Q90P - Chemistry: Molecular Nature Of Matter And Change

Question

Give the systematic names for the formulas or the formulas for the names: 

a.   $\mathbf{tin}\left(\mathrm{lv}\right)\mathbf{ }\mathbf{chorlde}$

b.   ${\mathbf{FeBr}}_{\mathbf{3}\mathbf{ }\mathbf{,}}$

c.    cuprous bromide;

d.    ${\mathbf{Mn}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$

Step-by-Step Solution

Answer

a.  The formula of the name tin(IV) chloride is ${\mathbf{Sn}}^{\mathbf{4}\mathbf{+}}{\mathbf{CI}}^{\mathbf{1}\mathbf{-}}\mathbf{ }\mathbf{\to }{\mathbf{SnCI}}_{\mathbf{4}}$

b.  The name of the formula ${\mathrm{FeBr}}_{3 ,}$ is Iron(III) bromide.

c.  The formula of the name cuprous bromide is ${\mathbf{Cu}}^{\mathbf{1}\mathbf{+}}{\mathbf{Br}}^{\mathbf{1}\mathbf{-}}\mathbf{\to }\mathbf{CuBr}$

d.  The name of the formula ${\mathrm{Mn}}_2{\mathrm{O}}_3$ is manganese (II) oxide.

Step 1: Definition of chemical formula

An expression that states the number and type of atoms present in a molecule of a substance is defined as a chemical formula

Step 2: Explanations

a. Tin has a charge of 4+ and chlorine has a 1- charge.

The charges will just interchange as subscripts of these elements. 

 ${\mathbf{Sn}}^{\mathbf{4}\mathbf{+}}{\mathbf{CI}}^{\mathbf{1}\mathbf{-}}\mathbf{ }\mathbf{\to }{\mathbf{SnCI}}_{\mathbf{4}}$

b. The compound is an ionic compound. The name does not depend on the number of elements.

Iron can have many forms So, the charge should be known.

The subscripts will just interchange as charges of these elements

 ${\mathbf{FeBr}}_{\mathbf{3}\mathbf{ }}\mathbf{\to }{\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}{\mathbf{Br}}^{\mathbf{1}\mathbf{-}}$

The cation is Iron (III) and the anion is bromide. So it is Iron(III) bromide.

 

C. Cuprous means a copper with +1 charge bromide has a 1- charge. 

The charges will just interchange as subscripts of these elements.

 ${\mathbf{Cu}}^{\mathbf{1}\mathbf{+}}{\mathbf{Br}}^{\mathbf{1}\mathbf{-}}\mathbf{\to }\mathbf{CuBr}$

 

d. The compound is an ionic compound. The name does not depend on the number of elements.

Manganese ions can have many forms so the charges should be known.

The subscripts will just interchange as charges of these elements

 ${\mathbf{Mn}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{\to }{\mathbf{Mn}}^{\mathbf{+}\mathbf{3}}{\mathbf{O}}^{\mathbf{2}\mathbf{-}}$

The cation is manganese (II) and the anion is oxide. So it is manganese (II) oxide.