The ratio of mass per cent of C and H of an organic compound (CxHyOz) is 6 : 1. If one molecule of t

Q: Why do we divide the mass ratio by atomic mass?

The mass ratio tells us the relative weights of the elements, but chemical formulas are based on the number of atoms (moles). Dividing mass by atomic mass gives us the mole ratio, which directly translates to the subscripts in the empirical formula.

Q: How did we get the coefficient $(x + y/4)$ for oxygen in the combustion equation?

In the combustion of $C_xH_y$, all carbon goes to $CO_2$ (requiring $2x$ oxygen atoms) and all hydrogen goes to $H_2O$ (requiring $y/2$ oxygen atoms). The total oxygen atoms needed are $2x + y/2$. Since oxygen gas is diatomic ($O_2$), we divide by 2 to get the coefficient: $x + y/4$.

Q: Why did we multiply the final ratio $1 : 2 : 1.5$ by 2?

Empirical formulas must have whole-number subscripts because you cannot have a fraction of an atom in a molecule. Multiplying the entire ratio by 2 converts the $1.5$ into $3$, giving us the simplest whole-number ratio of $2 : 4 : 3$.

Q: What is the difference between an empirical formula and a molecular formula?

An empirical formula shows the simplest whole-number ratio of atoms in a compound, while a molecular formula shows the exact number of each type of atom in a single molecule. This problem asks for the empirical formula.

Q: Could $x$ and $y$ be larger numbers like $2$ and $4$ initially?

Yes, the mole ratio $1:2$ is just the simplest ratio. If we used $x=2$ and $y=4$, then $z$ would be $2 + 4/4 = 3$. The ratio $x:y:z$ would be $2:4:3$, which leads to the exact same empirical formula!

Question

The ratio of mass per cent of C and H of an organic compound ($C_xH_yO_z$) is $6 : 1$. If one molecule of the above compound ($C_xH_yO_z$) contains half as much oxygen as required to burn one molecule of compound $C_xH_y$ completely to $CO_2$ and $H_2O$. The empirical formula of compound $C_xH_yO_z$ is

Options:

(A) $C_3H_6O_3$

(B) $C_2H_4O$

(C) $C_3H_4O_2$

(D) $C_2H_4O_3$

Sigma Percentile · Accepted Answer

Stoichiometry is often seen as just a balancing act, a mere accounting of atoms. But at its core, it is the language of the universe, describing exactly how matter transforms. Today, we are going to solve a beautiful puzzle that combines the mole concept with the stoichiometry of combustion.

Imagine you are a chemical detective. You are given an unknown organic compound, $C_xH_yO_z$, and a few cryptic clues about its composition and how it burns. Our mission? To deduce its empirical formula. Let's break this down step by step.

### Decoding the Mass Ratio

The first clue is the mass ratio of carbon to hydrogen, which is given as $6 : 1$.

Mass ratios are great, but chemical formulas are built on the number of atoms, not their weight. To bridge this gap, we need to convert this mass ratio into a mole ratio. We do this by dividing the mass of each element by its respective atomic mass.

For carbon, with an atomic mass of $12$, the number of moles is $\frac{6}{12} = 0.5$.

For hydrogen, with an atomic mass of $1$, the number of moles is $\frac{1}{1} = 1$.

So, our mole ratio of $C : H$ is $0.5 : 1$. To make this a simple whole-number ratio, we divide both by the smallest value, giving us $1 : 2$.

This tells us that for every carbon atom, there are two hydrogen atoms. We can confidently set $x = 1$ and $y = 2$.

### The Combustion Connection

Now, let's look at the second clue. The problem mentions the combustion of the hydrocarbon part, $C_xH_y$.

When a hydrocarbon burns completely in the presence of oxygen, it produces carbon dioxide ($CO_2$) and water ($H_2O$). The balanced chemical equation for this process is a fundamental tool in chemistry:

$$C_xH_y + \left(x + \frac{y}{4}ight) O_2 ightarrow x CO_2 + \frac{y}{2} H_2O$$

Take a close look at the stoichiometric coefficient of oxygen, which is $\left(x + \frac{y}{4}ight)$. This represents the number of $O_2$ **molecules** required for the complete combustion of one molecule of $C_xH_y$.

### The Oxygen Puzzle

The problem states a fascinating condition: the number of oxygen atoms in our original compound, $z$, is exactly half of the oxygen **atoms** required to burn $C_xH_y$.

First, let's find the total number of oxygen atoms required for combustion. Since each $O_2$ molecule contains two atoms, we multiply the number of molecules by $2$:

$$	ext{Oxygen atoms required} = 2 	imes \left(x + \frac{y}{4}ight) = 2x + \frac{y}{2}$$

According to the problem, $z$ is half of this value. Therefore:

$$z = \frac{1}{2} 	imes \left(2x + \frac{y}{2}ight) = x + \frac{y}{4}$$

### Bringing It All Together

We are now in the endgame. We have our expressions, and we have our values for $x$ and $y$. Let's substitute them in!

Plugging $x = 1$ and $y = 2$ into our equation for $z$:

$$z = 1 + \frac{2}{4} = 1 + 0.5 = 1.5$$

So, our ratio for $x : y : z$ is $1 : 2 : 1.5$.

However, an empirical formula must consist of the simplest **whole-number** ratio of atoms. We cannot have one and a half oxygen atoms in a formula! To fix this, we multiply the entire ratio by $2$:

$$x : y : z = 2 : 4 : 3$$

### The Final Reveal

With our whole-number ratio secured, we can finally write down the empirical formula of our mystery compound.

The empirical formula is **$C_2H_4O_3$**.

This problem is a fantastic demonstration of how different chemical concepts—mass percent, mole ratio, and reaction stoichiometry—intertwine to reveal the hidden structure of matter. Keep practicing, and soon you'll be decoding these chemical puzzles with ease!

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Visualized Solution

$M a ss t o M o l e R a t i o$

$S im pl es t R a t i o o f x an d y$

$C o mb u s t i o n o f H y d r oc a r b o n$

$O x y g e n A t o m s R e q u i r e d$

$C o n d i t i o n f or z$

$C a l c u l a t in g z$

$F ina l E m p i r i c a l F or m u l a$

$C o n c l u s i o n$

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Decoding the Mass Ratio

The Combustion Connection

The Oxygen Puzzle

Bringing It All Together

The Final Reveal

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass to Mole Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplest Ratio of x and y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Combustion of Hydrocarbon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Oxygen Atoms Required

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for z

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating z

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Empirical Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion