# Question 0557d

Aug 20, 2017

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

#### Explanation:

For starters, you know that because you're dealing with a hydrocarbon, i.e. a compound that only contains carbon and hydrogen, you can use the mass of the sample and the mass of carbon to determine the mass of hydrogen present in the sample.

overbrace("0.60 g")^(color(blue)("mass of carbon")) + overbrace(xcolor(white)(.)"g")^(color(blue)("mass of hydrogen")) = overbrace("0.70 g")^(color(blue)("mass of hydrocarbon"))

You can thus say that the sample contains

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

of hydrogen. To convert the masses to moles, use the molar masses of carbon and hydrogen, respectively.

$\text{For C: " 0.60 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12color(red)(cancel(color(black)("g")))) = "0.05 moles C}$

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

Now, to find the mole ratio that exists between carbon and hydrogen in the hydrocarbon, divide both values by the smallest one.

"For C: " (0.05 color(red)(cancel(color(black)("moles"))))/(0.05color(red)(cancel(color(black)("moles")))) = 1

"For H: " (0.10color(red)(cancel(color(black)("moles"))))/(0.05color(red)(cancel(color(black)("moles")))) = 2

This means that the hydrocarbon contains carbon and hydrogen in a $1 : 2$ mole ratio. Since this represents the smallest whole number ratio that can exist between the two elements in this hydrocarbon, you can say that the empirical formula of the hydrocarbon will be

${\text{C"_1"H"_2 implies "CH}}_{2} \to$ the empirical formula

Now, the molecular formula is always a multiple of the empirical formula.

$\text{molecular formula" = color(blue)(n) xx "empirical formula}$

This means that you have

("CH"_ 2)_ color(blue)(n) = "C"_ color(blue)(n)"H"_ (2color(blue)(n))

In order to find the value of $\textcolor{b l u e}{n}$, use the fact that the mass of $1$ mole of ${\text{C"_ color(blue)(n)"H}}_{2 \textcolor{b l u e}{n}}$ is equal to $\textcolor{b l u e}{n}$ times the mass of $1$ mole of ${\text{CH}}_{2}$.

The molar mass of ${\text{CH}}_{2}$ is equal to

${\text{12 g mol"^(-1) + 2 xx "1 g mol"^(-1) = "14 g mol}}^{- 1}$

Since you know that the molar mass of the hydrocarbon is equal to ${\text{70 g mol}}^{- 1}$, you can say that you have

$14 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))) * color(blue)(n) = 70 color(red)(cancel(color(black)("g mol}}^{- 1}}}}$

This will get you

$\textcolor{b l u e}{n} = \frac{70}{14} = 5$

Therefore, the molecular formula of the hydrocarbon is

color(darkgreen)(ul(color(black)(("CH"_ 2)_ 5 = "C"_ 5 "H"_ 10)))#