Using stoichiometry to determine empirical formula

[littlebearrrr]

Homework Statement

1.000 g of vanadium is mixed with 8.000 g of bromine. After the elements react, some bromine is left over, along with a single compound that contains the two elements. The excess bromine is removed and allowed to react with excess sodium sulfite and excess sodium hydroxide, producing a mixture that contains 2.93 g of sodium sulfate. The balanced equation for this second reaction is:

2NaOH + Na2SO3 + Br2 -> 2NaBr + Na2SO4 + H2O

Based on this info, determine the empirical formula of the compound that was formed when vanadium reacted with bromine.

Homework Equations

None.

The Attempt at a Solution

Well, I know that my endgoal here is to find the moles of vanadium and moles of bromine for the product in the first reaction. Once I get there, I'm pretty much all set to go.

To start, I used the given amount of vanadium/bromine to find moles of vanadium/bromine. Since I know vanadium is the limiting reactant, I used my calculated moles of bromine (left over) to find the amount of bromine left over from the reaction. At this point, I was unsure what to do next. Mostly, I am confused about what I need to do with the given amount of sodium sulfate (and how the second reaction is relevant to the problem).

 

[djh101]

The second equation tells you how much Br was left over after reacting completely with V. You can subtract the excess amount from the original amount to find the amount of Br that reacted. From here you can find molar ratio of V and Br using the amount of V reacted and the amount of Br reacted.

 

[littlebearrrr]

Thanks for the response! I got VBr3.

Steps:
grams of sodium sulfate -> moles of sodium sulfate -> moles of Br2 -> grams of leftover Br2

Subtract this from the original amount to obtain grams of Br2 that reacted in the first reaction. Now that we have amount of grams that reacted for both vanadium and bromine, we can find the moles for each to obtain the e. formula.

Thanks again :)