# Question #c0a3d

##### 2 Answers

#### 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,

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 **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

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

Now, you know that you're dealing with a **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

#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 **moles** of gas **molecular formula**, you need to figure out how many moles of each element you have in **mole** of gas

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)|)))#

#### Explanation:

Let the molecular formula of the Hydrocarbon be

So the ratio of masses of Carbon and Hydrogen present in HC as per its molecular formula is

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

This ratio being least the empirical formula of HC will be

Now let the molecular formula of HC following the empirical formula be

Again the vapor density of HC given is 35. So its molar mass =

Comparing the molar masses now we can write

Now we can write MF as

MF of HC is