Question

A 31.3-g sample of ammonium carbonate contains ________ mol of ammonium ions.

Answer 1

Answer:

$n_{NH_4}=0.652molNH_4$

Explanation:

Hello,

Ammonium carbonate has the following formula:

$(NH_4)_2CO_3$

In this manner, the moles are computed by knowing its molecular mass:

$M_{(NH_4)_2CO_3}=14*2+1*8+12*1+16*3=96g/mol$

Thus, by applying a mass-mole-ions relationship, one obtains:

$ions_{(NH_4)_2CO_3}=31.3g(NH_4)_2CO_3*\frac{1mol(NH_4)_2CO_3}{96g(NH_4)_2CO_3}*\frac{2molNH_4}{1mol(NH_4)_2CO_3}\\n_{NH_4}=0.652molNH_4$

Best regards.

Answer 2

We need to calculate the equivalent amount in units of moles of ammonium ions from the mass units. For this we need the molar mass of the substances involved. We calculate as follows: 

31.3 g (NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol ammonium ions