2.0 g of molybdenum (Mo) combines with oxygen to form 3.0 g of a molybdenum oxide. What is the equivalent weight of Mo in this compound?

1 Answer
Nov 28, 2015


#"16 g/equiv."#


The trick here is to recognize that you're dealing with a redox reaction in which molybdenum is being oxidized and oxygen is being reduced.

You don't actually have to know the formula of the oxide in order to be able to determine the equivalent weight of molybdenum in the oxide, you can use oxygen's known oxidation state.

More specifically, oxygen will go from an oxidation state of #0# as a reactant to an oxidation state of #(-2)# as part of the oxide.

This means that every mole of oxygen that takes part in the reaction will gain 2 moles of electrons.

As you know, equivalent weight in the context of a redox reaction is defined as the mass of a compound that supplies or reacts with one mole of electrons.

So, if one mole of oxygen has a mass of #"16.0 g"#, and you have one mole of oxygen gaining 2 moles of electrons, it follows that oxygen's equivalent weight in this reaction will be

#16.0"g"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole"))))/("2 moles electrons") = "8.00 g/equiv."#

Here #"1 equiv ." = " 1 mole electrons"#.

Now all you have to do is figure out exactly what mass of oxygen took part in the reaction. So, if you start with #"2.0 g"# of molybdenum and end up with #"3.0 g"# of oxide, it follows that the difference between these two masses will be the mass of oxygen that took part in the reaction.

#m_"oxygen" = "3.00 g" - "2.0 g" = "1.0 g O"#

If your reaction used #"1.0 g"# of oxygen, and its equivalent weight is #"8.00 g/equiv."#, it follows that your reaction needed

#1.0 color(red)(cancel(color(black)("g O"))) * "1 equiv."/(8.00color(red)(cancel(color(black)("g O")))) = "0.125 equiv. O"#

Since in any redox reaction the number of equivalents gained by the species that gets reduced must be equal to the number of equivalents lost by the species that gets oxidized, you know that molybdenum lost #"0.125 equiv."#.

Since #"2.0 g"# of molybdenum took part in the reaction, you can say that its equivalent weight will be

#"2.0 g"/"0.125 equiv." = "16 g/equiv."#