Question

Write the general chemical equation for this equation ?

A 1.50 g sample of hydrocarbon (CxHy) undergoes complete combustion to produce 4.40 g of C02 and 2.70 g of H20... A little bonus question - What is the empirical formula of this compound?

God bless your soul for trying to figure this one out! :)

Answer 1

See the answer below...

Explanation:

We know the atomic mass of the Carbon,Hydrogen and Oxyzen atoms.

Carbon`->color(red)(12`
Hydrogen`->color(red)(1`
Oxyzen`->color(red)(16`

We can write the common equation of the reaction in this form:-
`color(blue)(C_xH_y+` `color(red)((2x+y/2)/2)color(blue)(O_2`=`color(red)xcolor(blue)(CO_2+` `color(red)(y/2` `color(blue)(H_2O`

[WE HAVE TO DETERMINE THIS BY CHECKING THE MASS OF BOTH SIDE]

CASE no. 1:-

Number of moles of carbon present in `4.4`gram of `CO_2` is `4.4/44=1/10` mole. [No of moles=`"Given mass"/"Molecular weight"xx"Number of atoms in Molecule"`]

CASE no. 2:-

Number of moles of hydrogen present in `2.70` gram of `H_2O` is `2.70/18xx2=3/10` [Same as before...]

HENCE,we get that `x/y=(1/10)/(3/10)=1/3`

Hence the compound is `C_2H_6`...as `CH_3` does not exist...

Answer 2

Two options:

`"C"_x"H"_(3x)(g) + 7/4x"O"_2(g) -> x"CO"_2(g) + 3/2x"H"_2"O"(g)`

`"C"_x"H"_(y)(g) + (x + y/4)"O"_2(g) -> x"CO"_2(g) + y/2"H"_2"O"(g)`

The first one is more accurate, since we DO have the restriction that there are three mols of `"H"` for every mol of `"C"` from the mol ratio among the products.

---

Here's another way you can do it. It doesn't require trial and error with `x` and `y` as much.

From conservation of mass, then from the mass of the products, we can determine the mass of carbon and of hydrogen you should get.

`"4.40 g CO"_2 xx "1 mol"/("44.01 g CO"_2) xx "1 mol C"/"1 mol CO"_2 xx "12.011 g C"/"1 mol C"`

`= "1.20 g C" = "0.100 mol C"`

`"2.70 g H"_2"O" xx "1 mol"/("18.015 g H"_2"O") xx "2 mol H"/("1 mol H"_2"O") xx "1.008 g H"/"1 mol H"`

`= "0.302 g H" = "0.300 mol H"`

So, you would get

`"mol C"/"mol H" = 1/3`

And the empirical formula `"CH"_3`. To find the molar mass, you need the mols contained in that `"1.50 g"`. That's not so obvious, but the molecular formula has to be based on an integer multiple of `"CH"_3` fragments.

So it could only be `"C"_2"H"_6`, `"C"_3"H"_9`, etc. However, only the first choice makes physical sense. Simple hydrocarbons (only `"C"` and `"H"`) only consist of `"C"_n"H"_(2n)`, `"C"_n"H"_(2n+2)`, or `"C"_n"H"_(2n-2)`.

If you didn't know that, you can try balancing right now.

Postulate that an integer `x` amount of mols of `"CH"_3` atoms will multiply the fragment to give a real molecule to get:

`"C"_x"H"_(3x)(g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O"(g)`

where `x = 2, 3, . . .`.

So then we want to balance this knowing that.

`"C"_x"H"_(3x)(g) + (x + 3/4x)"O"_2(g) -> x"CO"_2(g) + 3/2x"H"_2"O"(g)`

`=> color(blue)("C"_x"H"_(3x)(g) + (7/4x)"O"_2(g) -> x"CO"_2(g) + 3/2x"H"_2"O"(g))`

The molar mass of the hydrocarbon would be `15.0347x`. There is no such thing as conservation of mols before you balance a reaction, so this is as general as we can get unless we redefine `3x` as `y`. That gives:

`=> color(blue)("C"_x"H"_(y)(g) + (x + y/4)"O"_2(g) -> x"CO"_2(g) + y/2"H"_2"O"(g))`

which is even more general, although you have a greater restriction in knowing that you must have a `3:1`, `"H":"C"` ratio.

[Note that `x + y/4 = (2x + y/2)/2`.]