Question #c0a3d

2 Answers
May 4, 2016

Answer:

#"C"_5"H"_10#

Explanation:

The idea here is that you can sue the vapor density of a gas to determine its molar mass.

Vapor density is calculated by looking at how many molecules of gas would occupy a given volume compared with the number of molecules of hydrogen gas, #"H"_2#, that would occupy the same volume under the same conditions for pressure and temperature.

Simply put, vapor density tells you the density of a gas relative to the density of hydrogen gas.

If you use the fact that the mass of a gas, i.e. how many molecules it contains, can be expressed using its molar mass, you can say that

#color(blue)(|bar(ul(color(white)(a/a)"vapor density" = "molar mass of the gas"/"molar mas of H"_2color(white)(a/a)|)))#

In your case, the unknown hydrocarbon is said to have a vapor density equal to #35#. This means that its molar mass will be

#35 = M_ "M gas"/M_ ("M H"_ 2) implies M_ "M gas" = 35 xx M_ ("M H"_2)#

If you take the molar mass of hydrogen gas to be equal to #"2 g mol"^(-1)#, you will end up with

#M_"M gas" = 35 * "2 g mol"^(-1) = "70 g mol"^(-1)#

Now, you know that you're dealing with a #"0.70-g"# sample of this gas #"X"#. Use its molar mass to determine how many moles you have in this sample

#0.70color(red)(cancel(color(black)("g"))) * "1 mole gas X"/(70color(red)(cancel(color(black)("g")))) = "0.010 moles X"#

This sample is said to contain #"0.60 g"# of carbon and

#m_"gas" = m_(C) + m_(H)#

#m_(H) = "0.70 g" - "0.60 g" = "0.10 g H"#

Use the molar masses of the elements to determine how many moles of each you have

#"For C: " 0.60 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12color(red)(cancel(color(black)("g")))) = "0.050 moles C"#

#"For H: " 0.10 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1color(red)(cancel(color(black)("g")))) = "0.10 moles H"#

This is how many moles of each element you have in #0.010# moles of gas #"X"#. In order to find the compound's molecular formula, you need to figure out how many moles of each element you have in #1# mole of gas #"X"#.

In your case, you will have

#1 color(red)(cancel(color(black)("mole X"))) * "0.050 moles C"/(0.010color(red)(cancel(color(black)("moles X")))) = "5 moles C"#

#1 color(red)(cancel(color(black)("mole X"))) * "0.10 moles H"/(0.010color(red)(cancel(color(black)("moles X")))) = "10 moles H"#

Therefore, the hydrocarbon's molecular formula is

#color(green)(|bar(ul(color(white)(a/a)"C"_5"H"_10color(white)(a/a)|)))#

May 4, 2016

Answer:

#C_5H_10#

Explanation:

Let the molecular formula of the Hydrocarbon be #C_xH_y# . ,Where x= no. Of C-atom and y = no of H-atom present in one molecule of HC.We know that the ”atomic mass of “#C =12g/(mol) # and “atomic mass of H”#=1g/(mol)#

So the ratio of masses of Carbon and Hydrogen present in HC as per its molecular formula is #(12x:y)#

Now by the problem
0.7g HC contains 0.6g Carbon , So the rest amount i.e o.1g is Hydrogen. Hence the ratio of masses of C and H is #0.6:0.1=6:1# Comparing this ratio with the raio obtained from formula we can write:
#(12x:y)=6:1=>(12x)/y=6/1=>x/y= 1/2#
This ratio being least the empirical formula of HC will be #CH_2#

Now let the molecular formula of HC following the empirical formula be
#(CH_2)_n# , where n = common multlple. As per this MF the molar mass of HC is #(12+1*2)*n=14n#
Again the vapor density of HC given is 35. So its molar mass = #2xx"Vapor Density" g/(mol)=2*35=70g/(mol)#

Comparing the molar masses now we can write
#14n=70=>n=5#
Now we can write MF as #(CH_2)n=(CH_2)*5=C_5H_10#

MF of HC is #C_5H_10#