What is the average mass and absolute uncertainty of calcium in each sample?

Question

Five calcite, CaCO3CaCO3 (MM 100.085 g/mol), samples of equal mass have a total mass of 10.1±0.1 g10.1±0.1 g. What is the average mass and absolute uncertainty of calcium in each sample? Assume that the relative uncertainties in atomic mass are small compared to the uncertainty of the total mass.

Answer 1

The question states you have 5 samples of Calcite (CaCO₃) of equal mass of 10.1 g, and equal uncertainties of 0.1 g, and asks you to calculate the average mass of Calcium (Ca) and its uncertainty.

You are provided with a molar mass for Calcite of 100.085 g/mol. You are asked to derive the average mass and absolute uncertainty of calcium in each sample, and they ask you to assume that the atomic masses of atomic weights (and thus the uncertainty of the molar mass) are small.

Seeing that the sample masses and uncertainties of the samples are all the same, you don't have to average anything, we can work with the 10.1 mass and 0.1 uncertainty. So you have to look up the atomic mass of Calcium in the periodic table, and use the ratio of that number to the molar mass to calculate the mass of calcium in the samples, and do the same thing with the uncertainty.

Sample mass of CO₃*(Ca Atomic Mass/Molar Mass CaCO3) => Mass of Ca

The periodic table gives Calcium's atomic mass as 40.078 ± 0.004 (so yeah, its uncertainty is really low, compared to that of the samples)

Ratio of Ca mass to CaCO₃ = 40.078/100.085 = 0.400439... The atomic mass of Ca has 5 significant digits, so that's all we can allow for this step, so our ratio is = 0.4004

10.1 g CaCO₃ * 0.4004 = 4.04404 g Ca. The sample mass only has 3 significant digits,

so we report the mass of Ca in our sample as 4.04 g.

The uncertainty in the sample's mass is 0.1

-so the uncertainty for just the mass of the element Ca in our sample should be 0.1* 0.4004 , which is 0.04 g. (we are only allowed 1 significant digit here)

So the average mass of Ca in our sample is 4.04±0.04 g