Question

9
An organic compound X, which contains carbon, hydrogen and oxygen only, has an M, of about 85. When 0.43
g of X are burned in excess oxygen, 1.10 g of carbon dioxide and 0.45 g of water are formed. Find the empirical
and molecular formulae of compound-X.​

Answer 1

Explanation:

Its empirical formula is CH2. One molecule of ethylene (molecular formula C2H4) contains two atoms of carbon and four atoms of hydrogen. Its empirical formula is CH2.

Answer 2

Given:

Molar mass of compound X $=85$

Mass of X burnt $=0.43 g$

Mass of carbon dioxide produced $=1.10 g$

Mass of water produced $=0.45g$

To Find: Empirical  and molecular formulae of compound X.

Solution:

Number of moles can be given as:

$\[No.\,of\,moles = \frac{{Mass}}{{Molar\,mass}}\]$

Therefore,

Number of moles of X burnt

$\[No.\,of\,moles = \frac{{0.43\,g}}{{85}}\]$

$\[No.\,of\,moles = \,0.005\]$

Number of moles of carbon dioxide produced

$\[No.\,of\,moles = \frac{{1.10\,g}}{{44}}\]$

$\[No.\,of\,moles = \,0.025\]$

Number of moles of water produced

$\[No.\,of\,moles = \frac{{0.45\,g}}{{18}}\]$

$\[No.\,of\,moles = \,0.025\]$

Since 0.005 moles of X gives 0.025 moles of carbon dioxide

Therefore,

One mole of X will give 0.025/0.005 = 5 moles of carbon dioxide

Similarly,

One mole of X will give 0.025/0.005 = 5 moles of water

The reaction of combustion of X can be given as:

$\[{C_x}{H_y}{O_z} + w{O_2} \to 5C{O_2} + 5{H_2}O\]$

On comparing both sides of the above equation, we have

$\[x = 5\]$

$\[y = 10\]$

As the molecular mass of the given compound X is 85, it can be represented as:

$\[5 \times 12 + 10 \times 1 + z \times 16 = 85\]$

Therefore, 'z' can be given as:

$\[60 + 10 + 16z = 85\]$

$\[16z = 85 - 70\]$

$\[z = 15/16\]$

$\[z = 0.9375 \approx 1\]$

Hence, the molecular formula of the compound X is $\[{C_5}{H_{10}}O\]$. the empirical formula is also $\[{C_5}{H_{10}}O\]$ as the ratio of C, H and O cannot be simplified further.