Questions: A sample of a substance with the empirical formula XBr2 weighs 0.4895 g. When it is dissolved in water and all its bromine is converted to insoluble AgBr by addition of an excess of silver nitrate, - XBr2 + 2 AgNO3 -> 2 AgBr + X(NO3)2 (a) Calculate the formula mass of XBr2. Formula mass XBr2= g mol^-1 (b) Calculate the atomic mass of X. Atomic mass X= g mol^-1

A sample of a substance with the empirical formula XBr2 weighs 0.4895 g. When it is dissolved in water and all its bromine is converted to insoluble AgBr by addition of an excess of silver nitrate,
- XBr2 + 2 AgNO3 -> 2 AgBr + X(NO3)2
(a) Calculate the formula mass of XBr2.

Formula mass XBr2= g mol^-1
(b) Calculate the atomic mass of X.

Atomic mass X= g mol^-1

Transcript text: A sample of a substance with the empirical formula $\mathrm{XBr}_{2}$ weighs 0.4895 g . When it is dissolved in water and all its bromine is converted to insoluble AgBr by addition of an excess of silver nitrate, \[ -\mathrm{XBr}_{2}+2 \mathrm{AgNO}_{3} \longrightarrow 2 \mathrm{AgBr}+\mathrm{X}\left(\mathrm{NO}_{3}\right)_{2} \] (a) Calculate the formula mass of $\mathrm{XBr}_{2}$. Formula mass $\mathrm{XBr}_{2}=$ $\qquad$ $\mathrm{g} \mathrm{mol}^{-1}$ (b) Calculate the atomic mass of $X$. Atomic mass $\mathrm{X}=$ $\qquad$ $\mathrm{g} \mathrm{mol}^{-1}$

Solution

Solution Steps

Step 1: Determine the Mass of AgBr Formed

The reaction given is:

\[ \mathrm{XBr}_{2} + 2 \mathrm{AgNO}_{3} \longrightarrow 2 \mathrm{AgBr} + \mathrm{X}\left(\mathrm{NO}_{3}\right)_{2} \]

From the stoichiometry of the reaction, 1 mole of $\mathrm{XBr}_{2}$ produces 2 moles of $\mathrm{AgBr}$. We need to find the mass of $\mathrm{AgBr}$ formed to proceed with the calculations.

Step 2: Calculate the Molar Mass of AgBr

The molar mass of $\mathrm{AgBr}$ is calculated as follows:

Atomic mass of Ag = 107.87 g/mol
Atomic mass of Br = 79.90 g/mol

\[ \text{Molar mass of } \mathrm{AgBr} = 107.87 + 79.90 = 187.77 \, \text{g/mol} \]

Step 3: Calculate the Moles of AgBr Formed

Assuming all bromine in $\mathrm{XBr}_{2}$ is converted to $\mathrm{AgBr}$, the mass of $\mathrm{AgBr}$ formed is equal to the mass of bromine in the original sample. Since the problem does not provide the mass of $\mathrm{AgBr}$, we will assume the entire mass of $\mathrm{XBr}_{2}$ is converted to $\mathrm{AgBr}$.

Step 4: Calculate the Molar Mass of $\mathrm{XBr}_{2}$

The molar mass of $\mathrm{XBr}_{2}$ can be calculated using the mass of the sample and the stoichiometry of the reaction. Since 1 mole of $\mathrm{XBr}_{2}$ produces 2 moles of $\mathrm{AgBr}$, we can set up the equation:

\[ \text{Molar mass of } \mathrm{XBr}_{2} = \frac{\text{mass of sample}}{\text{moles of } \mathrm{XBr}_{2}} \]

Given that the mass of the sample is 0.4895 g, and assuming all of it is converted to $\mathrm{AgBr}$, we can calculate the moles of $\mathrm{XBr}_{2}$ using the molar mass of $\mathrm{AgBr}$.

Step 5: Calculate the Atomic Mass of $X$

The molar mass of $\mathrm{XBr}_{2}$ is the sum of the atomic mass of $X$ and twice the atomic mass of bromine:

\[ \text{Molar mass of } \mathrm{XBr}_{2} = \text{Atomic mass of } X + 2 \times 79.90 \]

Rearranging to solve for the atomic mass of $X$:

\[ \text{Atomic mass of } X = \text{Molar mass of } \mathrm{XBr}_{2} - 2 \times 79.90 \]