Question

write the formula for the following ionic compounds: part 1 of 3 name empirical formula rubidium oxide part 2 of 3 name empirical formula cobalt(iii) sulfide (containing the co³⁺ ion) part 3 of 3 name empirical formula iron(ii) sulfate (containing the fe²⁺ ion and so₄²⁻ ion)

Explanation:

Step1: Determine ion charges

Rubidium (Rb) forms a +1 ion ($Rb^+$) and oxygen forms a - 2 ion ($O^{2 -}$). To balance the charges, we need 2 Rb⁺ ions for every 1 $O^{2 -}$ ion.

Step2: Write formula

The formula for rubidium oxide is $Rb_2O$.

Step3: For cobalt(III) sulfide

Cobalt(III) has a +3 charge ($Co^{3+}$) and sulfur forms a - 2 ion ($S^{2 -}$). The least - common multiple of 3 and 2 is 6. So we need 2 $Co^{3+}$ ions and 3 $S^{2 -}$ ions to balance the charges. The formula is $Co_2S_3$.

Step4: For iron(II) sulfate

Iron(II) has a +2 charge ($Fe^{2+}$) and the sulfate ion ($SO_4^{2 -}$) has a - 2 charge. The charges are already balanced, so the formula is $FeSO_4$.

Answer:

Part 1 of 3: $Rb_2O$
Part 2 of 3: $Co_2S_3$
Part 3 of 3: $FeSO_4$