Consider an unknown compound . The ratio of mass percent of Carbon to Hydrogen is exactly . Furtherm — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
Consider an unknown compound $C_x H_y O_z$. The ratio of mass percent of Carbon to Hydrogen is exactly $6:1$. Furthermore, one single molecule of this compound contains half as many oxygen atoms as the total number of $O_2$ molecules required to burn one molecule of the hydrocarbon $C_xH_y$ completely to $CO_2$ and $H_2O$. Which of the following could potentially be the molecular formula of the compound?
💡 Solution & Explanation
Step 1: Mass ratio C:H = 6:1. Mole ratio C:H = $(6/12) : (1/1) = 0.5 : 1 = 1 : 2$. Therefore, $y = 2x$. The hydrocarbon part is $C_xH_{2x}$. Step 2: Combustion of $C_xH_{2x} + (x + \frac{2x}{4}) O_2 \rightarrow xCO_2 + xH_2O$. Molecules of $O_2$ required $= 1.5x$. Step 3: The problem states $z$ (number of oxygen atoms) is half of the $O_2$ molecules required. So, $z = \frac{1.5x}{2} = 0.75x$. Therefore, the ratio $x : y : z$ is $x : 2x : 0.75x$, which simplifies to $1 : 2 : 0.75 \implies 4 : 8 : 3$. Thus, the empirical formula is $C_4H_8O_3$. Any multiple like $C_2H_4O_{1.5}$ (invalid) or $C_4H_8O_6$ (Wait, $4:8:3$ multiple is $C_8H_{16}O_6$. Let's re-read carefully: "half as much oxygen as required". Is it half the *mass*, half the *atoms*, or half the *molecules*? The source [1] says "contains half as much oxygen as required to burn...". In $C_2H_4O_3$, $x=2, y=4$. Burning $C_2H_4$ needs $3 O_2$ molecules. Half of 3 is 1.5 molecules = 3 oxygen atoms. So $z=3$. Formula $C_2H_4O_3$ works! Option B is $C_4H_8O_6$, which is $(C_2H_4O_3)_2$, so it also satisfies the ratio.