In an organic compound of molar mass 108 g mol^-1 C, H and N atoms are present in 9 : 1 : 3.5 by wei

Q: Why do we add the ratio parts together?

Adding the ratio parts gives us a 'total relative weight' that corresponds to the given molar mass. It allows us to express the mass of each element as a fraction of the whole.

Q: Can we find the empirical formula first instead of doing this?

Absolutely! You can find the empirical formula using the mass percentages or ratios, calculate the empirical formula mass, and then divide the molar mass by it to find the multiplying factor. Both methods are perfectly valid, but the direct method used here is often faster.

Q: What if the number of atoms comes out as a decimal?

In these types of problems, if you use the exact molar mass, the number of atoms should be very close to a whole number. If you get a significant decimal, you should recheck your calculations. Minor deviations (like 5.99) can be rounded to the nearest integer.

Q: Why did we multiply the fraction by 108?

108 g is the mass of one mole of the entire compound. Multiplying the weight fraction of an element by 108 gives the actual mass of that specific element present in one mole of the compound.

Q: Is the empirical formula always different from the molecular formula?

Not always. Sometimes the empirical formula mass is exactly equal to the molar mass. In such cases, the multiplying factor is 1, and the empirical formula is identical to the molecular formula.

Question

In an organic compound of molar mass $108 	ext{ g mol}^{-1}$ C, H and N atoms are present in $9 : 1 : 3.5$ by weight. Molecular formula can be

Options:

(A) $	ext{C}_6	ext{H}_8	ext{N}_2$

(B) $	ext{C}_7	ext{H}_{10}	ext{N}$

(C) $	ext{C}_5	ext{H}_6	ext{N}_3$

(D) $	ext{C}_4	ext{H}_{18}	ext{N}_3$

Sigma Percentile · Accepted Answer

### Decoding the Molecular Formula

Imagine you are a chemical detective, and you've just been handed a mysterious organic compound. You know its total molar mass is **$108 	ext{ g mol}^{-1}$**, and you have the weight ratio of its constituent elements: Carbon, Hydrogen, and Nitrogen, which is **$9 : 1 : 3.5$**. Your mission is to piece together its exact molecular formula. Let's embark on this analytical journey.

### The Power of Proportions

The weight ratio gives us the relative mass contribution of each element. To find the actual mass of each element in one mole of the compound, we first need to determine the total sum of these ratio parts.

By adding the parts together, we get:

$$ 	ext{Total parts} = 9 + 1 + 3.5 = 13.5 $$

This total of $13.5$ parts perfectly represents the entire molar mass of the compound, which is $108 	ext{ g}$. This is our master key. We can now unlock the exact mass of each element by finding what fraction of the total mass it occupies.

### Calculating Individual Masses and Atoms

Let's start with **Carbon (C)**. Carbon takes up $9$ parts out of the total $13.5$. To find its mass in one mole of the compound, we multiply its fraction by the total molar mass:

$$ 	ext{Mass of C} = \frac{9}{13.5} 	imes 108 = 72 	ext{ g} $$

Since the atomic mass of carbon is $12 	ext{ g mol}^{-1}$, we can find the number of carbon atoms by dividing its total mass by its atomic mass:

$$ 	ext{Number of C atoms} = \frac{72}{12} = 6 $$

Next, we move to **Hydrogen (H)**. It represents $1$ part out of $13.5$.

$$ 	ext{Mass of H} = \frac{1}{13.5} 	imes 108 = 8 	ext{ g} $$

With hydrogen's atomic mass being $1 	ext{ g mol}^{-1}$, the calculation is straightforward:

$$ 	ext{Number of H atoms} = \frac{8}{1} = 8 $$

Finally, let's analyze **Nitrogen (N)**. Its share is $3.5$ parts out of $13.5$.

$$ 	ext{Mass of N} = \frac{3.5}{13.5} 	imes 108 = 28 	ext{ g} $$

Nitrogen has an atomic mass of $14 	ext{ g mol}^{-1}$. Dividing its total mass by its atomic mass gives us the number of atoms:

$$ 	ext{Number of N atoms} = \frac{28}{14} = 2 $$

### The Final Assembly

We have successfully gathered all the pieces of our molecular puzzle. We have exactly **$6$ carbon atoms**, **$8$ hydrogen atoms**, and **$2$ nitrogen atoms**.

Putting them all together, the molecular formula of our mysterious organic compound is:

$$ \mathbf{C_6H_8N_2} $$

This elegant method of using weight ratios directly with the molar mass bypasses the need to calculate the empirical formula first, saving precious time during exams. However, it's always good to remember that the empirical formula here would be $	ext{C}_3	ext{H}_4	ext{N}$, which is the simplest whole-number ratio of the atoms.

Animated Solution for Chemistry - Basic Concepts in Chemistry: In an organic compound of molar mass $108 g mol^{- 1}$ C, H and N atoms are present in $9 : 1 : 3.5$ by weight. Molecular formula can be

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Visualized Solution

\text{Given Data}

\text{Total Parts by Weight}

\text{Mass and Atoms of Carbon}

\text{Mass and Atoms of Hydrogen}

\text{Mass and Atoms of Nitrogen}

\text{Molecular Formula}

\text{Empirical vs Molecular Formula}

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Decoding the Molecular Formula

The Power of Proportions

Calculating Individual Masses and Atoms

The Final Assembly

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