Hydrated Double salt Question

A sample of a hydrated double salt, Fe(NH4)x(SO4)2.6H2O, which contains 0.363g of iron. The sample was boiled with an excess of sodium hydroxide and the ammonia given off was absorbed in 20 cm3 of 0.800 mol dm–3 hydrochloric acid. The resulting solution required 50 cm3 of 0.060 mol dm–3 sodium hydroxide to neutralise the excess acid. Determine the value of x in the molecular formula and hence the molecular mass of the hydrated double salt.

NH_3+HCl\rightarrow NH_4Cl

Amount of HCl in 20 cm3 of 0.800M

1000cm3 = 0.8 moles

20cm3 = ?

\frac{20*0.8}{1000}=0.016\, \textup{moles}

Amount of NaOH in 50 cm3 of 0.060

1000cm3 = 0.06 moles

50cm3 = ?

\frac{50*0.06}{1000}=0.003\, \textup{moles}

Amount of HCl reacted with NH3

0.016 – 0.003 = 0.013 moles

Moles of NH3

The ratio of acid to ammonia is 1:1, therefore 0.013 moles of ammonia were produced.

Moles of Fe

1 mole = 56g

? = 0.363g

\frac{0.363}{56}=0.00648\, \textup{moles}

Comparing the ratio of NH3 to that of Fe

NH3 : Fe

0.013:0.00648

2:1

Therefore x = 2

Molecular mass of Fe(NH4)2(SO4)2.6H2O

56 +(14+4)*2+(32+16*4)*2+18*6 = 392g

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