# What mass of "calcium carbonate" is equivalent to "Avogadro's number" pf formula units of CaCO_3?

Apr 23, 2017

A mass of $100.09 \cdot g$ is specified.

#### Explanation:

$6.022 \times {10}^{23}$ units of ANYTHING, specifies 1 mole of that thing, i.e. ${N}_{A}$ individual units of that thing...........

And thus in $1 \cdot m o l$ of $C a C {O}_{3}$, ${N}_{A}$ formula units, there are: ${N}_{A}$ $C a$ atoms; ${N}_{A}$ $C$ atoms; and $3 \times {N}_{A}$ $O$ atoms.

Because ${N}_{A}$ ""^12C atoms have a mass of $12.00 \cdot g$, and ${N}_{A}$ ""^40Ca atoms have a mass of $40.00 \cdot g$, and ${N}_{A}$ ""^16O atoms have a mass of $16.0 \cdot g$, if we sum the masses appropriately we get a mass of approx. $100 \cdot g$, per molar quantity. Agreed?

From where did I get these molar masses? From where will YOU get the atomic mass if asked this question is asked of you in a test?