# Question 49fad

Jun 15, 2017

$1.08 \times {10}^{22}$ $\text{atoms}$

#### Explanation:

We're asked to find the total number of atoms in $145$ $\text{mg}$ of caffeine, with its given chemical formula.

Let's first do a simple conversion from milligrams to grams, because we'll be using molar mass calculations in a bit:

145cancel("mg")((1"g")/(10^3cancel("mg"))) = 0.145"g"

Now, let's calculate the molar mass of caffeine, using the molar masses of the individual elements and how many of each element is in the compound:

overbrace((8)(12.01"g/mol"))^"C" + overbrace((10)(1.01"g/mol"))^"H" + overbrace((4)(14.01"g/mol"))^"N" + overbrace((2)(16.00"g/mol"))^"O"

= color(red)(194.22 color(red)("g/mol"

Now, let's convert the given mass ($0.145$ $\text{g}$) to moles using this molar mass:

0.145cancel("g C"_8"H"_10"N"_4"O"_2)((1"mol C"_8"H"_10"N"_4"O"_2)/(color(red)(194.22)cancel(color(red)("g C"_8"H"_10"N"_4"O"_2))))

$= 7.466 \times {10}^{-} 4$ $\text{mol C"_8"H"_10"N"_4"O"_2}$

Now, using Avogadro's number, let's convert this mole number to the number of caffeine molecules:

$7.466 \times {10}^{-} 4$
cancel("mol C"_8"H"_10"N"_4"O"_2)((6.022xx10^23"molecules C"_8"H"_10"N"_4"O"_2)/(1cancel("mol C"_8"H"_10"N"_4"O"_2")))

$= 4.496 \times {10}^{20}$ ${\text{molecules C"_8"H"_10"N"_4"O}}_{2}$

In one molecule of caffeine, there are 8 ("C") + 10 ("H") + 4 ("N") + 2 ("O") = color(green)(24 sfcolor(green)("atoms", so therefore,

$4.496 \times {10}^{20}$ cancel("molecules C"_8"H"_10"N"_4"O"_2)((color(green)(24"atoms"))/(1cancel("molecule")))

 = color(blue)(1.08 xx 10^22 color(blue)("atoms"

rounded to $3$ significant figures, the amount given in the problem.

Therefore, there are color(blue)(1.08 xx 10^22# atoms in $145$ $\text{mg}$ of caffeine.