Question

A 0.1g sample of a compound when burnt completely in oxygen produced 0.191g of $CO_{2}$ and 0.1172g of $H_{2}O$. What is the empirical formula of the compound?

Answer 1

From the given,

Mass of ${ CO }_{ 2 }$  = 44 g

Therefore, 44 g of ${ CO }_{ 2 }$  contains 12 g of carbon.

Let's calculate how much weight of C present in 0.191 g of ${ CO }_{ 2 }$

$0.191\quad g\quad of\quad { CO }_{ 2 }\quad =\quad \frac { 12\quad \times \quad 0.191 }{ 44 } \quad =\quad 0.05209\quad g\quad of\quad C$

Given mass of ${ H }_{ 2 }O$ = 0.1172 g

Therefore, 18 g of ${ H }_{ 2 }O$ contains 2 g of Hydrogen

Let's calculate how much weight of H present in 0.1172 g of H.

$0.1172\quad g\quad of\quad { H }_{ 2 }O\quad =\quad \frac { 2\quad \times \quad 0.1172 }{ 18 } \quad =\quad 0.01302g\quad of\quad H$

Total mass of carbon and hydrogen = 0.05209 + 0.01302 = 0.06511 g

Mass of Oxygen = 0.1 - 0.06511 = 0.03489 g

Let's convert mass into moles of each element.

$For\quad carbon\quad =\quad \frac { Weight\quad of\quad carbon }{ Molar\quad mass } \quad =\quad \frac { 0.05209 }{ 12 } \quad =\quad 0.00434\quad mol$

$For\quad Hydrogen\quad =\quad \frac { Weight\quad of\quad Hydrogen }{ Molar\quad mass } \quad =\quad \frac { 0.01302 }{ 1 } \quad =\quad 0.01302\quad mol$

$For\quad Oxygen\quad =\quad \frac { Weight\quad of\quad oxygen }{ Molar\quad mass } \quad =\quad \frac { 0.03489 }{ 16 } \quad =\quad 0.00218\quad mol$

Simple ratio of each element will be

$C\quad =\quad \frac { 0.004 }{ 0.00218 } \quad =\quad 2;\quad \\ \\H\quad =\quad \frac { 0.01302 }{ 0.00218 } \quad =\quad 6;\quad \\ \\O\quad =\quad \frac { 0.00218 }{ 0.00218 } \quad =\quad 1$

Therefore, empirical formula of compound will be ${ C }_{ 2 }{ H }_{ 6 }O$