Question

Consider 1.00 g sample of copper (II) nitrate trihydrate.

How many moles of copper (II) nitrate trihydrate formula units are present in this sample?

How many moles of nitrate ion are present in the sample?

What mass in grams of water is present in this sample?

How many oxygen atoms are present in this sample?

Answer 1

1)

Molar mass of Cu(NO3)2.3H2O,

MM = 1*MM(Cu) + 2*MM(N) + 9*MM(O) + 6*MM(H)

= 1*63.55 + 2*14.01 + 9*16.0 + 6*1.008

= 241.618 g/mol

mass(Cu(NO3)2.3H2O)= 1.00 g

number of mol of Cu(NO3)2.3H2O,

n = mass of Cu(NO3)2.3H2O/molar mass of Cu(NO3)2.3H2O

=(1.0 g)/(241.618 g/mol)

= 4.14*10^-3 mol

Answer: 4.14*10^-3 mol

2)

1 mol of Cu(NO3)2.3H2O has 2 mol of nitrate ion

So,

mol of nitrate ions = 2*4.14*10^-3 mol

= 8.28*10^-3 mol

Answer: 8.28*10^-3 mol

3)

1 mol of Cu(NO3)2.3H2O has 3 mol of H2O

So,

mol of H2O = 3*4.14*10^-3 mol

= 0.01242 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O,

m = number of mol * molar mass

= 0.01242 mol * 18.016 g/mol

= 0.224 g

Answer: 0.224 g

4)

1 mol of Cu(NO3)2.3H2O has 9 mol of H2O

So,

mol of O atoms = 9*4.14*10^-3 mol

= 0.03726 mol

number of O atom = number of moles * Avogadro’s number

= 0.03726 mol * 6.022*10^23 atoms/moles

= 2.24*10^22 atoms

Answer: 2.24*10^22 atoms