Complete combustion of 750 g of an organic compound provides 420 of CO2 and 210 g of H2O. The percen

Q: Why do we ignore the oxygen in the organic compound?

In combustion analysis, oxygen is supplied in excess from the outside to burn the compound. Therefore, the oxygen in the products ($\text{CO}_2$ and $\text{H}_2\text{O}$) comes from both the compound and the external supply. We cannot directly use it to find the oxygen content of the original compound.

Q: Why is the fraction of hydrogen $\frac{2}{18}$ and not $\frac{1}{18}$?

A single water molecule ($\text{H}_2\text{O}$) contains *two* atoms of hydrogen. Since each hydrogen atom has an atomic mass of $1\text{ u}$, the total mass contributed by hydrogen is $2\text{ u}$ out of the total $18\text{ u}$ of the water molecule.

Q: Do I need to calculate the percentage of carbon to solve this question?

No, the question only asks for the percentage of hydrogen (the second blank). The carbon percentage ($15.3\%$) is given as extra information, though you can use it to verify your calculations!

Q: What if the compound contained other elements like nitrogen or halogens?

Liebig's method specifically estimates carbon and hydrogen. If other elements are present, they form different byproducts (like $\text{N}_2$ gas or halides) which are estimated using different techniques like Dumas or Carius methods.

Question

Complete combustion of $750	ext{ g}$ of an organic compound provides $420$ of $	ext{CO}_2$ and $210	ext{ g}$ of $	ext{H}_2	ext{O}$. The percentage composition of carbon and hydrogen in organic compound is $15.3$ and …… respectively (Round off to the nearest integer).

Sigma Percentile · Accepted Answer

### The Magic of Liebig's Method

Imagine you are handed a mysterious organic compound and asked to find out exactly how much hydrogen is hidden inside it. How do you do it? You burn it!

This brilliant technique is known as **Liebig's method of combustion analysis**. When we take our $750	ext{ g}$ of the unknown organic compound and burn it completely in the presence of excess oxygen, a chemical transformation occurs.

All the carbon atoms in the compound bond with oxygen to form carbon dioxide ($	ext{CO}_2$), and every single hydrogen atom bonds with oxygen to form water ($	ext{H}_2	ext{O}$).

### Tracking the Hydrogen

The law of conservation of mass is our greatest ally here. Atoms are neither created nor destroyed. This means that **all the hydrogen present in the $210	ext{ g}$ of water came exclusively from our original $750	ext{ g}$ compound**.

If we can figure out the mass of hydrogen inside that water, we have successfully found the mass of hydrogen in our compound!

Let's look at a single water molecule, $	ext{H}_2	ext{O}$. Its molar mass is $18	ext{ g/mol}$. Out of this total mass, the two hydrogen atoms contribute exactly $2	ext{ g/mol}$.

Therefore, the fraction of hydrogen in any sample of water is always $\frac{2}{18}$.

### Calculating the Mass

Since the combustion produced $210	ext{ g}$ of water, we can easily find the total mass of hydrogen by multiplying our fraction by the mass of the water.

$$ 	ext{Mass of H} = \frac{2}{18} 	imes 210 $$

Let's simplify this expression. The fraction $\frac{2}{18}$ reduces to $\frac{1}{9}$.

$$ 	ext{Mass of H} = \frac{1}{9} 	imes 210 = \frac{70}{3}	ext{ g} $$

Notice how we kept the mass as a fraction ($\frac{70}{3}	ext{ g}$) instead of converting it to a decimal like $23.33	ext{ g}$. This is a crucial pro-tip! Keeping intermediate values as fractions prevents rounding errors from accumulating and ruining your final answer.

### The Final Percentage

Now that we have the mass of hydrogen, finding its percentage composition in the original compound is straightforward. We divide the mass of hydrogen by the total mass of the organic compound and multiply by $100$.

$$ \%	ext{H} = \frac{	ext{Mass of H}}{	ext{Total Mass of Compound}} 	imes 100 $$

Let's substitute our values into the master equation.

$$ \%	ext{H} = \frac{\frac{70}{3}}{750} 	imes 100 $$

Time for the final calculation. A zero from the $750$ and the $100$ cancels out, leaving us with a simple division.

$$ \%	ext{H} = \frac{7000}{2250} = 3.11\% $$

The question asks us to round off to the nearest integer. Rounding down $3.11\%$, we get our final answer: **$3$**.

### A Bonus Verification

As a fun exercise, you can use the exact same logic to find the percentage of carbon. The $420	ext{ g}$ of $	ext{CO}_2$ contains all the carbon. Since carbon makes up $\frac{12}{44}$ of $	ext{CO}_2$'s mass, the mass of carbon is $\frac{12}{44} 	imes 420 = 114.54	ext{ g}$.

Dividing this by the total mass ($750	ext{ g}$) and multiplying by $100$ gives exactly **$15.3\%$**, which perfectly matches the data given in the question! This confirms our entire thought process is absolutely flawless.

Enter Numerical Value:

Visualized Solution

$Liebig’s Method of Estimation$

$Conservation of Atoms$

$Fraction of Hydrogen in Water$

$Mass of Hydrogen$

$Calculating Mass of Hydrogen$

$Percentage Composition Formula$

$Substituting Values$

$Final Calculation$

$Verification (Optional)$

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Magic of Liebig's Method

Tracking the Hydrogen

Calculating the Mass

The Final Percentage

A Bonus Verification

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Liebig’s Method of Estimation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fraction of Hydrogen in Water

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

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

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

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

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

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