A sample of metallic element X

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?


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


Thus, we know that the atomic mass of X is 107.88 amu. Looking at the periodic table, we can see that the element X is in fact silver (Ag).

This question can be found in General Chemistry, 9th edition, chapter 2, question 2.148

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