Two hydrides A and B of an element X contain 17.76% and 12.58% by mass of hydrogen. Identify A and B and comment on their structures.

Answer 1

Hydrogen has the lowest atomic mass. That would make the % of hydrogen in most hydrides very low since the H atoms would be linked to much heavier atoms. The two percents given are actually high indicating that the hydrides are linked to a relatively small atom, possibly from the 2nd row of the Periodic Table: LiH, BeH2, B2H6, CH4, NH3, H2O, HF. The % of hydrogen in one of these hydrides is very close to 17.76%. Find out which one. Example:
For BeH2,
Be/BeH2 = (1.008)(2)/11.03 = 0.1828 = 18.28% H.
The element we want is not Be. Try another one of the other hydrides.
Once you know which element forms a hydride with 17.76% H, look up its atomic mass. Lets call it x for now.
For the first compound:
100-17.76 = 82.24% X
Divide 82.24% by x to get m
Divide 17.76 by 1.008 = 17.62
Divide m and 17.62 by the smaller of the two. One ration is 1 and the other a small whole number. Those are your subscripts.
For the 2nd compound:
100-12.58= 87.42% X
Divide 87.42% by x to get n
Divide both, n and 12.58 by the smaller of the two. You should get 1 and another small whole number which are your subscripts for the 2nd compound.
This is a challenging percent composition problem. The challenge is for you.

Answer 2

Correction:
"Be/BeH2 = (1.008)(2)/11.03 = 0.1828 = 18.28% H."
should read:
2H/BeH2 = (1.008)(2)/11.03 = 0.1828 = 18.28% H.

Answer 3

I have worked out that the element is nitrogen and the first hydride is NH3. I calculated the second hydride to have the empirical formula NH2. Would the molecular formula be N2H6? I'm not too sure about the structures.