Answer:
The molecular formula is most likely to be first option, [tex]\rm C_{12} H_{18} O_{3}[/tex].
Explanation:
Both the molecular formula and the empirical formula of a molecule give the types of atoms in that molecule. However, the molecular formula of a molecule gives the exact number of atoms each molecule. On the other hand, the empirical formula gives only a simplified ratio.
For example, if the empirical formula of the molecule is [tex]\rm C_4H_6O[/tex], then the molecular formula would be [tex](\rm C_4H_6O)_k[/tex], or equivalently [tex]\mathrm{C}_{4\, k}\mathrm{H}_{6\, k} \mathrm{O}_{k}[/tex], where [tex]k[/tex] is a positive whole number ([tex]1,\, 2,\, \dots[/tex], etc.) The goal here is to find the value of
Look up the relative atomic mass data on a modern periodic table:
- C: 12.011.
- H: 1.008.
- O: 15.999.
The formula mass of [tex]\rm C_4 H_6 O[/tex] would be
[tex]\begin{aligned}&M(\rm C_4 H_6 O) \cr &\approx 4 \times 12.011 + 6 \times 1.008 + 15.999 \cr &= 70.091\;\rm g \cdot mol^{-1}\end{aligned}[/tex].
The formula mass of [tex]\mathrm{C}_{4\, k}\mathrm{H}_{6\, k} \mathrm{O}_{k}[/tex] would be
[tex]\begin{aligned}&M(\mathrm{C}_{4\, k}\mathrm{H}_{6\, k} \mathrm{O}_{k}) \cr &\approx (4\, k) \times 12.011 + (6\, k) \times 1.008 + k \times 15.999 \cr &= ( 4 \times 12.011 + 6 \times 1.008 + 15.999)\, k \\ &= (70.091\, k)\;\rm g \cdot mol^{-1}\end{aligned}[/tex].
On the other hand, since the molecule should have a molecular mass of [tex]\rm 212\; g \cdot mol^{-1}[/tex],
[tex]70.091\, k \approx 212[/tex].
[tex]k \approx 3[/tex]. (Round to the nearest whole number.)
Hence, the molecular formula would be [tex]\rm C_{(4 \times 3)} H_{(6 \times 3)}O_{3}[/tex], which simplifies to [tex]\rm C_{12} H_{18} O_{3}[/tex].
