Question

a). Calculate the molar mass in g/mol of the following compound (ammonium dichromate):

(NH₄)₂Cr₂O₇

b). Calculate the mass, in g, of the following: 0.158 mol IF₅ (iodine pentafluoride)

c). Calculate the amount, in moles, of the following: 25.4 g Pb(C₂H₃O₂)₂ (lead (II) acetate)

Professors Note: Make sure that you give your answer in a clear and well-reasoned manner, providing necessary explanation on how did you come up with it.

d). Calculate the percent by mass of C, H, and O in C₆H₁₂O₆ (glucose )

Answer 1

solution:

a). Calculate the molar mass in g/mol of the following compound (ammonium dichromate):

(NH₄)₂Cr₂O₇

_{Lets calculate individual mass of atoms;}

_{2 X N = 2 X 14 = 28}

_{8 X H = 8 X 1 = 8}

_{2 X Cr = 2 X 52 = 104}

7 X O = 7 X 14 = 112

Molar mass of (NH₄)₂Cr₂O₇ = 28 + 8 + 104 + 112 = 252.0 g/mol

Molar mass of (NH₄)₂Cr₂O₇ = 252.0 g/mol

b). Calculate the mass, in g, of the following: 0.158 mol IF₅ (iodine pentafluoride)

Solution:

# of mole = 0.158 mol

We know

# of mole = mass in g/molar mass.                 

                                                       (molar mass of IF5 = 221.89 g/mol)

mass in g = # of mole X molar mass = 0.158 mol X 221.89 g/mol

= 35.05 g.

Mass of 0.158 mol IF5 = 35.05 g

c). Calculate the amount, in moles, of the following: 25.4 g Pb(C₂H₃O₂)₂ (lead (II) acetate)

Professors Note: Make sure that you give your answer in a clear and well-reasoned manner, providing necessary explanation on how did you come up with it.

Given:

Mass = 25.4 g.

Molar mass = 325.28 g/mol

We know

# of mole = mass in g/molar mass =25.4 g/325.28 g/mol

# of mole = 0.0780 mol.

25.4 g Pb(C₂H₃O₂)₂ (lead (II) acetate) mole require =0.0780 mol

d). Calculate the percent by mass of C, H, and O in C₆H₁₂O₆ (glucose )

Solution:

mass of C6H12O6 = 180.0 g/mol

%mass of c = (mass of C in C6H12O6 / total mass of C6H12O6)X100%  

= (72/180.0) X 100% = 40.0%

%mass of c = 40.0% C in C6H12O6

Mass of H = (mass of H in C6H12O6 / total mass of C6H12O6)X100%  

= (12/180) x 100% = 6.66%

%mass of H = 6.66% H in C6H12O6

Mass of O = (mass of O in C6H12O6 / total mass of C6H12O6)X100%  

= (96/180) x 100% = 53.33%

%mass of O = 53.33% O in C6H12O6