MA.nH2O is a hydrated salt with unknown moles of water, a student tried to determine the
number of moles of water in this salt, so he took 1.023 grams of the hydrate, heated it in the oven
for 15 min, cooled and weighed it, he found that the hydrate lose 0.247 grams of its mass, he put
it again in the oven for extra 10 min, cooled and weighed again, he found that the hydrate lose
another 0.122 grams of its mass.
If you knew that the anhydrous salt has a molar mass of 159.55 g/mol, how many moles of water
are in one mole of this hydrate?
MA.nH2O is a hydrated salt with unknown moles of water, a student tried to determine the?
2020-05-02 8:14 pm
回答 (2)
2020-05-02 9:21 pm
Suppose that all the water of crystallization is removed after heating for extra 10 min.
Molar mass of H₂O = (1.008×2 + 16.00) g/mol = 18.016 g/mol
Mass of MA•nH₂O = 1.023 g
Mass of H₂O removed = (0.247 + 0.122) g = 0.369 g
Mass of MA left = (1.023 - 0.369) g = 0.654 g
Molar ratio MA : H₂O = 1 : n
(0.654/159.55) : (0.369/18.016) = 1 : n
(0.654/159.55)n = 0.369/18.016
n = (0.369/18.016) × (159.55/0.654)
n = 5
There are 5 mole of water in one mole of this hydrate.
Molar mass of H₂O = (1.008×2 + 16.00) g/mol = 18.016 g/mol
Mass of MA•nH₂O = 1.023 g
Mass of H₂O removed = (0.247 + 0.122) g = 0.369 g
Mass of MA left = (1.023 - 0.369) g = 0.654 g
Molar ratio MA : H₂O = 1 : n
(0.654/159.55) : (0.369/18.016) = 1 : n
(0.654/159.55)n = 0.369/18.016
n = (0.369/18.016) × (159.55/0.654)
n = 5
There are 5 mole of water in one mole of this hydrate.
2020-05-02 9:16 pm
grams of H2O = 0.247 g + 0.122 g = 0.369 g
moles of H2O = 0.369 g H2O x 1mole H2O/18.0 g H2O = 0.0205 moles H2O/ 1mole anhydrous salt
Mass of anhydrous salt:
1.023 g - 0.247 g - 0.122 g = 0.654 g anhydrous salt
0.654 g anhydrous salt x 1mole anhydrous salt 159.55 g = 0.004099 moles anhydrous salt
moles of H2O/1mole anhydrous salt :
0.0205 / 0.004099 = 5.00 moles of H2O
moles of H2O = 0.369 g H2O x 1mole H2O/18.0 g H2O = 0.0205 moles H2O/ 1mole anhydrous salt
Mass of anhydrous salt:
1.023 g - 0.247 g - 0.122 g = 0.654 g anhydrous salt
0.654 g anhydrous salt x 1mole anhydrous salt 159.55 g = 0.004099 moles anhydrous salt
moles of H2O/1mole anhydrous salt :
0.0205 / 0.004099 = 5.00 moles of H2O
收錄日期: 2021-05-01 09:37:38
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