# How many atoms of carbon are in "0.0235 g" of cocaine hydrochloride, "C"_17"H"_22"ClNO"_4?

May 29, 2017

$7.08 \cdot {10}^{20}$

#### Explanation:

First find the moles of carbon:

n(C)=("0.0235 g")/("339.8168 g/mol cocaine hydrochloride") xx "17 mols C"/"1 mol cocaine hydrochloride"

$n \left(C\right) = 1.176 \times {10}^{- 3}$ $\text{mol}$s

Then multiply the moles by avogadro's constant to tell you the number of carbon atoms in the sample.

$\left(1.176 \cdot {10}^{-} 3\right) \left(6.02 \cdot {10}^{23}\right)$

$= 7.08 \cdot {10}^{20}$ atoms of Carbon

May 29, 2017

There are color(blue)(7.08xx10^(20)color(white)(.)"atoms C" in $\text{0.0235 g C"_17"H"_22"ClNO"_4}$.

#### Explanation:

The molecular formula for cocaine hydrochloride, $\text{C"_17"H"_22"ClNO"_4}$, tells us that one mole of $\text{C"_17"H"_22"ClNO"_4}$ contains $\text{17 mol C}$.

$\text{0.0235 g C"_17"H"_22"ClNO"_4}$ represents a fraction of a mole.

In order to determine the mol $\text{C"_17"H"_22"ClNO"_34}$ in $\text{0.0235 g C"_17"H"_22"ClNO"_4}$, divide its given mass by its molar mass by multiplying by its inverse.

The molar mass of $\text{C"_17"H"_22"ClNO"_4}$ is $\text{339.816 g/mol}$.
https://www.ncbi.nlm.nih.gov/pccompound?term=%22COCAINE+HYDROCHLORIDE%22

color(blue)("Moles of Cocaine Hydrochloride"

Divide its given mass by its molar mass by multiplying by its inverse.

0.0235color(red)cancel(color(black)("g C"_17"H"_22"ClNO"_4))xx(1"mol C"_17"H"_22"ClNO"_4)/(339.816color(red)cancel(color(black)("g C"_17"H"_22"ClNO"_4)))="0.00006916 mol C"_17"H"_22"ClNO"_4

color(blue)("Atoms of Carbon"

Multiply mol ${\text{C"_17"H"_22"ClNO}}_{4}$ by $\text{17 mols C}$ divided by $\text{1 mol C"_17"H"_22"ClNO"_4}$, then multiply by $6.022 \times {10}^{23}$ $\text{atoms/mol}$.

0.000069161color(red)cancel(color(black)("mol C"_17"H"_22"ClNO"_4))xx(17"mol C")/(1color(red)cancel(color(black)("mol C"_17"H"_22"ClNO"_4)))="0.00117572 mol C"

0.00117572color(red)cancel(color(black)("mol C"))xx(6.022xx10^23"atoms C")/(1color(red)cancel(color(black)("mol C")))

$= 7.08 \times {10}^{20} \text{atoms C}$, rounded to three significant figures