Establish the formula for the limiting nitro compound containing 42.67% oxygen by weight.

Let the formula of the unknown limiting nitro compound be CnH2n + 1NO2, where CnH2n + 1 is an unknown alkyl radical.

Let’s find the molecular weight of the nitro compound:

M (CnH2n + 1NO2) = M (O2) * w (CnH2n + 1NO2) / w (O2) = 32 * 100 / 42.67 = 75 (amu).

Let’s find the molecular weight of the alkyl radical:

M (CnH2n + 1) = M (CnH2n + 1NO2) – M (NO2) = 75 – 46 = 29 (amu).

Let’s compose and solve the equation:

12 * n + 1 * (2n + 1) = 29,

n = 2, whence the molecular formula of this compound: C2H5NO2, which corresponds to nitroethane.

Answer: C2H5NO2.



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