Question

which pair of compounds below would have the same empirical formula
k₂cro₄ and k₂cr₂o₇
no₂ and n₂o₅
c₂h₄ and c₄h₈
fecl₃ and fecl₂
question 6
10 pts
a compound is found to consist of 94.08% oxygen and 5.92% hydrogen. what is the empirical formula of this compound? please enter your answers as numerical digits (i.e. 9). if any of your answers are 1, please write this even though the number 1 is generally assumed in chemical formulas.

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