Calculating the Formulas of Hydrated Salts

Calculating the Formulas of Hydrated Salts


Hey it’s professor Dave, let’s learn how
to calculate the formula of a hydrated salt. We commonly refer to sodium chloride as salt,
but this isn’t the only salt there is. Salt is actually just another term for an
ionic compound, so magnesium bromide, calcium fluoride, any compound consisting of formally
charged ions is called a salt, not just the one that we know of as table salt. And we also know that some salts exist as
hydrates, where some number of water molecules are actually incorporated into the lattice
structure of each formula unit in the solid. But let’s say we had some hydrate of a known
salt, and we were unsure of how many water molecules were involved in this hydrate. How might we find out? It’s actually quite simple, it just involves
an experiment and some simple calculations. For example, let’s say we have 15.67 grams
of a magnesium carbonate hydrate. We know that in this substance there is some
magnesium carbonate, and there is some water. It should be easy to know what mass there
is of each component. All we have to do is get rid of the water,
which is easy to do with gentle heating. The heat will cause the water molecules to
gain kinetic energy until they leave the lattice as water vapor. Once the salt is completely dry, or anhydrous,
meaning no more water is left, we measure the mass again, and this time we get 7.58
grams. That means that the dry magnesium carbonate
itself must have this mass, and the difference between this number and the original mass,
or 8.09 grams, must be the mass of the water that was vaporized. Now that we have the mass of each component,
let’s convert these values into moles. 7.58 grams of magnesium carbonate, divided
by the molar mass will give us around 0.09 moles of the salt, which still remains. And 8.09 grams of water, divided by its molar
mass, will give us around 0.45 moles of water molecules, which used to be inside the salt
but are now up in the air. Dividing both of these values by the smaller
of the two will give us their ratio, which we can clearly see is 1 to 5. That means that for every formula unit of
the magnesium carbonate in the lattice, there must have been precisely five water molecules. That will allow us to write the complete formula
for this hydrated salt, because where we may have had to include an X before, indicating
that we did not know the number of water molecules associated with one formula unit of this substance,
we can now accurately place a five here as the coefficient for water, because we have
empirically determined that there must be five water molecules for every formula unit
of the ionic solid. We can now use this information to infer certain
things about the geometry of this compound, or any number of other things. This is yet another powerful example of how
we can do simple, quantitative calculations to indirectly ascertain specific facts about
molecular structure that we could never see with our own eyes, such as the formula of
a hydrated salt. Let’s check comprehension.

24 thoughts on “Calculating the Formulas of Hydrated Salts

  1. how did scientists in the early 20th century accurately predict molecular structure. ie glucouse' s structure and so on

  2. Thanks for the video. I have a little question: Why do we divide by the smaller number of moles? Simple logic plz.

  3. How to find equivalent weight of salts ,reducing agent, oxidizing agent
    And ,how to identify salts,oxidizing agents, reducing agent
    Pls teach me sir

  4. Thank you, Professor Dave! You are wonderful at explaining! Please guys, help support Professor Dave on Patreon!

  5. My boy Professor Dave coming to the rescue to an AP Chem teacher that doesn't know how to teach. Your explanation is amazing, it would be almost impossible for people to not understand.

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