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.
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.
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.
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.
To identify the hydrides A and B and comment on their structures, we need to consider the percentage by mass of hydrogen in each hydride.
Let's assume the molar mass of hydride A is MA (g/mol) and the molar mass of hydride B is MB (g/mol).
Given that hydride A contains 17.76% by mass of hydrogen, the mass of hydrogen in hydride A can be calculated as follows:
Mass of hydrogen in hydride A = 17.76% × MA
Similarly, for hydride B:
Mass of hydrogen in hydride B = 12.58% × MB
We can compare the mass of hydrogen in each hydride to identify the possible elements.
Now, let's consider the atomic masses of hydrogen (H) and the element X. The atomic mass of hydrogen is approximately 1 g/mol.
If the mass of hydrogen in hydride A is greater than the mass of hydrogen in hydride B, it means that the molar mass of A is greater than the molar mass of B. This suggests that element X in hydride A has a higher atomic mass than in hydride B.
Comparing the masses of hydrogen in hydrides A and B will help us identify the element X and the structures of hydrides A and B.
Based on these calculations, we can determine the identity of A and B and comment on their structures.