what size sample which contains 12% chlorine should be taken for analysis if the chemist wishes to obtain a precipitate of silver chloride which weigh 0.50 gram

To determine the size of the sample containing 12% chlorine needed to obtain 0.50 grams of silver chloride (AgCl), we can start by understanding the relationship between chlorine and silver chloride. The key here is to use stoichiometry, which allows us to relate the mass of chlorine in the sample to the mass of silver chloride produced.

Understanding the Chemical Reaction

The formation of silver chloride can be represented by the following reaction:

Ag⁺ + Cl^- → AgCl

This means that one mole of silver ions reacts with one mole of chloride ions to produce one mole of silver chloride. To find out how much chlorine is needed, we first need to calculate the moles of silver chloride that correspond to 0.50 grams.

Calculating Moles of Silver Chloride

The molar mass of silver chloride (AgCl) can be calculated as follows:

• Molar mass of Ag = 107.87 g/mol
• Molar mass of Cl = 35.45 g/mol
• Total molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol

Now, we can calculate the number of moles of silver chloride in 0.50 grams:

Number of moles of AgCl = mass / molar mass

Number of moles of AgCl = 0.50 g / 143.32 g/mol ≈ 0.00349 moles

Relating Chlorine to Silver Chloride

Since one mole of AgCl requires one mole of Cl^-, we also need 0.00349 moles of chloride ions. Next, we can find the mass of chlorine required:

The molar mass of chlorine (Cl) is 35.45 g/mol, so:

Mass of Cl = moles × molar mass

Mass of Cl = 0.00349 moles × 35.45 g/mol ≈ 0.123 g

Determining the Sample Size

Now that we know we need approximately 0.123 grams of chlorine, we can calculate the total mass of the sample needed, given that the sample contains 12% chlorine by weight.

If the sample contains 12% chlorine, we can express this as:

Mass of Cl = 0.12 × mass of sample

Setting this equal to the mass of chlorine we calculated:

0.123 g = 0.12 × mass of sample

To find the mass of the sample:

mass of sample = 0.123 g / 0.12 ≈ 1.025 g

Final Result

Therefore, to obtain 0.50 grams of silver chloride, the chemist should take a sample weighing approximately 1.025 grams that contains 12% chlorine. This ensures that there is enough chloride available to react with silver ions to produce the desired amount of precipitate.