# Question 961df

Mar 27, 2016

$\text{17 g}$

#### Explanation:

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 $\text{8 g}$ of oxygen.

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

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

• $\text{32 g } \to$ oxygen
• $\text{68 g } \to$ metal

So, if $\text{32 g}$ of oxygen reacted with $\text{68 g}$ of metal, it follows that $\text{8 g}$ of oxygen reacted with

$8 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g oxygen"))) * overbrace("68 g metal"/(32color(red)(cancel(color(black)("g oxygen")))))^(color(purple)("given percent composition")) = color(green)(|bar(ul(color(white)(a/a)"17 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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, ${\text{O}}^{2 -}$.

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

The $\text{100.0-g}$ sample will have $\text{32 g}$ 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!

$1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mole e"^(-)))) * "68 g metal"/(4color(red)(cancel(color(black)("moles e"^(-))))) = color(green)(|bar(ul(color(white)(a/a)"17 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$