Question #79408

1 Answer
Feb 23, 2017

Answer:

The empirical formula is #"C"_6"H"_13"O"_3#.

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"# to #"O"#.

Your compound contains 54.53 % #"C"#, 9.15 % #"H"#, and 36.32 % #"O"#.

Assume that you have 100 g of sample.

Then it contains 54.53 g of #"C"#, 9.15 g of #"H"#, and 36.32 g of #"O"#.

#"Moles of C" = 54.53 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "4.540 mol B"#

#"Moles of H" = 9.15 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "9.077 mol H"#

#"Moles of O" = 36.32 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "2.270 mol O"#

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

#bb("Element"color(white)(m) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mm)"×3"color(white)(mm)"Integers")#
#color(white)(m)"C" color(white)(XXXmm)54.53 color(white)(Xm)4.540 color(white)(mmll)2.000color(white)(mml)6.000color(white)(mmml)6#
#color(white)(m)"H" color(white)(XXXXmll)9.15 color(white)(mm)9.077 color(white)(Xmll)4.304 color(white)(mll)12.91color(white)(mmml)13#
#color(white)(m)"O"color(white)(XXXmm)36.32 color(white)(Xm)2.270color(white)(mmll)1color(white)(mmmml)3color(white)(mmmmml)3#

The empirical formula is #"C"_6"H"_13"O"_3#.