Imperical Formula Question (1 Viewer)

[HQ_BUDDY]

Ok i have my chemistry topic test tmrw, i am in yr 11 at the moment and the teacher gave us one of the questions to try and wrk out at home then do in the exam.

Ok heres the question

A test tueb was washed, dryed and then weighed. A small quantity of mercury oxide was added then reweighed. The test tube was then heated strongly until the mercury oxide decomposed and only mercury was left. The test tube was left to cool, Then reweighed.

The original weight of test tube = 28.93g
Weight of Test tube + Mercury Oxide = 40.17g
Test tube + Mercury = 39.34g

Determine the imperical formula of Mercury Oxide?


ok our teacher did a bit of it in class.

Mercury Oxide = 11.24g
Mercury = 0.83g

Thats all we got up to. Can someone please explain to me how i can do this and also the correct answer, thx in advance guys, hope u can help.

FiFi

P.s. Sorry about the long post

 

[Will Hunting]

Hey man,

Can someone please explain to me how i can do this

First up, write a neutral species equation for this reaction. This will give you the theoretical formula for mercury oxide. Assuming the use of Hg(II), it should flow like this:

2HgO(s) --(heat)--> 2Hg(s) + O2(g) ... [1]

This tells you that the theoretical formula is HgO. You're looking for the empirical formula, though, meaning, in this context, that it's determined experimentally. Here's how it's broken down...

Mass of HgO = 11.24g
Mass of Hg - 10.41g
Hence, mass of O released = 11.24 - 10.41 = 0.83g

(figures assuming zero impurities)

(Seems like your teacher mixed up the masses of 2Hg and O2. You've gotta watch out for that. If you measured the mass of a tt with HgO, then that same tt with only Hg, the difference in mass = the mass of O released, not the mass of Hg left)

From this data, you can calculate the moles of each of the species involved in this reaction. The ratio of moles of Hg to moles of O will give you the empirical formula for HgO. Here's how it's done...

n = m/mm

For Hg:

n = 10.41/(2 x 200.6) = 2.59 x 10^-2 mol

For O:

n = 0.83/(2 x 16.0) = 2.59 x 10^-2 mol

Now, moles of Hg = moles of O = 2.59 x 10^-2 (all figures approximate)

From this, you have shown that Hg(II) and O combine (and decompose) in the ratio 1:1.

and also the correct answer

Hence, the empirical formula for Mercury Oxide is HgO (as predicted by [1])

Hope this helps, bro

 

[HQ_BUDDY]

thx mate ill have to read that a few times to understand it, we onyl just finished module one i dont know wat a mole is i guess ill have to look it up or something
thx again amte u have saved me a few marks