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?

1 Answer
Mar 31, 2018

Answer:

#4:3#

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 #8# parts by mass of oxygen.

Now, the equivalent weight of oxygen in oxides is #"8 g"#, so you know that the equivalent weight of the metal in this oxide is

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

This means that when this oxide is formed, every #"24 g"# of metal will combine with #"8 g"# of oxygen. In other words, if you use #"24 g"# of metal to produce this oxide, you know for a fact that the oxide will contain #"8 g"# of oxygen and have a total mass of

#"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 g"# of metal, you can say that the oxide will contain

#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#