Question

A student weighs out a 2.546 g sample of solid CoCO3●nH2O. Upon heating the following reaction occurs:

CoCO3●nH2O(s)CoO(s) + CO2(g) + n H2O(g) After the reaction has gone to completion the mass of solid

remaining is 0.840 g.

a. How many moles of CO2 gas were formed? You must show your work to receive any credit.

b. What is the molecular weight of CoCO3●nH2O?

c. What is the value of “n”? Round to the nearest whole number. You must show your work to receive any credit.

Answer 1

a) Initial mass of solid is reactant = 2.546 g

Final mass of solid after reaction is CoO = 0.84 g

Molar mass of CoO = 75 g/mol

Moles of CoO = mass/molar mass = 0.84/75 = 0.0112 moles.

According to reaction stoichiometry, when one mole of CoO is formed, one mole of CO₂ is formed.

When 0.0112 moles of CoO is formed, moles of CO₂ formed = 0.0112 moles

Thus, moles of CO₂ formed = 0.0112 moles. Mass of CO₂ = molar mass of CO₂ x moles of CO₂ = 44 x 0.0112 = 0.4928 g

b) Using mass conservation, mass of reactant = mass of product = 2.546 g.

for one mole of CoO to form, one mole of CoCO₃.nH₂O is required. Therefore, for 0.0112 moles of CoO to form, moles of CoO₃ required = 0.0112.

Molar mass of CoO₃.nH₂O = mass of CoO₃.nH₂O/moles of CoO₃.nH₂O = 2.546/0.0112 = 227.32 g/mol

Thus, molar mass/molecular weight of CoCO₃.nH₂O = 227.32 g/mol

c) Atomic weight of Co = 59 a.m.u, Atomic weight of C = 12 a.m.u and Atomic weight of O = 16 a.m.u

Therefore, molar mass of CoCO₃ = 59 + 12 + (3x16) = 119.

But molecular weight of CoCO₃.nH₂O = 227.32. The difference in weight is mass of nH₂O = 227.32 - 119 = 108.32.g

value of n = (difference in weight of CoCO₃.nH₂O and CoCO₃)/molar mass of H₂O = 108.32/18 = 6.017 ~ 6

Thus, value of n = 6