# Question c0a3d

May 4, 2016

${\text{C"_5"H}}_{10}$

#### 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, ${\text{H}}_{2}$, 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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{vapor density" = "molar mass of the gas"/"molar mas of H}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 ${\text{2 g mol}}^{- 1}$, you will end up with

${M}_{\text{M gas" = 35 * "2 g mol"^(-1) = "70 g mol}}^{- 1}$

Now, you know that you're dealing with a $\text{0.70-g}$ sample of this gas $\text{X}$. 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 $\text{0.60 g}$ of carbon and

${m}_{\text{gas}} = {m}_{C} + {m}_{H}$

${m}_{H} = \text{0.70 g" - "0.60 g" = "0.10 g H}$

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

$\text{For C: " 0.60 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12color(red)(cancel(color(black)("g")))) = "0.050 moles C}$

$\text{For H: " 0.10 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1color(red)(cancel(color(black)("g")))) = "0.10 moles H}$

This is how many moles of each element you have in $0.010$ moles of gas $\text{X}$. 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 $\text{X}$.

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

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{C"_5"H}}_{10} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

May 4, 2016

${C}_{5} {H}_{10}$

#### Explanation:

Let the molecular formula of the Hydrocarbon be ${C}_{x} {H}_{y}$ . ,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 “$C = 12 \frac{g}{m o l}$ and “atomic mass of H”$= 1 \frac{g}{m o l}$

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

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:
$\left(12 x : y\right) = 6 : 1 \implies \frac{12 x}{y} = \frac{6}{1} \implies \frac{x}{y} = \frac{1}{2}$
This ratio being least the empirical formula of HC will be $C {H}_{2}$

Now let the molecular formula of HC following the empirical formula be
${\left(C {H}_{2}\right)}_{n}$ , where n = common multlple. As per this MF the molar mass of HC is $\left(12 + 1 \cdot 2\right) \cdot n = 14 n$
Again the vapor density of HC given is 35. So its molar mass = $2 \times \text{Vapor Density} \frac{g}{m o l} = 2 \cdot 35 = 70 \frac{g}{m o l}$

Comparing the molar masses now we can write
$14 n = 70 \implies n = 5$
Now we can write MF as $\left(C {H}_{2}\right) n = \left(C {H}_{2}\right) \cdot 5 = {C}_{5} {H}_{10}$

MF of HC is ${C}_{5} {H}_{10}$