Calculate the amount of moles of Zn(NO3)2 in 163.97 g of zinc nitrate

Answer 1

The previous answer is a bit more complicated than need be, so I thought I'd add an easier approach.

Molar mass of Zn(NO3)2 = 189.36 g/mole.

163.97 g Zn(NO3)2 x 1 mole/189.36 g = 0.8659 moles Zn(NO3)2 (to 4 sig. figs.)

Answer 2

Zinc nitrate has 1 molecule of zinc (Zn), 2 molecules of nitrogen (N) and 6 molecules of oxygen (O).

Calculate the molecular mass of zinc nitrate by adding the molecular weights for all molecules in the compound.

Here are the molecular weights for each element:

Zn=65.38 g/mol

N=14.01 g/mol

O=16.00 g/mol

The molecular weight of zinc nitrate is then: 1*(65.38)+2*(14.01)+6*(16.00)=189.40 g/mol.

Using a conversion formula to calculate moles of zinc nitrate:

$163.97gZn(N{O}_3{)}_2\ast \frac{1molZn(N{O}_3{)}_2}{189.40gZn(N{O}_3{)}_2}=0.866molZn(N{O}_3{)}_2$

There are 0.866 moles of zinc nitrate.

Note: These numbers may vary slightly due to the source of the molecular weight data and the number of decimal places used.