Question

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

Answer 1

`7.08*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(C)= 1.176 xx 10^(-3)` `"mol"`s

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

`(1.176*10^-3)(6.02*10^23)`

`=7.08*10^20` atoms of Carbon

Answer 2

There are `color(blue)(7.08xx10^(20)color(white)(.)"atoms C"` in `"0.0235 g C"_17"H"_22"ClNO"_4"`.

Explanation:

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

`"0.0235 g C"_17"H"_22"ClNO"_4"` represents a fraction of a mole.

In order to determine the mol `"C"_17"H"_22"ClNO"_34"` in `"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 `"C"_17"H"_22"ClNO"_4"` is `"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 `"C"_17"H"_22"ClNO"_4` by `"17 mols C"` divided by `"1 mol C"_17"H"_22"ClNO"_4"`, then multiply by `6.022xx10^23` `"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")))`

`= 7.08xx10^20"atoms C"`, rounded to three significant figures