Question

asked Jul 23, 2020 148k views

1 Answer

Dave Poole · Answer 1 · 2020-07-29T02:37:30+0000

Answer: The empirical formula for the given organic 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=0.2829g

Mass of
H_2O=0.1159g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

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

So, in 0.2829 g of carbon dioxide,
(12)/(44)* 0.2829=0.077g of carbon will be contained.

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

So, in 0.1159 g of water,
(2)/(18)* 0.1159=0.013g of hydrogen will be contained.

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

Moles of Carbon =
$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=(0.077g)/(12g/mole)=0.0064moles$

Moles of Hydrogen =
$\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(0.013g)/(1g/mole)=0.013moles$

Moles of Oxygen =
$\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=(0.0105g)/(16g/mole)=0.0066moles$

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0064 moles.

For Carbon =
(0.0064)/(0.0064)=1

For Hydrogen =
$(0.013)/(0.0064)=2.03\approx 2$

For Oxygen =
$(0.0066)/(0.0064)=1.03\approx 1$

The ratio of C : H : O = 1 : 2 : 1

Hence, the empirical formula for the given compound is

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