A metal M forms the sulphate `M_(2)(SO_(4))_(3)`. A 0.596 gram sample of the sulphate reacts with excess `BaCl_(2)` to give 1.220 g `BaSO_(4)`. What is the atomic mass of M ?

Options

A. 26.9
B. 69.7
C. 55.8
D. 23

Answer

To find the atomic mass of the metal \( M \) in the sulfate \( M_2(SO_4)_3 \), we can follow these steps:

### Step 1: Write the balanced chemical reaction
The metal sulfate \( M_2(SO_4)_3 \) reacts with barium chloride \( BaCl_2 \) to form barium sulfate \( BaSO_4 \) and metal chloride \( MCl_3 \). The balanced equation can be written as:

\[ M_2(SO_4)_3 + 3BaCl_2 \rightarrow 3BaSO_4 + 2MCl_3 \]

### Step 2: Calculate the moles of barium sulfate produced
We know that 1.220 g of \( BaSO_4 \) is produced. The molar mass of \( BaSO_4 \) is calculated as follows:

- Atomic mass of Ba = 137.33 g/mol
- Atomic mass of S = 32.07 g/mol
- Atomic mass of O = 16.00 g/mol (4 oxygen atoms)

\[ \text{Molar mass of } BaSO_4 = 137.33 + 32.07 + (4 \times 16.00) = 233.39 \text{ g/mol} \]

Now, we can calculate the moles of \( BaSO_4 \):

\[ \text{Moles of } BaSO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{1.220 \text{ g}}{233.39 \text{ g/mol}} \approx 0.00523 \text{ mol} \]

### Step 3: Relate moles of \( M_2(SO_4)_3 \) to moles of \( BaSO_4 \)
From the balanced equation, we see that 1 mole of \( M_2(SO_4)_3 \) produces 3 moles of \( BaSO_4 \). Therefore, the moles of \( M_2(SO_4)_3 \) can be calculated as:

\[ \text{Moles of } M_2(SO_4)_3 = \frac{\text{Moles of } BaSO_4}{3} = \frac{0.00523}{3} \approx 0.00174 \text{ mol} \]

### Step 4: Calculate the molar mass of \( M_2(SO_4)_3 \)
We know the mass of the sulfate sample is 0.596 g. The molar mass of \( M_2(SO_4)_3 \) can be calculated using the formula:

\[ \text{Molar mass of } M_2(SO_4)_3 = \frac{\text{mass}}{\text{moles}} = \frac{0.596 \text{ g}}{0.00174 \text{ mol}} \approx 342.53 \text{ g/mol} \]

### Step 5: Set up the equation for molar mass
The molar mass of \( M_2(SO_4)_3 \) can also be expressed in terms of the atomic mass of \( M \):

\[ \text{Molar mass of } M_2(SO_4)_3 = 2 \times \text{atomic mass of } M + 3 \times \text{molar mass of } SO_4 \]

Where the molar mass of \( SO_4 \) is:

\[ \text{Molar mass of } SO_4 = 32.07 + (4 \times 16.00) = 96.07 \text{ g/mol} \]

So,

\[ \text{Molar mass of } M_2(SO_4)_3 = 2M + 3 \times 96.07 = 2M + 288.21 \]

### Step 6: Solve for the atomic mass of \( M \)
Setting the two expressions for the molar mass equal gives:

\[ 342.53 = 2M + 288.21 \]

Rearranging gives:

\[ 2M = 342.53 - 288.21 = 54.32 \]

Thus,

\[ M = \frac{54.32}{2} \approx 27.16 \text{ g/mol} \]

### Conclusion
The atomic mass of the metal \( M \) is approximately \( 27.16 \text{ g/mol} \).

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