# Question #b61a4

##### 1 Answer

#### Explanation:

Start by using the **ideal gas law equation** to find the **total number of moles** of gas present in the mixture.

#color(blue)(ul(color(black)(PV = nRT)))#

Here

#P# is the pressure of the gas#V# is the volume it occupies#n# is the number of moles of gas present in the sample#R# is theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

Rearrange to solve for

#PV = nRT implies n = (PV)/(RT)#

Plug in your values to find

#n = (2.46 color(red)(cancel(color(black)("atm"))) * 5.0color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (27 + 273.15)color(red)(cancel(color(black)("K"))))#

#n ~~ "0.5 moles"#

Now, you know that this mixture is **by mass**, which means that it contains

#28.4 color(red)(cancel(color(black)("g mixture"))) * overbrace("84.5 g C"/(100color(red)(cancel(color(black)("g mixture")))))^(color(blue)("= 84.5% carbon")) = "23.998 g C"#

Use the **molar mass** of carbon to determine how many *moles* of carbon are present in the sample

#23.998 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 1.998 ~~ "2 moles C"#

Notice that the two gases contain **equal numbers of moles** of carbon, since

#"C"_x"H"_8 implies "1 mole C"_x"H"_8 = xcolor(white)(.)"moles C"# #"C"_x"H"_10 implies "1 mole C"_x"H"_10 = xcolor(white)(.)"moles C"#

Now, if you take

#(0.5 - n)color(white)(.)"moles C"_x"H"_10#

You can also say that the sample contains

#overbrace((n * x)color(white)(.)"moles C")^(color(blue)("from C"_x"H"_8)) + overbrace([(0.5 - n) * x ]color(white)(.)"moles C")^(color(blue)("from C"_x"H"_10)) = "2 moles C"#

This is equivalent to

#color(red)(cancel(color(black)(n * x))) + 0.5 * x - color(red)(cancel(color(black)(n * x))) = 2#

which gets you

#0.5x = 2 implies x= 2/0.5 = 4#

Therefore, you can say that the two gases are

#"C"_4"H"_8" "# and#" " "C"_4"H"_10#