Question

The combustion of 40.17 g of a compound which contain s only 𝐶,𝐻,𝐶𝑙 and 𝑂 yields 58.67 g of CO2 and 15 g of 𝐻2𝑂. Another sample of the compound with a mass of 75.00 g is found to contain 22.06 g of 𝐶𝑙. What is the empirical formula ?

Answer 1

The empirical formula is `"C"_4"H"_5"ClO"_2`.

Explanation:

We must calculate the masses of `"C, H, Cl"` and `"O"` from the masses given.

`"Mass of C" = 58.67 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "16.01 g C"`

`"Mass of H" = 15 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "1.68 g H"`

`"Mass of Cl" = 40.17 color(red)(cancel(color(black)("g compd" )))× "22.06 g Cl"/(75.00 color(red)(cancel(color(black)("g compd")))) = "11.82 g Cl"`

`"Mass of O" = "Mass of compound -Mass of C - Mass of H - Mass of Cl" = "(40.17 - 16.01 - 1.68 - 11.82) g" = "10.67 g"`

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

`ulbb("Element"color(white)(X) "Mass/g"color(white)(X) "Amt/mol"color(white)(m) "Ratio"color(white)(m)"Integers")`
`color(white)(mm)"C"color(white)(mmmll)16.01color(white)(mmml)"1.333"color(white)(Xml)4.000color(white)(Xmmll)4`
`color(white)(mm)"H" color(white)(XXXml)1.68 color(white)(mmml)1.66 color(white)(mmll)4.995color(white)(Xmmm)5`
`color(white)(mm)"Cl" color(white)(XXXll)11.82 color(white)(mmm)0.3334 color(white)(mm)1color(white)(Xmmmmll)1`
`color(white)(mm)"O" color(white)(XXXll)10.67 color(white)(mmm)0.6669 color(white)(mm)2.001color(white)(Xmmll)2`

The empirical formula is `"C"_4"H"_5"ClO"_2`.