Question

a) How many moles of Cu are there in 35.7 grams of Cu?

b) How many MOLES of dinitrogen tetrafluoride are present in 3.33 grams of this compound ?

c) How many ATOMS of nitrogen are present in 3.29 grams of dinitrogen tetrafluoride ?  atoms of nitrogen .

d) How many GRAMS of fluorine are present in 6.41×1022 molecules of dinitrogen tetrafluoride ?  

e) How many GRAMS of dinitrogen tetrafluoride are present in 1.98 moles of this compound ?   

Answer 1

a)

Molar mass of Cu = 63.55 g/mol

mass(Cu)= 35.7 g

use:

number of mol of Cu,

n = mass of Cu/molar mass of Cu

=(35.7 g)/(63.55 g/mol)

= 0.5618 mol

Answer: 0.562 mol

b)

Molar mass of N2F4,

MM = 2*MM(N) + 4*MM(F)

= 2*14.01 + 4*19.0

= 104.02 g/mol

mass(N2F4)= 3.33 g

use:

number of mol of N2F4,

n = mass of N2F4/molar mass of N2F4

=(3.33 g)/(1.04*10^2 g/mol)

= 3.201*10^-2 mol

Answer: 3.20*10^-2 mol

c)

Molar mass of N2F4,

MM = 2*MM(N) + 4*MM(F)

= 2*14.01 + 4*19.0

= 104.02 g/mol

mass(N2F4)= 3.29 g

use:

number of mol of N2F4,

n = mass of N2F4/molar mass of N2F4

=(3.29 g)/(1.04*10^2 g/mol)

= 3.163*10^-2 mol

use:

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

number of molecules = 3.163*10^-2 * 6.022*10^23 molecules

number of molecules = 1.905*10^22 molecules

1 molecule of N2F4 has 2 atoms of N

So,

number of atoms of N = number of molecules * 2

= 1.905*10^22 * 2

= 3.81*10^22

Answer: 3.81*10^22

d)

1 molecule of N2F4 has 4 atoms of F

number of atoms of F = 4* number of molecules

   = 4*6.41*10^22

= 2.564*10^23 atoms

molar mass of F = 19.0 g/mol

This is mass of 6.022*10^23 atoms

use:

mass of 1 atom = molar mass /(6.022*10^23)

mass of 1 atom = 19.0/(6.022*10^23)

mass of 1 atom = 3.155*10^-23 g

mass of 2.564*10^23 atom = mass of 1 atom * total number of atoms

mass of 2.564*10^23 atom = 3.155*10^-23 * 2.564*10^23

mass of 2.564*10^23 atom = 8.09 g

Answer: 8.09 g

e)

Molar mass of N2F4,

MM = 2*MM(N) + 4*MM(F)

= 2*14.01 + 4*19.0

= 104.02 g/mol

use:

mass of N2F4,

m = number of mol * molar mass

= 1.98 mol * 1.04*10^2 g/mol

= 206 g

Answer: 206 g