# Question #20847

Jul 21, 2017

(a) 0.056 39 mol; (b) 0.1632 mol

#### Explanation:

(a) Ethanol

Step 1. Calculate the moles of ethanol

$\text{Moles of C"_2"H"_5"OH" = 1.299 color(red)(cancel(color(black)("g C"_2"H"_5"OH"))) × ("1 mol C"_2"H"_5"OH")/(46.07 color(red)(cancel(color(black)("g C"_2"H"_5"OH")))) = "0.028 196 mol C"_2"H"_5"OH}$

Step 2. Calculate the moles of $\text{C}$ atoms

$\text{Moles of C atoms" = "0.028 196" color(red)(cancel(color(black)("mol C"_2"H"_5"OH"))) × "2 mol C atoms"/(1 color(red)(cancel(color(black)("mol C"_2"H"_5"OH")))) = "0.056 39 mol C atoms}$

(b) 1,4-Diclorobenzene

Step 1. Calculate the moles of $\text{ClC"_6"H"_4"Cl}$

$\text{Moles of ClC"_6"H"_4"Cl" = 3.998 color(red)(cancel(color(black)("g ClC"_6"H"_4"Cl"))) × ("1 mol ClC"_6"H"_4"Cl")/(147.00 color(red)(cancel(color(black)("g ClC"_6"H"_4"Cl")))) = "0.027 197 mol ClC"_6"H"_4"Cl}$

Step 2. Calculate the moles of $\text{C}$ atoms

$\text{Moles of C atoms" = "0.027 197" color(red)(cancel(color(black)("mol ClC"_6"H"_4"Cl"))) × "6 mol C atoms"/(1 color(red)(cancel(color(black)("mol ClC"_6"H"_4"Cl")))) = "0.1632 mol C atoms}$