Question

fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:
some ionic compounds
cation | anion | empirical formula | name of compound
$ce{na+}$ | $ce{no^-_{2}}$ | $ce{nano_{2}}$ | sodium nitrite
$ce{k+}$ | $ce{bro^-_{3}}$ | $ce{kbro_{3}}$ | potassium bromate
$ce{na+}$ | $ce{io^-_{3}}$ | $ce{naio_{3}}$ | sodium iodate

Explanation:

Step1: Analyze the first ionic compound (Na⁺ and NO₂⁻)

To form a neutral ionic compound, the total positive charge must equal the total negative charge. The cation is \( \text{Na}^+ \) (charge +1) and the anion is \( \text{NO}_2^- \) (charge -1). So, the ratio of cation to anion is 1:1. Thus, the empirical formula is \( \text{NaNO}_2 \) (since \( 1\times(+1) + 1\times(-1) = 0 \)). The name is sodium nitrite (sodium for \( \text{Na}^+ \) and nitrite for \( \text{NO}_2^- \)).

Step2: Analyze the second ionic compound (K⁺ and BrO₃⁻)

The cation is \( \text{K}^+ \) (charge +1) and the anion is \( \text{BrO}_3^- \) (charge -1). The ratio of cation to anion is 1:1. So, the empirical formula is \( \text{KBrO}_3 \) ( \( 1\times(+1) + 1\times(-1) = 0 \) ). The name is potassium bromate (potassium for \( \text{K}^+ \) and bromate for \( \text{BrO}_3^- \)).

Step3: Analyze the third ionic compound (Na⁺ and IO₃⁻)

The cation is \( \text{Na}^+ \) (charge +1) and the anion is \( \text{IO}_3^- \) (charge -1). The ratio of cation to anion is 1:1. So, the empirical formula is \( \text{NaIO}_3 \) ( \( 1\times(+1) + 1\times(-1) = 0 \) ). The name is sodium iodate (sodium for \( \text{Na}^+ \) and iodate for \( \text{IO}_3^- \)).

Answer:

cation	anion	empirical formula	name of compound
\( \text{K}^+ \)	\( \text{BrO}_3^- \)	\( \text{KBrO}_3 \)	Potassium bromate
\( \text{Na}^+ \)	\( \text{IO}_3^- \)	\( \text{NaIO}_3 \)	Sodium iodate