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Chapter 3: Problem 101

\(\mathrm{MgCl}_{2}\) often occurs in table salt \((\mathrm{NaCl})\) and is responsible for caking of the salt. A 0.5200 g sample of table salt is found to contain \(61.10 \%\) Cl, by mass. What is the \(\% \mathrm{MgCl}_{2}\) in the sample? Why is the precision of this calculation so poor?

Short Answer

Expert verified
The percent of \( \mathrm{MgCl}_{2} \) in the sample is calculated to be 82.5%. The precision of this calculation is poor due to several reasons including averages used for atomic weights, possible alternate sources of Chlorine, and measurement errors.

Step by step solution

01

Calculate mass of chlorine

First, calculate the mass of Chlorine (Cl) in the sample based on the given percentage. As it's given that the Cl content is \(61.10\% \), the weight of Cl in the sample would be \(0.6110 \times 0.5200 \, g = 0.3177 \, g \)
02

Calculate mass of \( \mathrm{MgCl}_{2} \)

Now, calculate the mass of \( \mathrm{MgCl}_{2} \) containing this amount of Cl. In \( \mathrm{MgCl}_{2} \), Cl makes up \( (35.45 \times 2)/(24.31 + 35.45 \times 2) \times 100\% = 74.08\% \) of the mass. Hence, the mass of \( \mathrm{MgCl}_{2} \) is \(0.3177 \, g / 0.7408 = 0.429 \, g \).
03

Calculate percent of \( \mathrm{MgCl}_{2} \)

Next, calculate the percent of \( \mathrm{MgCl}_{2} \) in the total sample, which will be \( (0.429 \, g / 0.5200 \, g) \times 100\% = 82.5\% \)
04

Analyze precision

Finally, analyze the precision of the calculation. The percentage of \( \mathrm{MgCl}_{2} \) found is more than 100%, indicating an error. This could be due to several reasons, including: The atomic weights used are averages, not exact values; Chlorine could come from sources other than \( \mathrm{MgCl}_{2} \), and errors in measuring the mass of the Cl and salt. These reasons contribute to the poor precision of this calculation.

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