Question #8cd77

1 Answer
Apr 19, 2017

Answer:

The empirical formula is #"C"_5"H"_12#.

Explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of #"C"# to #"H"#.

#"Mass of C = 12 g"#

#"Mass of alkane = mass of C + mass of H"#

#"14.4 g = 12 g + mass of H"#

#"Mass of H = (14.4 – 12) g = 2.4 g"#

#"Moles of C" = 12 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01color(red)(cancel(color(black)( "g C")))) = "0.999 mol C"#

#"Moles of H "= 2.4 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "2.38 mol H"#

To get this into an integer ratio, we divide both numerator and denominator by the smaller value.

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

#bb("Element"color(white)(Ag) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(m)×5color(white)(m)"Integers")#
#color(white)(mll)"C" color(white)(XXXmm)12 color(white)(Xmml)0.999color(white)(Xmm)1color(white)(mmmm)5color(white)(mmmll)5#
#color(white)(mll)"H" color(white)(XXXXmll)2.4 color(white)(mml)"2.38 color(white)(Xmmll)2.38 color(white)(mm)11.9color(white)(mm)12#

There are 5 mol of #"C"# for 12 mol of #"H"#.

The empirical formula of the alkane is #"C"_5"H"_12#.

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