Question #0557d
1 Answer
Explanation:
For starters, you know that because you're dealing with a hydrocarbon, i.e. a compound that only contains carbon and hydrogen, you can use the mass of the sample and the mass of carbon to determine the mass of hydrogen present in the sample.
#overbrace("0.60 g")^(color(blue)("mass of carbon")) + overbrace(xcolor(white)(.)"g")^(color(blue)("mass of hydrogen")) = overbrace("0.70 g")^(color(blue)("mass of hydrocarbon"))#
You can thus say that the sample contains
#"0.70 g " - " 0.60 g = 0.10 g"#
of hydrogen. To convert the masses to moles, use the molar masses of carbon and hydrogen, respectively.
#"For C: " 0.60 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12color(red)(cancel(color(black)("g")))) = "0.05 moles C"#
#"For H: " 0.10 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1color(red)(cancel(color(black)("g")))) = "0.10 moles H"#
Now, to find the mole ratio that exists between carbon and hydrogen in the hydrocarbon, divide both values by the smallest one.
#"For C: " (0.05 color(red)(cancel(color(black)("moles"))))/(0.05color(red)(cancel(color(black)("moles")))) = 1#
#"For H: " (0.10color(red)(cancel(color(black)("moles"))))/(0.05color(red)(cancel(color(black)("moles")))) = 2#
This means that the hydrocarbon contains carbon and hydrogen in a
#"C"_1"H"_2 implies "CH"_2 -># the empirical formula
Now, the molecular formula is always a multiple of the empirical formula.
#"molecular formula" = color(blue)(n) xx "empirical formula"#
This means that you have
#("CH"_ 2)_ color(blue)(n) = "C"_ color(blue)(n)"H"_ (2color(blue)(n))#
In order to find the value of
The molar mass of
#"12 g mol"^(-1) + 2 xx "1 g mol"^(-1) = "14 g mol"^(-1)#
Since you know that the molar mass of the hydrocarbon is equal to
# 14 color(red)(cancel(color(black)("g mol"^(-1)))) * color(blue)(n) = 70 color(red)(cancel(color(black)("g mol"^(-1))))#
This will get you
#color(blue)(n) = 70/14 = 5#
Therefore, the molecular formula of the hydrocarbon is
#color(darkgreen)(ul(color(black)(("CH"_ 2)_ 5 = "C"_ 5 "H"_ 10)))#