Question

The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.751 g sample of ether was combusted in an oxygen rich environment to produce 4.159 g of CO 2 ( g ) and 2.128 g of H 2 O ( g ) .
Insert subscripts to complete the empirical formula of ether.

Answer 1

Answer: The empirical formula for the given compound is $C_{4}H_{10}O$

Explanation:

The chemical equation for the combustion of ether follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=4.159g$

Mass of $H_2O=2.128g$

Mass of sample = 1.751 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 4.159 g of carbon dioxide, $\frac{12}{44}\times 4.159=1.134g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.128 g of water, $\frac{2}{18}\times 2.128=0.236g$ of hydrogen will be contained.

Mass of oxygen in the compound = (1.751) - (1.134 + 0.236) = 0.381 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.134g}{12g/mole}=0.0945moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.236g}{1g/mole}=0.236moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.381g}{16g/mole}=0.0238moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0238 moles.

For Carbon = $\frac{0.0945}{0.0238}=3.97\approx 4$

For Hydrogen = $\frac{0.236}{0.0238}=9.91\approx 10$

For Oxygen = $\frac{0.0238}{0.0238}=1$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is $C_{4}H_{10}O$