If a pure compound is composed of `X_(2)Y_(3)` molecules and consists of `60%` X by weight what is the atomic weight of Y in term of atomic weight of X (Atomic mass of `X=M_(x)`)?

Options

A. `(9)/(4)M_(x)`
B. `(4)/(9)M_(x)`
C. `(2)/(3)M_(x)`
D. `(3)/(2)M_(x)`

Answer

To solve the problem step by step, we need to find the atomic weight of Y in terms of the atomic weight of X, given that the compound is composed of \(X_2Y_3\) and consists of 60% X by weight.

### Step 1: Understand the Composition of the Compound
The molecular formula of the compound is \(X_2Y_3\). This means there are 2 atoms of X and 3 atoms of Y in one molecule of the compound.

### Step 2: Set Up the Mass Percentage Equation
The mass percentage of X in the compound is given as 60%. The formula for mass percentage is:

\[ \text{Mass \% of X} = \frac{\text{Mass of X}}{\text{Total mass of compound}} \times 100 \]

### Step 3: Calculate the Mass of X
The mass of X in one molecule of \(X_2Y_3\) can be calculated as:
\[ \text{Mass of X} = 2 \times M_x \]
where \(M_x\) is the atomic weight of X.

### Step 4: Calculate the Total Mass of the Compound
The total mass of the compound \(X_2Y_3\) is the sum of the masses of X and Y:
\[ \text{Total mass} = \text{Mass of X} + \text{Mass of Y} = 2M_x + 3M_y \]
where \(M_y\) is the atomic weight of Y.

### Step 5: Substitute into the Mass Percentage Equation
Now we can substitute the mass of X and the total mass into the mass percentage equation:
\[ 60 = \frac{2M_x}{2M_x + 3M_y} \times 100 \]

### Step 6: Simplify the Equation
Dividing both sides by 100 gives:
\[ 0.6 = \frac{2M_x}{2M_x + 3M_y} \]

### Step 7: Cross Multiply
Cross multiplying gives:
\[ 0.6(2M_x + 3M_y) = 2M_x \]

### Step 8: Distribute and Rearrange
Distributing the 0.6:
\[ 1.2M_x + 1.8M_y = 2M_x \]
Rearranging gives:
\[ 1.8M_y = 2M_x - 1.2M_x \]
\[ 1.8M_y = 0.8M_x \]

### Step 9: Solve for \(M_y\)
Now, divide both sides by 1.8:
\[ M_y = \frac{0.8}{1.8}M_x \]
Simplifying this fraction:
\[ M_y = \frac{4}{9}M_x \]

### Conclusion
Thus, the atomic weight of Y in terms of the atomic weight of X is:
\[ M_y = \frac{4}{9}M_x \]