Question Video: Using Precipitation Gravimetry to Calculate the Percentage Mass of NaCl in a Mixed Sample | Nagwa Question Video: Using Precipitation Gravimetry to Calculate the Percentage Mass of NaCl in a Mixed Sample | Nagwa

# Question Video: Using Precipitation Gravimetry to Calculate the Percentage Mass of NaCl in a Mixed Sample Chemistry • Third Year of Secondary School

A 2.45 g 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 g. To the nearest integer, what is the percentage by mass of NaCl in the mixture? [Na = 23 g/mol, Cl = 35.5 g/mol, Ag = 108 g/mol]

05:56

### 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.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions