# Question 54f82

Apr 10, 2016

${\text{0.6777 moles C"_12"H"_22"O}}_{11}$

#### 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, ${\text{C"_12"H"_22"O}}_{11}$, 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 $4.897 \cdot {10}^{24}$ atoms of carbon, $\text{C}$.

In order to convert this to moles of carbon, use the fact that a mole of atoms is defined as a collection of $6.022 \cdot {10}^{23}$ atoms.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole atoms" = 6.022 * 10^(23)"atoms} \textcolor{w h i t e}{\frac{a}{a}} |}} \to}$ 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, $12 \times \text{C}$
• twenty two atoms of hydrogen, $22 \times \text{H}$
• eleven atoms of oxygen, $11 \times \text{O}$

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

$8.132 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles C"))) * "1 mole sucrose"/(12color(red)(cancel(color(black)("moles C")))) = color(green)(|bar(ul(color(white)(a/a)"0.6777 moles sucrose} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to four sig figs.