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 zn²⁺ oh⁻ ☐ ☐ sn⁴⁺ clo₃⁻ ☐ ☐ ca²⁺ po₄³⁻ ☐ ☐

Explanation:

For $\boldsymbol{\ce{Zn^{2+}}}$ and $\boldsymbol{\ce{OH^-}}$:

Step1: Balance charges

The cation is $\ce{Zn^{2+}}$ (charge +2) and the anion is $\ce{OH^-}$ (charge -1). To balance the charges, we need 2 $\ce{OH^-}$ ions for each $\ce{Zn^{2+}}$ ion (since $2\times(-1) + (+2) = 0$).

Step2: Write empirical formula

The empirical formula is $\ce{Zn(OH)2}$.

Step3: Name the compound

The cation is zinc (II) (since $\ce{Zn}$ has a +2 charge here), and the anion is hydroxide. So the name is zinc hydroxide.

For $\boldsymbol{\ce{Sn^{4+}}}$ and $\boldsymbol{\ce{ClO3^-}}$:

Step1: Balance charges

The cation is $\ce{Sn^{4+}}$ (charge +4) and the anion is $\ce{ClO3^-}$ (charge -1). We need 4 $\ce{ClO3^-}$ ions for each $\ce{Sn^{4+}}$ ion (since $4\times(-1) + (+4) = 0$).

Step2: Write empirical formula

The empirical formula is $\ce{Sn(ClO3)4}$.

Step3: Name the compound

The cation is tin (IV) (since $\ce{Sn}$ has a +4 charge here), and the anion is chlorate. So the name is tin (IV) chlorate.

For $\boldsymbol{\ce{Ca^{2+}}}$ and $\boldsymbol{\ce{PO4^{3-}}}$:

Step1: Balance charges

The cation is $\ce{Ca^{2+}}$ (charge +2) and the anion is $\ce{PO4^{3-}}$ (charge -3). To balance the charges, we find the least common multiple of 2 and 3, which is 6. So we need 3 $\ce{Ca^{2+}}$ ions (total charge $3\times(+2) = +6$) and 2 $\ce{PO4^{3-}}$ ions (total charge $2\times(-3) = -6$).

Step2: Write empirical formula

The empirical formula is $\ce{Ca3(PO4)2}$.

Step3: Name the compound

The cation is calcium (which has a fixed +2 charge, so we don't need to specify the charge in the name), and the anion is phosphate. So the name is calcium phosphate.

Filled Table:

cation	anion	empirical formula	name of compound
$\ce{Sn^{4+}}$	$\ce{ClO3^-}$	$\ce{Sn(ClO3)4}$	tin (IV) chlorate
$\ce{Ca^{2+}}$	$\ce{PO4^{3-}}$	$\ce{Ca3(PO4)2}$	calcium phosphate

Answer:

For $\boldsymbol{\ce{Zn^{2+}}}$ and $\boldsymbol{\ce{OH^-}}$:

Step1: Balance charges

The cation is $\ce{Zn^{2+}}$ (charge +2) and the anion is $\ce{OH^-}$ (charge -1). To balance the charges, we need 2 $\ce{OH^-}$ ions for each $\ce{Zn^{2+}}$ ion (since $2\times(-1) + (+2) = 0$).

Step2: Write empirical formula

The empirical formula is $\ce{Zn(OH)2}$.

Step3: Name the compound

The cation is zinc (II) (since $\ce{Zn}$ has a +2 charge here), and the anion is hydroxide. So the name is zinc hydroxide.

For $\boldsymbol{\ce{Sn^{4+}}}$ and $\boldsymbol{\ce{ClO3^-}}$:

Step1: Balance charges

The cation is $\ce{Sn^{4+}}$ (charge +4) and the anion is $\ce{ClO3^-}$ (charge -1). We need 4 $\ce{ClO3^-}$ ions for each $\ce{Sn^{4+}}$ ion (since $4\times(-1) + (+4) = 0$).

Step2: Write empirical formula

The empirical formula is $\ce{Sn(ClO3)4}$.

Step3: Name the compound

The cation is tin (IV) (since $\ce{Sn}$ has a +4 charge here), and the anion is chlorate. So the name is tin (IV) chlorate.

For $\boldsymbol{\ce{Ca^{2+}}}$ and $\boldsymbol{\ce{PO4^{3-}}}$:

Step1: Balance charges

The cation is $\ce{Ca^{2+}}$ (charge +2) and the anion is $\ce{PO4^{3-}}$ (charge -3). To balance the charges, we find the least common multiple of 2 and 3, which is 6. So we need 3 $\ce{Ca^{2+}}$ ions (total charge $3\times(+2) = +6$) and 2 $\ce{PO4^{3-}}$ ions (total charge $2\times(-3) = -6$).

Step2: Write empirical formula

The empirical formula is $\ce{Ca3(PO4)2}$.

Step3: Name the compound

The cation is calcium (which has a fixed +2 charge, so we don't need to specify the charge in the name), and the anion is phosphate. So the name is calcium phosphate.

Filled Table:

cation	anion	empirical formula	name of compound
$\ce{Sn^{4+}}$	$\ce{ClO3^-}$	$\ce{Sn(ClO3)4}$	tin (IV) chlorate
$\ce{Ca^{2+}}$	$\ce{PO4^{3-}}$	$\ce{Ca3(PO4)2}$	calcium phosphate