# The equivalent weight of metal is triple to that of oxygen. What is the ratio of weight of its oxide to that of metal?

##
**Options:**

#"(1) 1 : 2"#

#"(2) 1 : 3"#

#color(blue)("(3) 4 : 3"#

#"(4) 3 : 4"#

#bb("My approach")#

#"E"_"m" = 3 × "E"_"o" = 3 × 8 = 24#

#("M"_"m")/"n" = 24 color(white)(...)[∵ "Equivalent weight" = "Molecular weight" / "n-factor"]#

Formula of metal oxide: #"M"_2"O"_"n"#

#"Weight of metal oxide" = 2"M"_"m" + ("n" × 16) = 2"M"_"m" + "16n"#

#"Ratio" =("2M"_"m" + "16n")/("M"_"m") = 2 + (16 × "n"/"M"_"m") = 2 + (16 × 1/24) = 8 : 3#

But correct option given is #4 : 3#

Is the formula of metal oxide correct?

**Options:**

Formula of metal oxide:

But correct option given is

Is the formula of metal oxide correct?

##### 1 Answer

#### Answer:

#### Explanation:

You don't need the formula of the oxide to find this ratio, all you really to know here is the **equivalent weight** of the metal.

As you know, when it comes to metals that react with oxygen to form metal oxides, the equivalent weight of the metal is the mass of metal that combines with **parts by mass** of oxygen.

Now, the equivalent weight of oxygen in oxides is

#3 xx "8 g" = "24 g"#

This means that when this oxide is formed, **every**

#"24 g + 8 g = 32 g"#

Therefore, the ratio between the mass of the oxide and the equivalent weight of the metal will be

#(32 color(red)(cancel(color(black)("g"))))/(24color(red)(cancel(color(black)("g")))) = 4/3#

The result will be the same *regardless* of the mass of the metal. For example, if you use

#5 color(red)(cancel(color(black)("g metal"))) * "8 g oxygen"/(24 color(red)(cancel(color(black)("g metal")))) = 5/3 quad "g oxygen"#

Once again, the ratio between the mass of the oxide and the equivalent weight of the metal will be

#( (5/3 + 5) color(red)(cancel(color(black)("g"))))/(5 color(red)(cancel(color(black)("g")))) = 4/3#