# How many grams of Cr(NO_3)_2 are in 8.4 mole?

Mar 24, 2016

Chromous nitrate, $C r {\left(N {O}_{3}\right)}_{2}$, has a formula weight of $176.01$ $g \cdot m o l$

#### Explanation:

You have an $8.4$ $m o l$ quantity, which has a mass of $8.4 \cdot \cancel{m o l} \times 176.01 \cdot g \cdot \cancel{m o {l}^{-} 1}$ $\cong$ $1.5$ $k g$.

Hi there!

With $C r {\left(N {O}_{3}\right)}_{2}$, there is approximately 1460.97 g in 8.3 moles.

#### Explanation:

To figure this out, note the relationship between mass, molar mass and moles whereby:

n (moles) = mass/molar mass...

or if you're more of a visual oriented learner, I will show the unit conversion/dimensional analysis method! Either will work!

Let's start off with what you're given:

n = 8.3
m (mass in grams) = ?
M (molar mass) = $C r {\left(N {O}_{3}\right)}_{2} = \left(52.00 \frac{g}{m o l} +\right) 2 \left(\left(14.01 \frac{g}{m o l}\right) + 3 \left(16.00 \frac{g}{m o l}\right)\right)$

$M = 176.02 \frac{g}{m o l}$

Now you know n and M, you can find m using the aforementioned expression! Substituting in you get:

$8.3 = \frac{m}{176.02}$

$m = \left(8.3\right) \left(176.02\right)$

$m = 1460.97 g$

You can also use unit conversion/dimensional analysis whereby:

$8.3 m o l \cdot 176.02 \frac{g}{m o l} \to$ Moles cancel and you're left with:

$1460.97 g$

Therefore, in 8.3 moles of $C r {\left(N {O}_{3}\right)}_{2}$, there is approximately 1460.97 g! Hopefully this was helpful! If you have any questions, let me know! :)