Here is the question: If, after heating, the solid has a molar mass of 201 g/mol and a formula of XY, what is the formula of the hydrate? Use Lab Group 1's results.
Here are the group's results:
mass before heating = 1.48g
mass after heating = 1.26g
Please help me understand how to work this.
Here are the group's results:
mass before heating = 1.48g
mass after heating = 1.26g
Please help me understand how to work this.
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1.48 - 1.26 = 0.22 g
0.22 g / 18.015 g/mol = 0.012212 mol (of water in the hydrate)
1.26 g / 201 g/mol = 0.0062686 mol (of the anhydrate)
What we want to know is the smallest whole-number ratio of water to anhydrate:
0.012212 mol / 0.0062686 mol = 1.948
Within experimental error, 1.948 equals 2 waters of hydration:
XY · 2H2O
0.22 g / 18.015 g/mol = 0.012212 mol (of water in the hydrate)
1.26 g / 201 g/mol = 0.0062686 mol (of the anhydrate)
What we want to know is the smallest whole-number ratio of water to anhydrate:
0.012212 mol / 0.0062686 mol = 1.948
Within experimental error, 1.948 equals 2 waters of hydration:
XY · 2H2O