Question

A natural gas has the following composition by volume CH4 = 83.5%, C2 H6 =12.5% and N2=4%. Estimate the following: Estimate the Composition in mole % and Composition in weight %?? .

Answer 1

Here's what I get.

Explanation:

This is a long question, so I will show only the calculations for methane. You can use them as a pattern for the ethane and nitrogen calculations.

The results are in the table below.

`ulbb(color(white)(l)"Item"color(white)(mm)"CH"_4color(white)(m)"C"_2"H"_6color(white)(m) "N"_2color(white)(ml)"Total")`
`"Vol. %"color(white)(mml)83.6color(white)(mll)12.5color(white)(mml)4color(white)(mml)100`
`ul("Mole %" color(white)(mlll)83.6color(white)(mll)12.5color(white)(mml)4color(white)(mml) 100)`

`"Moles" color(white)(mmll)83.6color(white)(mll)12.5color(white)(mml)4color(white)(mml) 100`
`ul("Mass/g"color(white)(m) 1341color(white)(mm)375.8color(white)(m)110color(white)(mm)1829)`
`"Mass %" color(white)(mlll)73.3color(white)(mll)20.6color(white)(mml)6color(white)(mml) 100`

Mole percent

Since `n∝V`, the volume percent is the same as the mole percent,

Mass percent

Assume that we have 100 mol of gas. Then we have

`"83.6 mol CH"_4, "12.5 mol C"_2"H"_6`, and `"4 mol N"_2`

`"Mass of CH"_4 = 83.6 color(red)(cancel(color(black)("mol CH"_4))) × ("16.04 g CH"_4)/(1color(red)(cancel(color(black)("mol CH"_4)))) = "1341 g CH"_4`

Calculate the masses of ethane and nitrogen in the same way.

`"Total mass = (1341 + 375.8 + 110) g = 1829 g"`

`"% CH"_4 = ("Mass of CH"_4)/("Total mass") × 100 % = (1341 color(red)(cancel(color(black)("g"))))/(1828 color(red)(cancel(color(black)("g")))) × 100 % = 73.3 %`

Calculate the mass percent of ethane and nitrogen in the same way.