1 Answer

Gurjit · Answer 1 · 2021-10-23T06:31:43+0000

Answer: The empirical formula for the given compound is

Step-by-step explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen 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=13.86g

Mass of
H_2O=3.926g

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 13.86 g of carbon dioxide,
(12)/(44)* 13.86=3.78g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 3.926 g of water,
(2)/(18)* 3.926=0.436g of hydrogen will be contained.

Mass of oxygen in the compound = (5.00) - (3.78 + 0.436) = 0.784 g

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

Moles of Carbon =
$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=(3.78g)/(12g/mole)=0.315moles$

Moles of Hydrogen =
$\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(0.436g)/(1g/mole)=0.436moles$

Moles of Oxygen =
$\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=(0.784g)/(16g/mole)=0.049moles$

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.049 moles.

For Carbon =
$(0.315)/(0.049)=6.42\approx 6$

For Hydrogen =
$(0.436)/(0.049)=8.89\approx 9$

For Oxygen =
(0.049)/(0.049)=1

The ratio of C : H : O = 6 : 9 : 1

Hence, the empirical formula for the given compound is

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