Question #a782a

1 Answer
Mar 7, 2016

Answer:

The empirical formula is #"CH"_2#.

Explanation:

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

#"Mass of C" = 2.64 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.7204 g C"#

#"Mass of H" = 1.08 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.1208 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)(mll)"Integers"#
#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.7204 color(white)(mll)"0.05998" color(white)(Xl)1color(white)(Xmmmm)1#
#color(white)(ll)"H" color(white)(XXXXl)0.1208 color(white)(mll)"0.1198" color(white)(mlll)1.998 color(white)(XXX)2#

The empirical formula is #"CH"_2"#.