Question

Question #54f82

Answer 1

`"0.6777 moles C"_12"H"_22"O"_11`

Explanation:

Your strategy here will be to use Avogadro's number to convert the number of atoms of carbon to moles of carbon, then use the molecular formula of sucrose, `"C"_12"H"_22"O"_11`, to find how many moles of this compound would contain that many moles of carbon.

So, you know that you're dealing with a sample of sucrose that is known to contain `4.897 * 10^(24)` atoms of carbon, `"C"`.

In order to convert this to moles of carbon, use the fact that a mole of atoms is defined as a collection of `6.022 * 10^(23)` atoms.

`color(blue)(|bar(ul(color(white)(a/a)"1 mole atoms" = 6.022 * 10^(23)"atoms"color(white)(a/a)|)) ->)` Avogadro's number

You can say that the sample of sucrose contains

`4.897 * 10^(24)color(red)(cancel(color(black)("atoms"))) * "1 mole C"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms")))) = "8.132 moles C"`

Now that you know how many moles of carbon you have in your sample, use the fact that one molecule of sucrose contains

• twelve atoms of carbon, `12 xx "C"`
• twenty two atoms of hydrogen, `22 xx "H"`
• eleven atoms of oxygen, `11 xx "O"`

The carbon atoms are shown here in black, the oxygen atoms in red, and the hydrogen atoms in white.

So, if one molecule of sucrose contains `12` atoms of carbon, it follows that one mole of sucrose will contain `12` moles of carbon.

This means that the number of moles of carbon you have here will be enough to make up

`8.132 color(red)(cancel(color(black)("moles C"))) * "1 mole sucrose"/(12color(red)(cancel(color(black)("moles C")))) = color(green)(|bar(ul(color(white)(a/a)"0.6777 moles sucrose"color(white)(a/a)|)))`

The answer is rounded to four sig figs.