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! :)

2 Answers
Nov 14, 2017

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...

Nov 15, 2017

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#.]