Question

Five mineral samples of equal mass of calcite, CaCO3 (MM 100.085) , had a total mass of 12.4 ± 0.1 g. What is the average mass of calcium in each sample? (Assume that the relative uncertainties in atomic mass are small compared the uncertainty of the total mass.)

Answer 1

number of moles of CaCO3 = mass / molar mass = 12.4/100.085 = 0.124 mol
1 mol of CaCO3 has 1 mol of Ca
so,
number of moles of Ca = 0.124 mol

Molar Mass of Ca = 40.078 g/mol
mass of Ca = number of moles * molar mass
                       = 0.124 *40.078
                       = 4.97 g

average mass = 4.97 / 5 = 0.994 g

uncertainty will just propagate as we are just dividing by a number

uncertainty in average mass = 0.1 g

Answer:
0.994 ± 0.1 g