Question

Five calcite, CaCO3 (MM 100.085 g/mol), samples of equal mass have a total mass of 12.2±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 the uncertainty of the total mass.

Answer 1

Answer:

• The average mass of calcium in each sample is: 0.978 g

• The absolute uncertainty is: 0.008 g

Explanation:

The absolute uncertainty of the total samples indicated in the statement is ± 0.1 g.

When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the relative uncertainties together.

The relative uncertainty is the absolute uncertainty divided by the quantity:

• Relative uncertainty = 0.1g / 12.2 g = 0.008

The average mass of calcium is calculated using proportions, along with the molar masses:

• Molar mass of calcium: 40.078 g/ mol (from a periodic table)

• Molar mass of calcite: 100.085 g/mol (given)

Proportion:

• 40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite

• x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium

So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:

• Average mass = total mass of five samples / number of samples

• Average mass = 4.89 g / 5 = 0.978 g of calcium

So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as  you have only used multiplication and division).

The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:

• Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g

The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.