# How do you calculate the number of moles in 0.135 grams of HNO_3?

Apr 18, 2016

$= 0.00214 m o l H N {O}_{3}$

or

$2.14 x {10}^{-} 3 m o l H N {O}_{3}$

#### Explanation:

In order to calculate the moles in 0.135 grams of $H N {O}_{3}$

We begin by finding the molar mass of $H N {O}_{3}$ using the atomic masses from the periodic table.

$H = 1 x 1.01 = 1.01$
$N = 1 x 14.01 = 14.01$
$O = 3 x 15.99 = 47.97$

$1.01 + 14.01 + 47.97 = 62.99 \frac{g}{m o l}$

Now we can use a conversion factor to convert
grams to moles

0.135 cancel(g HNO_3) x (1 mol)/(62.99cancel(g HNO_3 )

$= 0.00214 m o l H N {O}_{3}$

or

$2.14 x {10}^{-} 3 m o l H N {O}_{3}$

May 1, 2016

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

$n = m \div M$

0.135 grams (m) of $H N {O}_{3}$ has been provided for you.

Now you need to find the molar mass (M). If you look closely at the compound formula, you can see that Hydrogen, Nitrogen and Oxygen are present. The molar mass of: Hydrogen is 1.0 g/mol, Nitrogen is 14.0 g/mol and Oxygen 16.0 g/mol.
Therefore, the molar mass of $H N {O}_{3}$ is:
$\left[1 \times 1.0 + 1 \times 14.0 + 3 \times 16.0\right]$ = 63.0 g/mol.

Looking back at the $n = m \div M$ formula,
the number of moles (n) present is:
$\left[n = 0.135 \div 63.0\right]$ = 0.00214 moles.
Note: Answer rounded to 3 significant figures.