Question

Question #68096

Answer 1

`92.26% ->` regardless of the mass of the sample.

Explanation:

For starters, you know that a molecule of benzene, `"C"_6"H"_6`, contains

• six atoms of carbon, `6 xx "C"`
• six atoms of hydrogen, `6 xx "H"`

This implies that `1` mole of benzene will contain

• six moles of carbon, `6 xx "C"`
• six moles of hydrogen, `6 xx "H"`

Now, you know that benzene has a molar mass of `"78.11 g mol"^(-1)` , which implies that `1` mole of benzene has a mass of `"78.11 g"`.

You also know that carbon has a molar mass of `"12.011 g mol"^(-1)`, which tells you that `1` mole of carbon has a mass of `"12.011 g"`.

This means that if you take a sample of `"78.11 g"` of benzene, the equivalent of `1` mole of benzene, you know for a fact that it contains

`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 g"` of benzene.

`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 `"100 g"` of benzene contain `"92.26 g"` of carbon, you can say that benzene is

`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.