Video Transcript
An aqueous solution contains an
unknown mass of MgSO4. Upon adding excess BaCl2, a
precipitate of BaSO4 is formed. After being filtered, washed, and
dried, the precipitate contains 0.012 moles of BaSO4. What mass, in grams, of MgSO4 is
present in the original solution? Take the molar mass of MgSO4 to be
120 grams per mole.
In this question, we are asked to
determine the total mass of magnesium sulfate present in a solution if 0.012 moles
of barium sulfate precipitate is formed during a reaction. This is an example of precipitation
gravimetry, which is an analytical technique that uses the formation and mass of a
precipitate to determine the mass of an analyte. Let’s get started by writing a
chemical equation to represent the reaction described in the question.
We are told that excess barium
chloride, or BaCl2 aqueous, is added to a solution of magnesium sulfate, or MgSO4
aqueous. One of the products formed is the
barium sulfate precipitate, or BaSO4 solid. The ions that remain in the
solution after the reaction has taken place are magnesium ions and chloride
ions. We can show this by writing the
chemical formula MgCl2 aqueous.
Magnesium sulfate, or MgSO4, is the
analyte. We want to know how much magnesium
sulfate, in grams, was dissolved in the magnesium sulfate solution before the
precipitation reaction occurred. We know that the precipitate
contains 0.012 moles of BaSO4. If we balance our chemical
equation, we can determine the molar ratio of magnesium sulfate to barium
sulfate. Then we can use the molar ratio to
convert from moles of barium sulfate to moles of magnesium sulfate. Lastly, we can convert moles of
magnesium sulfate to grams of magnesium sulfate.
When looking at the chemical
equation, we can see the magnesium atoms, the sulfur atoms, the oxygen atoms, the
barium atoms, and the chlorine atoms are all already balanced. Therefore, we have a balanced
chemical equation. The molar ratio of MgSO4 to BaSO4
is, therefore, one to one. Now, let’s convert moles of BaSO4
to moles of MgSO4 using the molar ratio. Since the molar ratio is one to
one, this means that for every 0.012 moles of BaSO4 produced, 0.012 moles of MgSO4
reacted.
Finally, we’re ready to convert
0.012 moles of MgSO4 to grams. We can make use of the following
equation, where 𝑛 is the number of moles of magnesium sulfate, lowercase 𝑚 is the
mass in grams of magnesium sulfate, and uppercase 𝑀 is the molar mass of magnesium
sulfate in grams per mole. Let’s start by rearranging our
equation to solve for mass.
We get mass equals number of moles
multiplied by molar mass. We can substitute 0.012 moles for
the number of moles and 120 grams per mole for molar mass, which was provided in the
problem. After multiplying, the result is
1.44 grams.
In conclusion, the mass in grams of
MgSO4 that is present in the original solution is 1.44 grams.
