22.74 g = xble, cover, salt
-19.74 g = xble, cover
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...xxxxxx = weight of the hydrated salt.
The rest of it isn't that easy because the data boggles my mind.
First weighing ok. Second weighing--I don't see how it could GAIN weight.
Third weighing--It went down, which is good, but since it changed so much you should have heated it again. The idea is to heat to constant weight.
I would calculate the weight of the anhydrous salt this way; it's the best data you have.
21.54 g xble, cover, anhyd salt
-19.74 = xble, cover.
---------
xxxxxxx = weight anhydrous salt
I did a chem lab and have to figure out some stuff off of some data I found during the lab
Weight of empty crucible and cover: 19.74 g
Weight of crucible, cover and hydrated salt: 22.74 g
Weight of crucible, cover and hydrated salt after 1st heating: 22.13 g
Weight of crucible, cover and hydrated salt after 2nd heating: 22.56 g
Weight of crucible, cover and hydrated salt after 3rd heating: 21.54 g
Calculate the weight of the hydrated salt.
Calculate the weight of the anhydrous salt.
2 answers
Calculate the weight of the water lost by heating.
These are the other post lab questions
Calculate the percentage of water in the hydrate.
Calculate the number of moles of H2O lost by heating.
Calculate the number of moles of anhydrous salt.
Enter the formula of the hydrated salt.(how do i find this)
Calculate the theoretical percentage of the water in the hydrate.
Calculate the experimental error.
Calculate the percentage error.
These are the other post lab questions
Calculate the percentage of water in the hydrate.
Calculate the number of moles of H2O lost by heating.
Calculate the number of moles of anhydrous salt.
Enter the formula of the hydrated salt.(how do i find this)
Calculate the theoretical percentage of the water in the hydrate.
Calculate the experimental error.
Calculate the percentage error.
3 answers