Question

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. heating 0.6498 g of one of the compounds leaves a residue of 0.6018 g. heating 0.4172 g of the other compound results in a mass loss of 0.016 g. determine the empirical formula of each compound.

Answer 1

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$.

Assuming complete decomposition of both samples,

• $m(\text{Hg}) = m(\text{residure})$
• $m(\text{O}) = m(\text{loss})$

First compound:

• $m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
• $m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$; $0.6498 \; g$ of the first compound would contain

• $n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
• $n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

• $m(\text{O}) = m(\text{loss}) = 0.016 \; g$
• $m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$; $0.4172 \; g$ of the first compound would contain

• $n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
• $n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$.