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

First we work out the molar quantity:

`"Moles"="Mass"/"Molar mass"=(450*g)/(100.09*g*mol^-1)=4.50*mol`.

Now, by definition, `1*mol` of stuff specifies `6.022xx10^23` individual items of stuff; we can also use the abbreviation `N_A=6.022xx10^23*mol^-1`.

Now calcium carbonate is NOT MOLECULAR. And thus we say that there are `4.50*molxxN_A` `"formula units"` of calcium carbonate.

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`:

`Rightarrow N = frac(m)(M) times L`

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

`Rightarrow N = frac(450 " g")(M) times 6.022 times 10^(23)` `"mol"^(- 1)`

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

We now need to calculate the molar mass `M` of `"CaCO"_(3)`:

`Rightarrow M("CaCO"_(3)) = (40.078 + 12.011 + 3 times 15.999)` `"g mol"^(- 1)`

`Rightarrow M("CaC)"_(3)) = 100.086` `"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))`

`Rightarrow N approx 2.71 times 10^(24)`

Therefore, there are around `2.71 times 10^(24)` ions of `"CaCO"_(3)`.