Question

What is the mass percent of sulfur in the \(\left[\mathrm{Ag}(\mathrm{SCN})_{2}^{-}\right]\) complex ion?

Short answer

Answer: The mass percent of sulfur in the \(\left[\mathrm{Ag}(\mathrm{SCN})_{2}^{-}\right]\) complex ion is approximately \(27.39\%\).

Step-by-step solution

Identify the elements and their quantities in the complex ion

In the complex ion, we have: 1 silver(Ag) atom 2 sulfur(S) atoms 4 carbon(C) atoms 2 nitrogen(N) atoms

Find the molar mass of the complex ion

To find the molar mass of the entire complex ion, sum the product of the molar mass of each element and their quantities: Molar mass of Ag = 107.87 g/mol Molar mass of S = 32.07 g/mol Molar mass of C = 12.01 g/mol Molar mass of N = 14.01 g/mol Molar mass of the complex ion: \(107.87 + 2 * 32.07 + 4 * 12.01 + 2 * 14.01 = 234.05 \, g/mol\)

Find the molar mass of sulfur in the complex ion

There are 2 sulfur atoms in the complex ion, so the molar mass of sulfur in the complex ion is: \(2 * 32.07 = 64.14 \, g/mol\)

Calculate the mass percent of sulfur

Divide the molar mass of sulfur by the molar mass of the entire complex ion and multiply by 100% to find the mass percent of sulfur: Mass percent of sulfur: \(\dfrac{64.14}{234.05} * 100\% = 27.39\%\) The mass percent of sulfur in the \(\left[\mathrm{Ag}(\mathrm{SCN})_{2}^{-}\right]\) complex ion is approximately \(27.39\%\).