 Solution

Take dry air as unity, and compare each gas at the same temperature and pressure. In the ideal gas approximation, $\rho =pM/(RT)$, so the relative density isHere $M$ is molar mass. The following values use modern approximate molar masses; rounding and historical atomic weight conventions can shift the last figures.

“Olefiant gas” is confirmed as a historical synonym of Ethylene in the NIST Chemistry WebBook, SRD 69. In this gas list, “carbonic acid” means carbon dioxide, not an isolated vapour of aqueous ${\mathrm{H}}_2{\mathrm{CO}}_3$. “Alcohol” and “ether” refer to ethanol and diethyl ether, rather than their entire modern chemical classes.^[ref]

The order is therefore Ammonia, then carbon monoxide and Ethylene nearly tied, then air, then carbon dioxide and nitrous oxide nearly tied, then ethanol vapour, then diethyl ether vapour. Nearly equal molar masses do not imply equal chemical behaviour.

If instead molecular hydrogen is chosen as unity, divide each molar mass by about $2.016\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and state that convention. In the older nominal ${\mathrm{H}}_2=2$ approximation, the corresponding values are about $22,14,22,14.5,8.5,14,23,37$ in the source order. These are a different reference scale, not different substances.

The vapour comparisons are not comparisons of the liquid densities. Pure ethanol and diethyl ether must be considered at conditions where they remain gaseous, or by a low-pressure gas approximation. Their liquids can be lighter than water while their vapours are heavier than air.