Question

question 8 (3 points) calculate the percentage of chlorine in 41.7 g of disulfur dichloride (s₂cl₂), which contains 21.9 g of sulfur. verify the formula. do not type anything, please upload your file below

Explanation:

Step1: Find the mass of chlorine

The mass of disulfur dichloride is 41.7 g and the mass of sulfur is 21.9 g. So the mass of chlorine $m_{Cl}=41.7 - 21.9=19.8$ g.

Step2: Calculate the percentage of chlorine

The percentage of chlorine $=\frac{m_{Cl}}{m_{total}}\times100\%=\frac{19.8}{41.7}\times 100\%$.
\[

$$\begin{align*} \frac{19.8}{41.7}\times 100\%&=\frac{1980}{41.7}\%\\ &\approx 47.5\% \end{align*}$$

\]

Step3: Verify the formula (using molar - mass method)

The molar mass of $S_2Cl_2$: The molar mass of $S$ is $M_S = 32.07$ g/mol and the molar mass of $Cl$ is $M_{Cl}=35.45$ g/mol. The molar mass of $S_2Cl_2$ is $M = 2\times32.07+2\times35.45=64.14 + 70.9=135.04$ g/mol.
The mass - percentage of $Cl$ in $S_2Cl_2$ from the formula is $\frac{2\times35.45}{135.04}\times 100\%=\frac{70.9}{135.04}\times 100\%\approx 52.5\%$. There is a difference likely due to experimental error in the given masses.

Answer:

The percentage of chlorine is approximately 47.5%