# Question 06d81

Dec 25, 2017

$a . 2 m o l \text{C} C {l}_{4}$
$b . 308.0 g \text{C} C {l}_{4}$
$c . 1.2 \times {10}^{24} \text{C"Cl_4 " molecules}$

#### Explanation:

1. Given the $\eta C {l}_{2}$, this value can be the basis to find the $\eta \text{C} C {l}_{4}$ through molar conversion. Refer to the provided balanced equation for the mole ratio.
$= 4.0 \cancel{m o l C {l}_{2}} \times \frac{1 m o l \text{C} C {l}_{4}}{2 \cancel{m o l C {l}_{2}}}$
$= 2 m o l \text{C} C {l}_{4}$
2. Find the molar mass of the involved compound required in the problem. Relative atomic masses of the elements composing $\text{C} C {l}_{4}$ are obtainable from the periodic table; i.e.,
$\text{Molar mass "C} C {l}_{4} = \frac{154 g}{m o l}$
3. Then, find $m \text{C"Cl_4}$ that can be computed as
=2cancel(mol"C"Cl_4)xx(154g"C"Cl_4)/(1cancel(mol"C"Cl_4))
$= 308 g \text{C} C {l}_{4}$
4. Lastly, find the number of molecules of $\text{C} C {l}_{4}$. Knowing that the value in $\eta \text{C} C {l}_{4} = 2 m o l$ and from the relationship that $1 m o l \text{C"Cl_4=6.02xx10^23"C"Cl_4" molecules}$, the number of molecules can be computed as follows:
=2cancel(mol"C"Cl_4)xx(6.02xx10^23"C"Cl_4" molecules")/(1cancel(mol"C"Cl_4))#
$= 12.04 \times {10}^{23} \text{C"Cl_4" molecules}$
5. Make sure to express the resulting value in the standard scientific notation. Knowing the fact that moving the decimal point to the left corresponds to a positive exponent (from the mnemonics LIP-Left Is Positive); that is,
$= 1.2 \times {10}^{24} \text{C"Cl_4" molecules}$