# What is the mass of "0.475 mol" of calcium arsenate, "Ca"_3"(AsO"_4)_2"?

Oct 30, 2014

In order to determine the mass of 0.475 mole of calcium arsenate, you must first determine the mass of one mole, which is its molar mass. This is done by multiplying the atomic weight of each element times the number of atoms of each element in the formula, adding the numbers for each element, and recording the answer in g/mol. Then multiply the known moles times the molar mass with grams on top and mol on the bottom.

Ca: 40.078 x 3 = 120.234 g/mol
As: 74.92160 x 2 = 149.8432 g/mol
O: 15.999 x 8 = 127.992 g/mol
The molar mass of ${\text{Ca"_3"(AsO"_4)}}_{2}$ = 398.0692 g/mol

${\text{0.475 mol Ca"_3"(AsO"_4)}}_{2}$ x $\text{398.0692 g"/"1 mol}$ =

${\text{189.08287 g Ca"_3"(AsO"_4)}}_{2}$= ${\text{189.08287 g Ca"_3"(AsO"_4)}}_{2}$ = ${\text{189 g Ca"_3"(AsO"_4)}}_{2}$ (rounded to three significant figures in accordance with the rounding rules for multiplication and division)

Note that moles cancel, leaving grams.

The mass of 0.475 mol ${\text{Ca"_3"(AsO"_4)}}_{2}$ is $\text{189 g}$.