What is the empirical formula of this compound?

A. [tex]\( \text{C}_8 \text{H}_{11} \text{NO}_2 \)[/tex]
B. [tex]\( \text{C}_{16} \text{H}_{22} \text{N}_2 \text{O}_4 \)[/tex]
C. [tex]\( \text{C}_4 \text{H}_5 \text{NO} \)[/tex]
D. [tex]\( \text{C}_8 \text{H}_{11} \text{NO}_2 \)[/tex]



Answer :

To find the empirical formula of the given compounds, we need to identify the simplest whole-number ratio of atoms in each compound. Here are the compounds provided:

1. [tex]\( C_8H_{11}NO_2 \)[/tex]
2. [tex]\( C_{16}H_{22}N_2O_4 \)[/tex]
3. [tex]\( C_4H_5NO \)[/tex]
4. [tex]\( C_8H_{11}NO_2 \)[/tex]

The empirical formula is the simplest ratio of the atoms in a compound.

- For the first compound [tex]\( C_8H_{11}NO_2 \)[/tex], reducing it to the simplest ratio would involve dividing by the greatest common divisor of the subscripts, which in this case is 1. Thus, the empirical formula remains [tex]\( C_8H_{11}NO_2 \)[/tex].
- For the second compound [tex]\( C_{16}H_{22}N_2O_4 \)[/tex], reducing it by dividing by the greatest common divisor, which is 2, we get the empirical formula:
[tex]\( \left(\frac{C_{16}}{2}\right)H_{11} \left(\frac{N_2}{2}\right)O_2= C_8H_{11}NO_2 \)[/tex].
- For the third compound [tex]\( C_4H_5NO \)[/tex], there is no further simplification required because the subscripts are already in their simplest form. Thus, the empirical formula is [tex]\( C_4H_5NO \)[/tex].
- The fourth compound again is [tex]\( C_8H_{11}NO_2 \)[/tex], which is already in its simplest form as explained above.

Comparing all the compounds, the third compound [tex]\( C_4H_5NO \)[/tex] is distinct because it cannot be further reduced. The given analysis shows that the empirical formula that represents a distinct and simplest form among the given compounds is:

[tex]\[ C_4H_5NO \][/tex]

Hence, the empirical formula of the given compound is [tex]\( C_4H_5NO \)[/tex].