Question #961df

Start by making sure that you understand what it is you're looking for here.

In this context, a substance's equivalent weight will tell you the mass of said substance that will combine with #8# parts by mass of oxygen.

Simply put, the equivalent weight will tell you the mass of the substance that reacts with #"8 g"# of oxygen.

This tells you that in order to find the metal's equivalent weight, you must determine what mass of metal reacted with #"8 g"# of oxygen to produce that oxide.

Since the oxide is said to be #32%"w/w"# oxygen, you can say that a #"100.0-g"# sample will contain

• #"32 g " -># oxygen
• #"68 g " -># metal

So, if #"32 g"# of oxygen reacted with #"68 g"# of metal, it follows that #"8 g"# of oxygen reacted with

#8 color(red)(cancel(color(black)("g oxygen"))) * overbrace("68 g metal"/(32color(red)(cancel(color(black)("g oxygen")))))^(color(purple)("given percent composition")) = color(green)(|bar(ul(color(white)(a/a)"17 g"color(white)(a/a)|)))#

ALTERNATIVE APPROACH

You can double-check your answer by using another definition given for equivalent weight.

In the context of a redox reaction, like you have here, the equivalent weight of a substance is the mass of that substance that accepts or supplies one mole of electrons.

When a metal oxide is formed, a neutral oxygen atom picks up two electrons to form the oxide anion, #"O"^(2-)#.

That means that one mole of oxygen atoms must pick up two moles of electrons.

The #"100.0-g"# sample will have #"32 g"# of oxygen. Use oxygen's molar mass to determine how many moles of oxygen you get in that sample

#32color(red)(cancel(color(black)("g"))) * "1 mole oxygen"/(16.0color(red)(cancel(color(black)("g")))) = "2 moles oxygen"#

This means that the metal must have supplied a total of

#2 color(red)(cancel(color(black)("moles oxygen"))) * "2 moles e"^(-)/(1color(red)(cancel(color(black)("mole oxygen")))) = "4 moles e"^(-)#

This means that the metal's equivalent weight will once again be - remember, you're looking for the mass of the metal that will supply #1# mole of electrons to the reaction!

#1color(red)(cancel(color(black)("mole e"^(-)))) * "68 g metal"/(4color(red)(cancel(color(black)("moles e"^(-))))) = color(green)(|bar(ul(color(white)(a/a)"17 g"color(white)(a/a)|)))#