Question

Five mineral samples of equal mass of calcite, caco3 (mm 100.085) , had a total mass of 10.9 ± 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

The molecular mass of calcite is 100.085 (given).

Atomic mass of calcium = 40.078 amu

Percent amount of calcium in calcite = $\frac{40.078}{100.085}\times 100 = 40.044$%

The total mass of five mineral samples = $10.9\pm 0.1 g$  (given)

So, the mass of one sample = $\frac{10.9}{5} = 2.18 g$

and $\frac{0.1}{5} = 0.02 g$

The average mass of calcium in sample = $\frac{2.18 g\times 40.044}{100} = 0.873 g$

and $\frac{0.02 g\times 40.044}{100} = 0.008 g$

Hence, the average mass of calcium in each sample is $0.873 g\pm 0.008 g$.