# 2.00 grams of a straight-chained hydrocarbon contains 1.68grams of carbon, the rest hydrogen. 1. Calculate the empirical formula of this hydrocarbon.?

Jun 27, 2018

The empirical formula is ${\text{C"_4"H}}_{9}$.

#### Explanation:

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

We must convert the masses of $\text{C}$ and $\text{H}$ to moles and then find the ratio.

Step 1. Calculate the mass of $\text{H}$

$\text{Mass of H" = "mass of hydrocarbon - mass of C" = "2.00 g - 1.68 g = 0.32 g}$

Step 2. Calculate the moles of $\text{C}$ and $\text{H}$

$\text{Moles of C" = 1.68 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "0.1399 mol C}$

$\text{Moles of H" = 0.32 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "0.317 mol H}$

From this point on, I like to summarize the calculations in a table.

$\boldsymbol{\underline{\text{Element"color(white)(m) "Mass/g"color(white)(m) "Moles"color(white)(m) "Ratio"color(white)(m)×4color(white)(m)"Integers}}}$
$\textcolor{w h i t e}{m m} \text{C} \textcolor{w h i t e}{X X X m} 1.68 \textcolor{w h i t e}{X m l l} 0.1399 \textcolor{w h i t e}{X m} 1 \textcolor{w h i t e}{m m m l} 4 \textcolor{w h i t e}{m m m m l} 4$
$\textcolor{w h i t e}{m m} \text{H} \textcolor{w h i t e}{X X X m} 0.32 \textcolor{w h i t e}{m m l l} 0.317 \textcolor{w h i t e}{X m l} 2.27 \textcolor{w h i t e}{m m} 9.08 \textcolor{w h i t e}{m m m} 9$

The molar ratio is $\text{C:H = 4:9}$.

The empirical formula is ${\text{C"_4"H}}_{9}$.

Here is a video that illustrates how to determine an empirical formula.