Question

An organic substance contains 62.06% C and 10.35% H. If its vapor density is 29, find the atomic formula of that organic matter. ​

Answer 1

Answer:

The atomic formula of the given organic matter is  $C_3H_6O$·

Explanation:

Given that,

The % of Carbon in the organic compound $=$ $62.06$ %

The % of Hydrogen in the organic compound  $=$ $10.35$ %

The rest of % will always be taken as oxygen·

Therefore,

The % of Oxygen in the organic compound  $=$ $100\ -\ 72.41$ $=\ 27.59$ %

The vapour density of the organic compound  $=$  $29$

From this vapour density, we can find out the molecular mass of the organic compound·

The relation that connects molecular mass and vapour density is,

Molecular mass  $=$ $2$ × vapour density

 ∴     Molecular mass  $=$ $2$ × $29$

 ⇒    Molecular mass  $=$ $58\ g\ mol^-^1$

To find out the atomic formula,

First,

Convert these mass percentages into grams·

For that, we assume the starting material as $100$ g of the compound·

Thus,

$62.06$ % of carbon means  $62.06$ g of carbon in $100$ g of the compound·

$10.35$ % of hydrogen means $10.35$ g of hydrogen in $100$ g of the compound·

$27.59$ % of oxygen means $27.59$ g of oxygen in  $100$ g of the compound·

Then,

Find out the no of moles of each element·

No of moles of carbon $=$ $\frac{62.06}{12}$ $=\ 5.17$

No of moles of hydrogen $=$ $\frac{10.35}{1}$ $=\ 10.35$

No of moles of oxygen $=$ $\frac{27.59}{16}$ $=\ 1.72$

Then,

Divide the no of moles by the smallest number to get the atomic ratio·

$\frac{5.17}{1.72} \ =\ 3$

$\frac{10.35}{1.72} \ =\ 6$

$\frac{1.72}{1.72} \ =\ 1$

Therefore,

The ratio of C $:$ H  $:$ O  $=\ \ 3\ :\ 6\ :\ 1$

Therefore,

The empirical formula of the organic compound is $C_3H_6O$·

The empirical formula mass $=\ (\ 3\ *\ 12\ )\ +\ (6\ *\ 1)\ +\ (16)\ \ =\ \ 58$

So,

The empirical formula mass  $=$ Molecular formula mass

 The empirical formula  $=$ molecular formula

Hence,

The molecular formula  $=$  $C_3H_6O$·