which pair of compounds below would have the same empirical formula\nk₂cro₄ and k₂cr₂o₇\nno₂ and n₂o₅\nc₂h₄…

Explanation:

Step1: Recall empirical formula concept

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

Step2: Analyze each option for first question

• For $\mathrm{K_2CrO_4}$ and $\mathrm{K_2Cr_2O_7}$, the ratios of elements are different.
• For $\mathrm{NO_2}$ and $\mathrm{N_2O_5}$, the ratios of $\mathrm{N}$ and $\mathrm{O}$ are different.
• For $\mathrm{C_2H_4}$, dividing by 2 gives $\mathrm{CH_2}$. For $\mathrm{C_4H_8}$, dividing by 4 gives $\mathrm{CH_2}$. They have the same empirical formula.
• For $\mathrm{FeCl_3}$ and $\mathrm{FeCl_2}$, the ratios of $\mathrm{Fe}$ and $\mathrm{Cl}$ are different.

Step3: Solve second question

Assume 100 g of the compound. So, there are 94.08 g of $\mathrm{O}$ and 5.92 g of $\mathrm{H}$. The number of moles of $\mathrm{O}$, $n_O=\frac{94.08\ g}{16\ g/mol}=5.88\ mol$ The number of moles of $\mathrm{H}$, $n_H=\frac{5.92\ g}{1\ g/mol}=5.92\ mol$ The mole - ratio of $\mathrm{H}$ to $\mathrm{O}$ is $\frac{n_H}{n_O}=\frac{5.92}{5.88}\approx1:1$ The empirical formula is $\mathrm{HO}$.

Answer:

1. C. $\mathrm{C_2H_4}$ and $\mathrm{C_4H_8}$
2. HO