Question

Question #66091

Answer 1

Here's what I got.

Explanation:

Two things to keep in mind here

• the definition of a mole is given by Avogadro's constant
• we use subscripts to show how many atoms of a given element are present in one molecule or formula unit of a compound

So, you should know that in order to have exactly `1` mole of propane molecules, your sample must contain `6.022 * 10^(23)` molecules of propane.

This implies that in order to have `color(blue)(2)` moles of propane, you need to have `color(blue)(2)` times as many molecules of propane as you have in `1` mole of this compound.

You can thus say that your sample contains

`color(blue)(2) color(red)(cancel(color(black)("moles C"_3"H"_8))) * overbrace((6.022 * 10^(23)color(white)(.)"molecules C"_3"H"_8)/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))))^(color(purple)("Avogadro's constant"))`

`= color(darkgeeen)(ul(color(black)(1.2 * 10^(24)color(white)(.)"molecules C"_3"H"_8)))`

Now, a molecule of propane contains

• three atoms of carbon, `3 xx "C"`
• eight atoms of hydrogen, `8 xx "H"`

You know that this is the case because of the two subscripts used in the chemical formula of propane.

`"C"_3"H"_8 implies {(3 xx "C"), (8 xx "H") :}`

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You can thus say that your sample contains

`1.2 * 10^(24) color(red)(cancel(color(black)("molecules C"_3"H"_8))) * "3 atoms C"/(1color(red)(cancel(color(black)("molecule C"_3"H"_8))) )`

`= color(darkgreen)(ul(color(black)(3.6 * 10^(24)color(white)(.)"atoms C")))`

and

`1.2 * 10^(24) color(red)(cancel(color(black)("molecules C"_3"H"_8))) * "8 atoms H"/(1color(red)(cancel(color(black)("molecule C"_3"H"_8))) )`

`= color(darkgreen)(ul(color(black)(9.6 * 10^(24)color(white)(.)"atoms H")))`

I'll leave the answers rounded to two sig figs, but don't forget that you only have one significant figure for the number of moles of propane.