# Question c2141

Jan 8, 2017

Here's what I got.

#### Explanation:

In order to determine how many atoms of carbon you have in your sample, you must use a series of conversion factors to take you from

grams of sucrose $\to$ moles of sucrose $\to$ molecules of sucrose $\to$ atoms of carbon

To go from grams of sucrose to moles of sucrose, use the molar mass of the compound. Your sample will contain

1.0 color(red)(cancel(color(black)("g"))) * ("1 mole C"_12"H"_22"O"_11)/(342.3color(red)(cancel(color(black)("g")))) = "0.0029214 moles C"_12"H"_22"O"_11

Now, you can go from moles of sucrose to number of molecules of sucrose by using Avogadro's constant, which is essentially the definition of a mole.

0.0029214 color(red)(cancel(color(black)("moles C"_12"H"_22"O"_11))) * (6.022 * 10^(22)"molecules C"_12"H"_22"O"_11)/(1color(red)(cancel(color(black)("mole C"_12"H"_22"O"_11))))

$= 1.7593 \cdot {10}^{21} {\text{molecules C"_12"H"_22"O}}_{11}$

Now, as its chemical formula suggest, every 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, $11 \times \text{O}$

This means that your sample will contain

1.7593 * 10^(21) color(red)(cancel(color(black)("molecules C"_12"H"_22"O"_11))) * "12 atoms C"/(1color(red)(cancel(color(black)("molecule C"_12"H"_22"O"_11))))#

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.1 \cdot {10}^{22} \text{atoms of C}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of sucrose.