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Chapter 3: Problem 98

A sample of the compound MSO \(_{4}\) weighing \(0.1131 \mathrm{g}\) reacts with barium chloride and yields \(0.2193 \mathrm{g} \mathrm{BaSO}_{4}\) What must be the atomic mass of the metal M? [Hint: All the \(\mathrm{SO}_{4}^{2-}\) from the MSO \(_{4}\) appears in the \(\mathrm{BaSO}_{4}\) ]

Short Answer

Expert verified
The atomic mass of M is \(24.4 g/mol\).

Step by step solution

01

Calculate Mass of \(\mathrm{SO}_{4}^{2-}\) in \(\mathrm{BaSO}_{4}\)

To find the mass of \(\mathrm{SO}_{4}^{2-}\), the molecular mass of \(\mathrm{BaSO}_{4}\), which is 233.39 g/mol, is used. Of that, 96.06 g/mol is attributable to \(\mathrm{SO}_{4}^{2-}\). By calculating the percentage, it is found that \(\mathrm{SO}_{4}^{2-}\) accounts for \(96.06/233.39 = 0.4116 = 41.16\%\) of the mass of \(\mathrm{BaSO}_{4}\). Therefore, the mass of \(\mathrm{SO}_{4}^{2-}\) in 0.2193g of \(\mathrm{BaSO}_{4}\) is \(0.2193g * 0.4116 = 0.0902g\).
02

Determine the Mass of M

Since the mass of \(\mathrm{SO}_{4}^{2-}\) in MSO4 equals the mass of \(\mathrm{SO}_{4}^{2-}\) in \(\mathrm{BaSO}_{4}\), the mass of the metal M in MSO4 is simply the total mass of MSO4 minus the mass of \(\mathrm{SO}_{4}^{2-}\). Therefore, the mass of M is \(0.1131g - 0.0902g = 0.0229g\).
03

Calculate the Atomic Mass of M

The atomic mass of M will be the mass of M divided by the number of moles of M. The number of moles of \(\mathrm{SO}_{4}^{2-}\) can be calculated from the mass of \(\mathrm{SO}_{4}^{2-}\) using its molar mass \(96.06g/mol\), thereby giving \(0.0902g/96.06g/mol = 0.00094mol\). Hence, the atomic mass of M equals \(0.0229g / 0.00094mol = 24.4g/mol\).

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