MOLE : Q.A.E 2

Q1: How many oxygen atoms are in ${\text{120.23 g Fe}}_3{\left({\text{PO}}_4\right)}_2$?

Answer:

$\text{120.23g}$ ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$ contains ${\text{1.6203 x 10}}^{24}$ $\text{atoms O}$.

Explanation:

There are eight moles of oxygen atoms in each mole of ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$. To answer this question, you must convert the given mass of ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$ to moles of ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$. Then multiply the moles of ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$ times $\left(8{\text{mol O)/(1 mol Fe}}_3{\left({\text{PO}}_4\right)}_2\right)$ to determine the number of moles of oxygen. Then multiply the moles of oxygen times $6.022\times {\text{10}}^{23}$ $\text{atoms O}$.

Given/Known:
mass ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$$=$$\text{120.23 g}$
molar mass ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$$=$$\text{357.478g/mol}$
1 mole ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$ contains 8 moles oxygen atoms
1 mole O atoms = $6.022\times {\text{10}}^{23}$ $\text{atoms}$

Unknown:
number of oxygen atoms in ${\text{120.23g Fe}}_3{\left({\text{PO}}_4\right)}_2$

Solution:

1. Convert given mass of ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$ to moles.

$120.23{\text{g Fe}}_3{\left({\text{PO}}_4\right)}_2\times \frac{1{\text{mol Fe}}_3{\left({\text{PO}}_4\right)}_2}{357.478{\text{g Fe}}_3{\left({\text{PO}}_4\right)}_2}={\text{0.336328 mol Fe}}_3{\left({\text{PO}}_4\right)}_2$

2. Convert moles ${\text{Fe}}_3{\left({\text{PO}}_4\right)}_2$ to moles O.

$0.336328{\text{mol Fe}}_3{\left({\text{PO}}_4\right)}_2\times \frac{8\text{mol O}}{1{\text{mol Fe}}_3{\left({\text{PO}}_4\right)}_2}=\text{2.690624 mol O}$

3. Convert moles O to atoms O.

$2.690624\text{mol O}\times \frac{6.022\times {10}^{23}\text{atoms O}}{1\text{mol O}}=1.6203\times {10}^{24}\text{atoms O}$ (answer rounded to five significant figures)

SOURCE : https://socratic.org/questions/5473b86c581e2a14bf83682d



Q2: How many atoms of carbon are in $\text{0.0235 g}$of cocaine hydrochloride, ${\text{C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$?

Answer:

There are $7.08\times {10}^{20}.\text{atoms C}$ in ${\text{0.0235 g C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$.

Explanation:

The molecular formula for cocaine hydrochloride, ${\text{C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$, tells us that one mole of ${\text{C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$ contains $\text{17 mol C}$.

${\text{0.0235 g C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$ represents a fraction of a mole.

In order to determine the mol ${\text{C}}_{17}{\text{H}}_{22}{\text{ClNO}}_{34}$ in ${\text{0.0235 g C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$, divide its given mass by its molar mass by multiplying by its inverse.

The molar mass of ${\text{C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$ is $\text{339.816 g/mol}$.
https://www.ncbi.nlm.nih.gov/pccompound?term=%22COCAINE+HYDROCHLORIDE%22

$\text{Moles of Cocaine Hydrochloride}$

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

$0.0235{\text{g C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4\times \frac{1{\text{mol C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4}{339.816{\text{g C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4}={\text{0.00006916 mol C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$

$\text{Atoms of Carbon}$

Multiply mol ${\text{C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$ by $\text{17 mols C}$ divided by ${\text{1 mol C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4$, then multiply by $6.022\times {10}^{23}$ $\text{atoms/mol}$.

$0.000069161{\text{mol C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4\times \frac{17\text{mol C}}{1{\text{mol C}}_{17}{\text{H}}_{22}{\text{ClNO}}_4}=\text{0.00117572 mol C}$

$0.00117572\text{mol C}\times \frac{6.022\times {10}^{23}\text{atoms C}}{1\text{mol C}}$

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

SOURCE :https://socratic.org/questions/592c925811ef6b2834ec19cb