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 =
`(40.078)/(100.085)* 100 = 40.044`%

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

So, the mass of one sample =
`(10.9)/(5) = 2.18 g`

and
`(0.1)/(5) = 0.02 g`

The average mass of calcium in sample =
`(2.18 g* 40.044)/(100) = 0.873 g`

and
`(0.02 g* 40.044)/(100) = 0.008 g`

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