Animated Solution for Chemistry - Basic Concepts in Chemistry: The ratio of the mass percentages of 'C' and 'H' and 'C' and 'O' of a saturated acyclic organic compound 'X' are $4:1$ and $3:4$ respectively. Then, the moles of oxygen gas required for complete combustion of two moles of organic compound 'X' is ......... .

The journey to solving this problem is like being a chemical detective. We are given a mystery compound, 'X', and a few clues about its composition and structure. Our mission? To find out exactly how much oxygen is needed to completely burn two moles of it. Let's dive into the investigation!

Our first set of clues comes in the form of mass percentage ratios. We know that the compound contains Carbon (C), Hydrogen (H), and Oxygen (O).

The problem states: Mass ratio of $C:H=4:1$ Mass ratio of $C:O=3:4$

To make sense of these ratios, we need to unify them. Notice that Carbon is common to both ratios. If we can make the mass of Carbon the same in both expressions, we can combine them into a single $C:H:O$ ratio.

Conveniently, the atomic mass of Carbon is 12. Let's scale both ratios so that the Carbon part equals 12: Multiply the $C:H$ ratio by 3: $C:H=12:3$ Multiply the $C:O$ ratio by 4: $C:O=12:16$

Now, we have a beautifully unified mass ratio: Mass ratio $C:H:O=12:3:16$

Mass ratios are great, but chemical formulas are based on the number of atoms, or moles. To convert our mass ratio into a molar ratio, we divide each mass value by the respective atomic mass of the element ($C=12$, $H=1$, $O=16$).

Moles of C = $\frac{12}{12}=1$ Moles of H = $\frac{3}{1}=3$ * Moles of O = $\frac{16}{16}=1$

This gives us a molar ratio of $1:3:1$. Therefore, the simplest integer ratio of atoms in our compound—the empirical formula—is $C{H}_{3}O$.

Here is where the structural hint becomes crucial. The problem states that compound 'X' is a saturated acyclic organic compound.

For a saturated acyclic hydrocarbon (an alkane), the general formula is $C_n{H}_{2n+2}$. Adding oxygen atoms (like in alcohols or ethers) does not change the required ratio of Carbon to Hydrogen for saturation. Thus, the general formula for our compound must be $C_n{H}_{2n+2}{O}_z$.

Let's test our empirical formula, $C{H}_{3}O$, against this rule. If $n=1$, the number of Hydrogen atoms should be $2(1)+2=4$. However, our empirical formula only has 3 Hydrogen atoms. This means $C{H}_{3}O$ cannot be the true molecular formula.

Let's scale it up by multiplying by 2, giving us $C_2{H}_6{O}_2$. Let's test this new formula. If $n=2$, the number of Hydrogen atoms should be $2(2)+2=6$. This matches perfectly! We have successfully identified compound 'X' as $C_2{H}_6{O}_2$ (which happens to be ethylene glycol).

Now that we know the exact identity of compound 'X', we can write its complete combustion reaction. Complete combustion means reacting the compound with Oxygen ($O_2$) to produce Carbon Dioxide ($C{O}_2$) and Water ($H_{2}O$).

The unbalanced equation is: $C_2{H}_6{O}_2+O_2⟶C{O}_2+H_{2}O$

Let's balance it step-by-step: 1. Carbon: We have 2 Carbons on the left, so we need 2 $C{O}_2$ on the right. 2. Hydrogen: We have 6 Hydrogens on the left, so we need 3 $H_{2}O$ on the right. 3. Oxygen: On the right side, we now have $(2\times 2)+(3\times 1)=7$ Oxygen atoms. On the left side, our compound already provides 2 Oxygen atoms. This means we need 5 more Oxygen atoms from the $O_2$ gas. To get 5 atoms from diatomic $O_2$, we need $\frac{5}{2}$ molecules.

The perfectly balanced equation is: $C_2{H}_6{O}_2+\frac{5}{2}{O}_2⟶2C{O}_2+3H_{2}O$

The balanced equation tells us the fundamental stoichiometry: 1 mole of compound 'X' requires $\frac{5}{2}$ moles of $O_2$ gas for complete combustion.

The question asks for the moles of oxygen required for two moles of compound 'X'. We simply multiply our stoichiometric requirement by 2: Moles of $O_2=2\times \frac{5}{2}=5$

Exactly 5 moles of oxygen gas are required. The mystery is solved!