molecule • 5H2O
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molecule • 5H2O
How do I approach a molarity/solution problem when the solute is specified as “molecule•#H2O”? (there was an instance in the Week 1 textbook problems)
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Re: molecule • 5H2O
This might seem like a product, but you just add the molar mass of the molecule by the molar mass of the water. That expression represents water molecules bound to a molecule.
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Re: molecule • 5H2O
Hi!
For these kinds of problems, all of the calculations are the same except you need to include the mass of the hydrate (the nH2O part). You would take the molar mass of the first part of the compound + 18.01(n) [this represents the molar mass of the hydrating part] and use that as your total molar mass in your stoichiometry. Hope that helps!
For these kinds of problems, all of the calculations are the same except you need to include the mass of the hydrate (the nH2O part). You would take the molar mass of the first part of the compound + 18.01(n) [this represents the molar mass of the hydrating part] and use that as your total molar mass in your stoichiometry. Hope that helps!
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Re: molecule • 5H2O
Just to clarify, does this mean that we should treat the mass of the molecule throughout the problem as being its mass plus the mass of the nH20?
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Re: molecule • 5H2O
Theo Teske 3B wrote:Just to clarify, does this mean that we should treat the mass of the molecule throughout the problem as being its mass plus the mass of the nH20?
yup! it would technically be considered a "joint molecule" when solving the equation. For instance, magnesium sulfate heptahydrate would include the molar mass of MgS04 and 7*18.01 for the hydrate.
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Re: molecule • 5H2O
You include the mass of the H2O (times whatever coefficient) in the molar mass of the entire molecule because it is the hydrated form of the molecule. You can look up the hydrated forms such as CuSO4 • 5H2O and you'll see that it is one substance and also looks different from just CuSO4.
Re: molecule • 5H2O
All you need to do is incude the mass of water with the mass of the molecule given! This will account for that compound.
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