Video Transcript
A 2.45-gram sample of white
crystals is known to contain NaCl and KNO3. The sample is fully dissolved in
deionized water and an excess of AgNO3 is then added, forming a precipitate of
AgCl. Upon being filtered, washed, and
dried, the precipitate is found to have a mass of 1.54 grams. To the nearest integer, what is the
percentage by mass of NaCl in the mixture? The molar mass of sodium is 23
grams per mole, chlorine is 35.5 grams per mole, and silver is 108 grams per
mole.
In this question, we are given a
sample that contains two different salts: NaCl and KNO3. Initially, we do not know the
amount of each salt in the sample. But when a precipitation gravimetry
experiment is conducted, a precipitate of AgCl forms. Precipitation gravimetry is an
analytical technique that uses the formation and mass of a precipitate to determine
the mass of an analyte.
Let’s begin by writing a chemical
equation to represent the precipitation reaction that takes place. First, solid sodium chloride and
potassium nitrate are dissolved in water to form an aqueous solution. So, we should write both of these
chemical formulas using the state symbol aq. Excess AgNO3, or silver nitrate, is
then added to the solution. Silver nitrate is the precipitating
agent, and it causes a chemical reaction to occur, producing a solid silver chloride
precipitate.
But what are the other
products? When sodium chloride reacts with
silver nitrate, solid silver chloride is produced. The sodium and nitrate ions remain
dissolved in the solution, so we can write the chemical formula of this product as
NaNO3 aqueous. Potassium nitrate does not react
with the silver nitrate. It remains in the solution as
dissolved ions. Since KNO3 does not participate in
the chemical reaction, let’s remove it from our equation.
In this experiment, NaCl is the
analyte. We know the mass of the silver
chloride precipitate is 1.54 grams. We need to use this information to
determine the mass of sodium chloride that reacted. Before we start our calculations,
let’s make sure that the chemical equation is balanced. After counting up each type of atom
on both sides of the equation, we can see that the equation is already balanced.
Now, we’re ready to carry out the
following steps to determine the percentage by mass of sodium chloride.
First, we’ll convert grams of
silver chloride to moles of silver chloride. Then, we’ll convert moles of silver
chloride to moles of sodium chloride. Next, we’ll convert moles of sodium
chloride to grams of sodium chloride. And finally, we will find the
percentage by mass of sodium chloride in the sample.
To convert grams of silver chloride
to moles of silver chloride, we need to make use of the following equation, where 𝑛
is the number of moles, lowercase 𝑚 is the mass in grams, and uppercase 𝑀 is the
molar mass in grams per mole. We need to calculate the molar mass
of AgCl by adding together the average molar masses of silver and chlorine provided
in the problem. Therefore, the molar mass of silver
chloride is 143.5 grams per mole.
Now we can substitute the mass of
silver chloride precipitate, which is 1.54 grams, and the molar mass of silver
chloride, which is 143.5 grams per mole, into our equation. After dividing, the result is
approximately 0.010 moles of silver chloride.
Now, looking back at the balanced
chemical equation, we can see that for every one mole of sodium chloride that
reacts, one mole of silver chloride is produced. Therefore, 0.010 moles of sodium
chloride must have reacted.
Now we can convert the number of
moles of NaCl to grams by making use of the same equation we used in step one. However, since we are trying to
solve for mass, let’s rearrange our equation.
First, we will need to calculate
the molar mass of sodium chloride by adding together the average molar masses of
sodium and chlorine provided in the problem. Therefore, the molar mass of sodium
chloride is 58.5 grams per mole. We can substitute approximately
0.010 moles for the number of moles and 58.5 grams per mole for the molar mass. After multiplying, we find that the
mass of sodium chloride is approximately 0.627 grams.
Finally, let’s calculate the
percentage by mass of sodium chloride in the original sample. We can use the following
equation. We need to divide approximately
0.627 grams of sodium chloride by the total mass of the original sample, which is
2.45 grams, and then multiply by 100 percent. The result is approximately 25.6246
percent. However, we need to round our
answer to the nearest integer. Therefore, the percentage by mass
of sodium chloride in the mixture is 26 percent.
