[sanado]

Hey guys, was just hoping you could explain this question to me.

5.00g of nitrogen is completely converted into an oxide of nitrogen. The mass of oxide formed is 19.3g. The empirical formula of the oxide would be?

Okay, so i got no issues with solving the problem. Convert mass of nitrogen into number of mole, find mass of oxygen by subtracting mass of nitrogen from oxygen. From there, find the empirical formula. Only question i found my self asking was at the start, do i assume that the nitrogen is in the form of N or in N2. If it is in N2, the answer becomes completely different. So my question is, do i assume that an element in gas form is being used only if stated?

[ARGOS++]

Dear Sanado;

In Chemistry, if not explicitly noted, all Compounds/Elements are “handled” in their most stabile “Form”/State by default.


Good Luck!
                    ARGOS⁺⁺

[Borek]

Show how you are getting different answers depending on whether it is N or N₂. I am getting the same.

It should be N₂, but it doesn't matter.

[DrCMS]

I agree with Borek.

If you start with 5g of nitrogen as N₂ or N and react it with oxygen to get 19.3g N_xO_y there is still only 5g of N and 14.3g O in the product.

[sanado]

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH 

I see now, even if i use N2 gas or O2 gas, i will still get the same answer. This is because when i found how many mole of N2 i have, i would then double it cause i have twice as many nitrogen. Cheerz guys!!!