Organic Compound Formula from Combustion Requirement

❓ Question

The ratio of mass percent of C and H of an organic compound

$$(C_xH_yO_z)$$

is:

$$6:1$$

If one molecule of the compound contains half as much oxygen as required to burn one molecule of compound

$$C_xH_yO_z$$

completely to:

$$CO_2 \text{ and } H_2O$$

then the empirical formula of the compound is ______.

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🖼 Question Image

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✍️ Short Explanation

This question combines:

👉 Percentage composition
👉 Combustion concept
👉 Empirical formula calculation.

We first use mass ratio of C and H, then use oxygen balance during combustion.

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🔷 Step 1 — Use Given Mass Ratio 💯

Given:

$$\text{Mass \% of C : H}=6:1$$

For compound:

$$C_xH_yO_z$$

Mass contribution:

$$12x : y = 6:1$$

So:

$$12x=6y$$
$$2x=y$$

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🔷 Step 2 — Oxygen Required for Combustion

Combustion reaction:

$$C_xH_yO_z + O_2 \rightarrow xCO_2+\frac y2 H_2O$$

Total oxygen atoms needed in products:

$$2x+\frac y2$$

Oxygen atoms already present in compound:

$$z$$

Extra oxygen atoms required from atmosphere:

$$2x+\frac y2-z$$

So oxygen molecules required:

$$\frac{2x+\frac y2-z}{2}$$

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🔷 Step 3 — Apply Given Condition

Question says:

Compound already contains half as much oxygen as required for combustion.

So:

$$z=\frac12\left(2x+\frac y2-z\right)$$

Multiply by 2:

$$2z=2x+\frac y2-z$$
$$3z=2x+\frac y2$$

Using:

$$y=2x$$
$$3z=2x+\frac{2x}{2}$$
$$3z=2x+x$$
$$3z=3x$$
$$z=x$$

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🔷 Step 4 — Find Simplest Ratio

We have:

$$y=2x$$

and

$$z=x$$

So:

$$x:y:z=1:2:1$$

Empirical formula:

$$\boxed{CH_2O}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Oxygen atoms aur oxygen molecules confuse kar lena

❌ Combustion oxygen balance galat likh dena

❌ Mass ratio ko mole ratio directly maan lena

Remember:

$$\text{Mass ratio} \neq \text{atomic ratio}$$

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✅ Final Answer

$$\boxed{CH_2O}$$

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