A question about gravimetric analysis

[chem_tr]

Hello, I have been studying undergraduate level chemistry courses for PhD proficiency exam, and I have very recently been started Analytical Chemistry.

I have a simple problem and I solved it in a different way, could you please show me how I can solve it by using the known "gravimetric factor"? Or I maybe have used it without noticing

The question is as follows:

An organic sample, weighing 0.2795 grams and known to include $\displaystyle C_6H_6Cl_6~(290.83~g/mol)$ and $\displaystyle C_{14}H_9Cl_5~(354.49~g/mol)$, is ignited with oxygen gas in a quartz tube, and products obtained (CO₂, H₂, and HCl) are passed from a NaHCO₃ solution. After acidification, the total chloride ions are found to precipitate 0.7161 grams of AgCl (143.22 g/mol). Calculate the halogen percentage of each chlorinated organic compounds.

Okay, my solution goes like this:

290.83x+354.49y=0.2795 g
6*35.5x+5*35.5y=0.005*35.5 g (0.005 is the number of moles of AgCl)
from there, rearrangement gives the following:

290.83x+354.49y=0.2795 g
-425.388x-354.49y=-0.35448 g

adding and dividing reveals that x=0.00055723 and from the second (or first) equation, y is found to be 0.0003313.

The gram equivalents of these x and y are 0.1620592 grams and 0.11744254 grams, respectively.

Their percentages are 57.98% and 42.02%, respectively.

The textbook says that the correct answers are 57.80% and 42.20%, resp.

My questions:

1) What is the cause I am getting different answers? I am suspecting from using 35.5 grams/mol for chlorine atom; maybe I have to find more precise number from elsewhere; it is not given to me in this question.

2) Can you please show the solution of this problem by using gravimetric factor, along with its description? I have been unable to comprehend it so far.

Thank you for your interest.

 

[GCT]

I'm not too familiar with this version of gravimetric analysis. Your answer does not seem too far off from the text's, so I'm not quite sure...perhaps the small deviation is because you neglected to use the stoichiometric ratio of molecule/chlorine. In the first organic compound it seems to be 1/6 and the second 1/5. Thus you're second equation may need to appear as 6*35.5(6moles/1mole)x+5*35.5(5moles/1mole)y=0.005*35.5 g (0.005 is the number of moles of AgCl).

 

[ravenprp]

Dear student,

Thank you for your question about gravimetric analysis. It seems like you have approached the problem correctly, but there may be some small errors in your calculations. Let's go through the problem step by step and see where the discrepancy may be coming from.

First, we need to determine the moles of each compound present in the sample. We can do this by dividing the mass of each compound by its molar mass. This gives us:

\displaystyle n_{C_6H_6Cl_6} = \frac{0.2795~g}{290.83~g/mol} = 0.000962 moles

\displaystyle n_{C_{14}H_9Cl_5} = \frac{0.2795~g}{354.49~g/mol} = 0.000788 moles

Next, we need to determine the moles of chlorine present in each compound. We can do this by multiplying the number of moles of each compound by the number of chlorine atoms in each compound. This gives us:

\displaystyle n_{Cl~in~C_6H_6Cl_6} = 0.000962~moles \times 6~atoms = 0.005772~moles

\displaystyle n_{Cl~in~C_{14}H_9Cl_5} = 0.000788~moles \times 5~atoms = 0.003940~moles

Now, we can use the gravimetric factor to calculate the mass of AgCl that should be formed from each compound. The gravimetric factor is the mass of AgCl that is formed from 1 mole of chlorine. In this case, it is 143.22~g/mol. Therefore, we can calculate the expected mass of AgCl from each compound as follows:

\displaystyle m_{AgCl~from~C_6H_6Cl_6} = 0.005772~moles \times 143.22~g/mol = 0.8269~g

\displaystyle m_{AgCl~from~C_{14}H_9Cl_5} = 0.003940~moles \times 143.22~g/mol = 0.5642~g

Now, we can compare this to the actual mass of AgCl that was precipitated