# How many molecules are present in 1296 g of dinitrogen pentoxide N_2O_5?

Apr 21, 2016

Approx $12 \times {N}_{A}$ where ${N}_{A}$ $=$ $\text{Avogadro's number, } {N}_{A} = 6.02214 \times {10}^{23}$.

#### Explanation:

Moles of ${N}_{2} {O}_{5}$ $=$ $\frac{1296 \cdot g}{108.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ ??mol.

Number of ${N}_{2} {O}_{5}$ molecules $=$ $6.02214 \times {10}^{23} \cdot \text{molecules } \cancel{m o {l}^{-} 1}$ $\times$ $\frac{1296 \cdot \cancel{g}}{108.01 \cdot \cancel{g} \cdot \cancel{m o {l}^{-} 1}}$ $=$ ??" molecules".

May 1, 2016

Assume that:
n - number of moles
m - mass of substance
M - molar mass

$n = m \div M$

1296 grams (m) of ${N}_{2} {O}_{5}$ has been provided for you.

Now you have to find the molar mass (M) of ${N}_{2} {O}_{5}$. If you break the formula up, you can see that Nitrogen and Oxygen are present.

Refer to your periodic table, just so you know that the molar mass (M) of Nitrogen is 14.0 g/mol and Oxygen 16.0 g/mol.

Therefore the molar mass (M) of ${N}_{2} {O}_{5}$ is:
$\left[2 \times 14.0 + 5 \times 16.0\right]$ = 108 g/mol.

To find out the number of moles (n), look back at the $n = m \div M$ formula and follow it. So, the number of moles (n) present in ${N}_{2} {O}_{5}$ is: [n = 1296 grams $\div$ 108 g/mol] = 12 moles.

To find the number of molecules, use this formula:
Number of molecules = number of moles $\times$ Avogadro's number

Avogadro's number = $6.02 \times {10}^{23}$
No of molecules in ${N}_{2} {O}_{5}$
= $12 \times \left[6.02 \times {10}^{23}\right]$ = ??