a 1.750-g sample of mercury metal ws combined with 1.394 g of bromine. Determine the empirical formula of the mercury bromide that is produced

HgBr₂.

Explanation:

To determine the empirical formula of the mercury bromide that is produced, we should calculate the no. of moles of Hg (1.750 g) and Br (1.394 g) used to prepare the mercury bromide.

no. of moles of Hg = mass/atomic mass = (1.75 g) / (200.59 g/mol) = 0.00872 mol.

no. of moles of Br = mass/atomic mass = (1.394 g) / (79.904 g/mol) = 0.01744 mol.

We divide by the lowest no. of moles (0.00872) to get the mole ratio:

The mole ratio of (Br: Hg) is: (2: 1).

So, the empirical formula is: HgBr₂.