Complete combustion of 1.80 g of an oxygen containing compound (CxHyOz) gave 2.64 g of CO2 and 1.08

Q: Why can't we just calculate the mass of oxygen from the $CO_2$ and $H_2O$ produced?

Because the combustion process requires adding external oxygen gas ($O_2$) to burn the compound. The oxygen in the $CO_2$ and $H_2O$ comes from BOTH the original compound and the external air. It's impossible to tell which oxygen atom came from where, so we must find the oxygen mass by subtracting the C and H masses from the total initial mass.

Q: Why do we multiply by 2 when calculating the moles of Hydrogen?

Look at the formula for water: $H_2O$. For every 1 mole of water molecules produced, there are 2 moles of hydrogen atoms inside them. Since we want the mass of individual hydrogen atoms that came from the original compound, we must account for both of them.

Q: What is the Principle of Atom Conservation (POAC)?

POAC is a fundamental rule stating that atoms are neither created nor destroyed in a chemical reaction. It allows us to equate the number of moles of a specific atom in the reactants directly to the number of moles of that same atom in the products, even without writing a fully balanced chemical equation.

Q: Could I have solved this by finding the empirical formula first?

You could, but it would be a longer, more tedious route. To find the empirical formula, you still need the individual masses of C, H, and O first. Once you have the mass of O, you already have what you need to find its percentage, making the empirical formula step unnecessary for this specific question.

Q: What if the compound also contained Nitrogen?

If the compound contained Nitrogen, the problem would have to provide additional data, such as the volume of $N_2$ gas produced (often measured using Dumas method) or ammonia produced (Kjeldahl method). You would calculate the mass of N, C, and H, and then subtract all three from the total mass to find the remaining Oxygen.

Question

Complete combustion of $1.80	ext{ g}$ of an oxygen containing compound ($C_xH_yO_z$) gave $2.64	ext{ g}$ of $CO_2$ and $1.08	ext{ g}$ of $H_2O$. The percentage of oxygen in the organic compound is

Options:

(A) $50.33$

(B) $53.33$

(C) $63.53$

(D) $51.63$

Sigma Percentile · Accepted Answer

### The Setup: Burning the Unknown

Imagine you are a chemical detective handed a mysterious white powder. You know it contains carbon, hydrogen, and oxygen, but you have no idea in what proportions. How do you find out? You burn it! This process, known as combustion analysis, is a classic and elegant technique in chemistry.

When we take our $1.80	ext{ g}$ sample of $C_xH_yO_z$ and ignite it in a stream of pure oxygen, a violent but perfectly predictable reaction occurs. The carbon atoms are ripped away and bond with oxygen to form carbon dioxide ($CO_2$). Simultaneously, the hydrogen atoms are oxidized to form water vapor ($H_2O$).

By carefully trapping and weighing these products, we find we have generated $2.64	ext{ g}$ of $CO_2$ and $1.08	ext{ g}$ of $H_2O$. The stage is set. Now, we must trace the atoms back to their source.

### Tracking the Carbon

Here is where the Principle of Atom Conservation (POAC) becomes our most powerful tool. Think of it as an absolute accounting rule: atoms cannot be created or destroyed. Therefore, **every single atom of carbon in the $CO_2$ must have come from our original organic compound**.

First, we determine how many moles of $CO_2$ we have. We divide the given mass by the molar mass of $CO_2$ ($44	ext{ g/mol}$):

$$n_{CO_2} = \frac{2.64}{44} = 0.06	ext{ mol}$$

Since each molecule of $CO_2$ contains exactly one carbon atom, the moles of carbon are also $0.06	ext{ mol}$. To find the mass of this carbon, we multiply by its atomic mass ($12	ext{ g/mol}$):

$$W_C = 0.06 	imes 12 = 0.72	ext{ g}$$

We have successfully isolated the mass of carbon in our original sample!

### Tracking the Hydrogen

We apply the exact same logic to hydrogen. All the hydrogen in the water vapor originated from our mystery compound. We start by finding the moles of water ($H_2O$), which has a molar mass of $18	ext{ g/mol}$:

$$n_{H_2O} = \frac{1.08}{18} = 0.06	ext{ mol}$$

But wait, don't make a silly mistake here! Look closely at the chemical formula of water: $H_2O$. **Every single molecule of water contains TWO atoms of hydrogen.** Therefore, the moles of hydrogen atoms will be twice the moles of water:

$$n_H = 2 	imes 0.06 = 0.12	ext{ mol}$$

Multiplying by the atomic mass of hydrogen ($1	ext{ g/mol}$), we find the mass of hydrogen:

$$W_H = 0.12 	imes 1 = 0.12	ext{ g}$$

### The Catch: Where Did the Oxygen Come From?

Now, you might be tempted to calculate the mass of oxygen directly from the $CO_2$ and $H_2O$. **Stop right there!** This is a classic trap.

Remember the setup? We burned the compound in a stream of *excess oxygen* from the air. The oxygen in the products is a mixture of the oxygen originally in the compound AND the oxygen we pumped into the furnace. We cannot easily separate them.

Instead, we must use a clever workaround. We know the total mass of the original compound was $1.80	ext{ g}$. We also know that this mass is composed entirely of carbon, hydrogen, and oxygen. Since we have already calculated the masses of carbon and hydrogen, the oxygen is simply whatever is left over!

$$W_O = W_{	ext{total}} - (W_C + W_H)$$

$$W_O = 1.80 - (0.72 + 0.12) = 1.80 - 0.84 = 0.96	ext{ g}$$

### The Final Calculation

We have cracked the case. Out of the original $1.80	ext{ g}$ sample, exactly $0.96	ext{ g}$ was oxygen. The final step is to express this as a mass percentage, which is what the question demands.

$$\% O = \frac{	ext{Mass of Oxygen}}{	ext{Total Mass}} 	imes 100$$

$$\% O = \frac{0.96}{1.80} 	imes 100 = 53.33\%$$

The percentage of oxygen in the organic compound is $53.33\%$. This elegant application of mass conservation not only solves the problem but also lays the groundwork for determining the full empirical formula of the compound.

Animated Solution for Chemistry - Basic Concepts in Chemistry: Complete combustion of $1.80 g$ of an oxygen containing compound ( $C_{x} H_{y} O_{z}$ ) gave $2.64 g$ of $C O_{2}$ and $1.08 g$ of $H_{2} O$ . The percentage of oxygen in the organic compound is

Select Answer:

Visualized Solution

\text{Combustion Reaction}

\text{Conservation of Carbon}

\text{Mass of Carbon}

\text{Conservation of Hydrogen}

\text{Mass of Hydrogen}

\text{Mass of Oxygen}

\text{Percentage of Oxygen}

\text{The Way Forward}

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Setup

Tracking the Carbon

Tracking the Hydrogen

The Catch

The Final Calculation

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