AS Chemistry need help

Hey
Need help with a few questions to do with hydrated salts
formula is XSO4.5H20
the mean amount of H20 in mol that was removed after the samples were heated is the mass/18 in my case 0.84/18 = 0.046666666?
So the mean amount (in mol) of the anhydrous salt is the mol of the H20 divided by 5? because of the ratio 1:5
this means the mol of the anhydrous salt is 0.00933333333
If thats so the molar mass of the anhydrous salt is the mass/mol so 1.21/0.00933333? which is roughly 130?
so the molar mass of X should be 130-96 (96 is the molar mass of the SO4
I get all this but what confuses me is when doing the molar mass of the water, why is the coefficient not included? How do you know when to include the coefficient and when not to? because you dont include it then but then you do when doing the ratio?
could someone explain it to me and tell me whether the steps I did were correct? thanks
Also whats the mean % by mass of H20 in the hydrated salt?
is it the (mass of the H20/mass of hydrated salt)*100 or is it the molar mass of water divided by the molar mass of the entire formula? if the second how would this work because I don't know the molar mass of X

Reply 1

LordPhylogeny

Original post by 76584

The steps were all correct you just need to use the coefficient for the molar mass because when working out the moles of a substance, the number of atoms within a specific molar value is irrespective of how many molecules of that substance are involved. In the ratio however this takes in to account how many molecules can bond to the anhydrous salt and therefore gives you your ratio.

The percentage by mass is exactly as you say, the percentage of the overall molar mass is taken up by that of water.

Reply 2

76584

Original post by LordPhylogeny

The steps were all correct you just need to use the coefficient for the molar mass because when working out the moles of a substance, the number of atoms within a specific molar value is irrespective of how many molecules of that substan+ce are involved. In the ratio however this takes in to account how many molecules can bond to the anhydrous salt and therefore gives you your ratio.

The percentage by mass is exactly as you say, the percentage of the overall molar mass is taken up by that of water.

thanks so much, that really cleared up a lot
But I have 2 more questions
, in mine when I deduced the identity of X i did the molar mass of SO4 takeaway the molar mass of the entire anhydrous salt so I did 1.21(mass of anhydrous salt) divided by 0.0093333(mole of anhydrous salt) which gave me 129. When I took the molar mass of SO4 from 129 it gave me the molar mass of X however the element I got from this was not the correct one according to my teacher? Where did I go wrong then because I'm pretty sure the mass's were correct?
and 2nd could you explain this "The percentage by mass is exactly as you say, the percentage of the overall molar mass is taken up by that of water."
that confused me, lol I get confused easily.

Reply 3

LordPhylogeny

I will start with the percentage one cause that is the simplist so for example if XSO4.5H20 was a thing then you would have to work out the combined molar mass of the water. This value as a percentage of the entire molecules molar mass would be the answer to your question.

For the first question however I have to ask, were these experimental results or theoretical results?

Reply 4

76584

Original post by LordPhylogeny

nah I worked out the mass's in class through an experiment.
Does it mean my results were just inaccurate then?

so for the mean % by mass of H20 would I do 18/114.1? Or would I include the 5 infront of the H20? Also, how would I do the molar mass of X? since that is an unknown element
would I be able to do the mass of H20 that I worked out divided by the mass of the hydrated salt that I worked out? or would that not work?
to clarify by mass i mean in grams. I worked it out by heating the hydrated salt

Reply 5

LordPhylogeny

Original post by 76584

I would suggest that your results inaccuracys were down to error in the experiment yes.

The former of your calculations 18/114.1 is correct and the percentage by mass can only be done with a known element the X was just there cause I couldn't think of an element to put there.

Reply 6

snikutsmullac

chemistry, why u no easy?

Reply 7

76584

Original post by LordPhylogeny

For my question in class "calculate the mean % by mass of H20 in the hydrated salt" and all we were given was XSO4.5H20
Firstly, why dont I include the 5? doesnt that contribute to molar mass?
and 2nd does that mean I would have to skip the question and come back to it after figuring out what x was? Or would I be able to use the mass(g) instead of molar mass(gmol-1)? so the mass of the H20 in my experiment divided by the mass I got for the entire formula?

Reply 8

LordPhylogeny

Skip the question come back to it when you know what the molar mass of X is.

You do include the five in the calculation of percnetage mass

Reply 9

76584

Original post by LordPhylogeny

Skip the question come back to it when you know what the molar mass of X is.

You do include the five in the calculation of percnetage mass

Awesome,
sorry for all the questions, promise this is the last
You said include the 5 in percentage mass and you said 18/114.1 is right but i didnt include the 5 in that. By that I mean i didnt times 18 by 5.