Problem 4

A mixture contains two organic compounds, A and B, both containing oxygen and miscible in all proportions. Controlled oxidation of this mixture yields a single compound C, which forms an addition compound with

\ce{NaHSO3}

. The molar mass ratio of this bisulfite adduct to compound C is 2.7931. Burning the mixture of A and B with a stoichiometric amount of air (20 %

\ce{O2}

, 80 %

\ce{N2}

by volume) yields a gas mixture of total volume 5.432 dm

^3

at STP. When bubbled through

\ce{Ba(OH)2}

solution, the volume decreases by 15.46 %. (a) Identify compounds A, B and C. (b) Calculate the molar ratio of A and B in the initial mixture. (

A_r(\text{C}) = 12

A_r(\text{O}) = 16

A_r(\text{S}) = 32

A_r(\text{Na}) = 23

)

Step 2 of 4: Gas split from Ba(OH)2 absorption

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V(\ce{CO2}) = 5.432 \times 0.1546 = 0.840\ \text{dm}^3; \quad V(\ce{N2}) = 5.432 - 0.840 = 4.592\ \text{dm}^3

Analysis

After steam condenses, only CO2 and N2 remain. Ba(OH)2 absorbs CO2, so $V(\ce{CO2}) = 0.840$ dm $^3$ and the remaining gas is $V(\ce{N2}) = 4.592$ dm $^3$ .