{12.50}{12.01}$}\\\\\large \textsf{$\phantom{n(\rm C)}=1.0408\ \rm mol$}\\\\\large \textsf{$n(\rm O) = \frac{49.9}{16.00}$}\\\\\large \textsf{$\phantom{n(\rm O)}=3.1188\ \rm mol$}\\[/tex]
[tex]\large \text{$\therefore $ the ratio of N : H : C : O}\\\\ \large \text{$\Rightarrow$2.0807 : 8.3433 : 1.0408 : 3.1188}[/tex]
Simplifying this ratio by dividing all parts by 2.0807:
[tex]\large \text{$\therefore$ 1 : 4.0098 : 0.5002 : 1.4989}\\\\\large \text{$\implies$ 1 : 4 : 0.5 : 1.5}[/tex]
Since the mole ratio is displayed in integers, multiply this result by 2:
[tex]\large \text{$\therefore$ 2 : 8 : 1 : 3 is the final mole ratio.}\\\\\\ \large \text{$\boxed{\boxed{\implies \rm N_2H_8CO_3$ or $\rm (NH_4)_2CO_3}}$}[/tex]
Note: the compound found, is a common ionic compound known as ammonium carbonate.
To learn more about empirical formula:
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