Question

How many moles of cobalt(III) bromide are in `"9.94 g"`?

Answer 1

There is `"0.0333 mol CoBr"_3"` in `"9.94 g"`.

Explanation:

The chemical formula for cobalt (III) chloride is `"CoBr"_3`. The molar mass of `"CoBr"_3` is determined by multiplying the subscript of each element by its atomic weight on the periodic table in g/mol.

Determine molar mass of compound.

`M_"CoBr3"=(1xx"58.933 g/mol Co") + (3xx"79.904 g/mol Br")="298.645 g/mol"`

Convert given mass to moles.

Divide the given mass by the molar mass. Since molar mass is a fraction, I prefer to divide by multiplying by its reciprocal (mol/g).

`9.94color(red)cancel(color(black)("g CoBr"_3))xx(1"mol CoBr"_3)/(298.645color(red)cancel(color(black)("g CoBr"_3)))= "0.0333 mol CoBr"_3"` (rounded to three significant figures)