Question #fbbce

1 Answer
Jun 17, 2017

Answer:

The empirical formula is #"C"_6"H"_7#.

Explanation:

We can calculate the masses of #"C"# and #"H"# from the masses of their oxides.

#"Mass of C" = 3.02 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.8241 g C"#

#"Mass of H" = 0.72 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0806 g H"#

Now, we must convert these masses to moles and find their ratios.

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

#"Element"color(white)(X) "Mass/g"color(white)(Xm) "Moles"color(white)(mll) "Ratio"color(white)(m)×6 color(white)(mll)"Integers"#
#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.8241 color(white)(mll)"0.06862" color(white)(Xl)1color(white)(mmm)6color(white)(Xmmmm)6#
#color(white)(ll)"H" color(white)(XXXXl)0.0806 color(white)(mll)"0.0800" color(white)(mlll)1.165color(white)(m)6.990 color(white)(XXX)7#

The empirical formula is #"C"_6"H"_7"#.