Question #68096
1 Answer
Explanation:
For starters, you know that a molecule of benzene,
- six atoms of carbon,
#6 xx "C"# - six atoms of hydrogen,
#6 xx "H"#
This implies that
- six moles of carbon,
#6 xx "C"# - six moles of hydrogen,
#6 xx "H"#
Now, you know that benzene has a molar mass of
You also know that carbon has a molar mass of
This means that if you take a sample of
#6 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "72.066 g"#
of carbon. Now, it's important to realize that the percentage of carbon in benzene is the same regardless of the mass of the sample.
To find the percent composition of carbon in benzene, calculate the mass of carbon present in
#100 color(red)(cancel(color(black)("g C"_6"H"_6))) * "72.066 g C"/(78.11color(red)(cancel(color(black)("g C"_6"H"_6)))) = "92.26 g C"#
So, if every
#color(darkgreen)(ul(color(black)("% composition C = 92.26%")))#
carbon. I'll leave the asnwer rounded to four sig figs.
Consequently, can use the percent composition of benzene to calculate the mass of carbon present in your sample
#0.2 color(red)(cancel(color(black)("g C"_6"H"_6))) * "92.26 g C"/(100color(red)(cancel(color(black)("g C"_6"H"_6)))) = "0.18 g C"#
I'll leave this value rounded to four sig figs, but keep in mind that you only have one significant figure for the mass of benzene.