[tex] \text{Hg} \text{O} [/tex] and [tex] \text{Hg}_{2} \text{O} [/tex].
Assuming complete decomposition of both samples,
- [tex] m(\text{Hg}) = m(\text{residure}) [/tex]
- [tex] m(\text{O}) = m(\text{loss}) [/tex]
First compound:
- [tex] m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g [/tex]
- [tex] m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g [/tex]
[tex] n = m/M [/tex]; [tex] 0.6498 \; g [/tex] of the first compound would contain
- [tex] n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol [/tex]
- [tex] n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol [/tex]
Oxygen and mercury atoms seemingly exist in the first compound at a [tex] 1:1 [/tex] ratio; thus the empirical formula for this compound would be [tex] \text{Hg} \text{O} [/tex] where the subscript "1" is omitted.
Similarly, for the second compound
- [tex] m(\text{O}) = m(\text{loss}) = 0.016 \; g [/tex]
- [tex] m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g [/tex]
[tex] n = m/M [/tex]; [tex] 0.4172 \; g [/tex] of the first compound would contain
- [tex] n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol [/tex]
- [tex] n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol [/tex]
[tex] n(\text{Hg}) : n(\text{O}) \approx 2:1 [/tex] and therefore the empirical formula
[tex] \text{Hg}_{2} \text{O} [/tex].
