Question

Question #c2141

Answer 1

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 `->` moles of sucrose `->` molecules of sucrose `->` 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.

Your sample will contain

`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 * 10^(21)"molecules C"_12"H"_22"O"_11`

Now, as its chemical formula suggest, every 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"`

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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))))`

`= color(darkgreen)(ul(color(black)(2.1 * 10^(22)"atoms of C")))`

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