# Question b61a4

Sep 7, 2017

${\text{C"_4"H}}_{8}$ and ${\text{C"_4"H}}_{10}$.

#### Explanation:

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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

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 the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Rearrange to solve for $n$

$P V = n R T \implies n = \frac{P V}{R T}$

Plug in your values to find

$n = \left(2.46 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 5.0color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (27 + 273.15)color(red)(cancel(color(black)("K}}}}\right)$

$n \approx \text{0.5 moles}$

Now, you know that this mixture is 84.5% carbon 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

• $\text{C"_x"H"_8 implies "1 mole C"_x"H"_8 = xcolor(white)(.)"moles C}$
• $\text{C"_x"H"_10 implies "1 mole C"_x"H"_10 = xcolor(white)(.)"moles C}$

Now, if you take $n$ to be the number of moles of ${\text{C"_x"H}}_{8}$ and present in the sample, you can say that the sample contains

$\left(0.5 - n\right) \textcolor{w h i t e}{.} {\text{moles C"_x"H}}_{10}$

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

$\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot x}}} + 0.5 \cdot x - \textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot x}}} = 2$

which gets you

$0.5 x = 2 \implies x = \frac{2}{0.5} = 4$

Therefore, you can say that the two gases are

$\text{C"_4"H"_8" }$ and ${\text{ " "C"_4"H}}_{10}$