Answer
0.0244 mole of $Ca(NO_3)_2$
Work Step by Step
1. Determine the molar mass of this compound ($Ca(NO_3)_2$), and setup the conversion factors:
$Ca: 40.08g $
$N: 14.01g * 2= 28.02g $
$O: 16.00g * 3 * 2= 96.00g $
40.08g + 28.02g + 96.00g = 164.10g
$ \frac{1 \space mole \space (Ca(NO_3))}{ 164.10 \space g \space (Ca(NO_3))}$ and $ \frac{ 164.10 \space g \space (Ca(NO_3))}{1 \space mole \space (Ca(NO_3))}$
2. Calculate the number of moles $(Ca(NO_3))$
$ 4.00g \times \frac{1 mole}{ 164.1g} = 0.0244 mole$
