# How many formula units constitute a mass of 450*g of "calcium carbonate"?

May 31, 2017

Approx. $2.7 \times {10}^{24}$ $\text{formula units.....}$

#### Explanation:

First we work out the molar quantity:

$\text{Moles"="Mass"/"Molar mass} = \frac{450 \cdot g}{100.09 \cdot g \cdot m o {l}^{-} 1} = 4.50 \cdot m o l$.

Now, by definition, $1 \cdot m o l$ of stuff specifies $6.022 \times {10}^{23}$ individual items of stuff; we can also use the abbreviation ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

Now calcium carbonate is NOT MOLECULAR. And thus we say that there are $4.50 \cdot m o l \times {N}_{A}$ $\text{formula units}$ of calcium carbonate.

May 31, 2017

$2.71 \times {10}^{24}$ ions

#### Explanation:

The number of particles $N -$ or in this case molecules $-$ of a substance is given by the equation $N = n L$; where $n$ is the number of mole and $L$ is Avogrado's constant.

Also, the definition of $n$ is given by the equation $n = \frac{m}{M}$; where $m$ is the mass and $M$ is the molar mass.

Let's substitute the definition of $n$ into the equation for $N$:

$R i g h t a r r o w N = \frac{m}{M} \times L$

Then, let's substitute the value of $m$ and $L$ into the equation:

$R i g h t a r r o w N = \frac{450 \text{ g}}{M} \times 6.022 \times {10}^{23}$ ${\text{mol}}^{- 1}$

$R i g h t a r r o w N = \frac{2.7099 \times {10}^{26} {\text{ g mol}}^{- 1}}{M}$

We now need to calculate the molar mass $M$ of ${\text{CaCO}}_{3}$:

$R i g h t a r r o w M \left({\text{CaCO}}_{3}\right) = \left(40.078 + 12.011 + 3 \times 15.999\right)$ ${\text{g mol}}^{- 1}$

$R i g h t a r r o w M \left({\text{CaC)}}_{3}\right) = 100.086$ ${\text{g mol}}^{- 1}$

Now, let's substitute this value into the equation:

Rightarrow N = frac(2.7099 times 10^(26) " g mol"^(- 1))(100.086 " g mol"^(- 1))

$R i g h t a r r o w N \approx 2.71 \times {10}^{24}$

Therefore, there are around $2.71 \times {10}^{24}$ ions of ${\text{CaCO}}_{3}$.