Question

In 23.8 ng (nanograms) of solid ammonium sulfate, (NH4)2SO4, determine (a) the number of hydrogen atoms, (b) the number of NH4 + ions, and (c) the atomic ratio of hydrogen to nitrogen

Answer 1

1st find the number of molecules of (NH4)2SO4

Molar mass of (NH4)2SO4 = 2*MM(N) + 8*MM(H) + 1*MM(S) + 4*MM(O)

= 2*14.01 + 8*1.008 + 1*32.07 + 4*16.0

= 132.154 g/mol

mass of (NH4)2SO4 = 23.8*10^-9 g

we have below equation to be used:

number of mol of (NH4)2SO4,

n = mass of (NH4)2SO4/molar mass of (NH4)2SO4

=(2.38*10^-8 g)/(132.154 g/mol)

= 1.801*10^-10 mol

we have below equation to be used:

number of molecules = number of mol * Avogadro’s number

number of molecules = 1.801*10^-10 * 6.022*10^23 molecules

number of molecules = 1.085*10^14 molecules

a)

1 molecule of (NH4)2SO4 has 8 H atom

So,

number of H atom = 8*1.085*10^14

= 8.68*10^14 atoms

Answer: 8.68*10^14 atoms

b)

1 molecule of (NH4)2SO4 has 2 NH4+ ions

So,

number of NH4+ ions = 2*1.085*10^14

= 2.17*10^14 ions

Answer: 2.17*10^14 ions

c)

1 molecule of (NH4)2SO4 has 8 H atom

1 molecule of (NH4)2SO4 has 2 H atom

atomic ratio = 8/2 = 4

Answer: 4:1