A sample of metallic element X, weighing 4.315 g, combines with 0.4810 L of Cl_{2} gas (at normal pressure and 20.0^{o}C) to form the metal chloride with the formula XCl. If the density of Cl_2 gas under these conditions is 2.948 g/L, what is the mass of the chlorine? The atomic mass of chlorine is 35.453 amu. What is the atomic mass of X? What is the identity of X?

#### Solution:

### We will begin by writing down what we know.

**Weight of sample metallic element X = 4.315 g**

**Amount of Cl_2 gas = 0.4810 L**

**Density of gas = 2.948 g/L**

**Atomic mass of Cl = 35.453 amu**

### To find the mass of Cl, we will multiply the amount of Cl gas by the density.

### Mass of Cl= 0.4810 L \times \frac{2.948 g}{1 L}

### Mass of Cl= 1.41799 g

### To find the atomic mass, we can multiply the atomic mass of the Cl by the ratio of the weight of the metallic element X to the mass of Cl as follows:

### 35.453 \text{amu Cl}\times \frac{4.315 g}{1.41799 g} =107.88 amu