Question

A. Five mineral samples of equal mass of calcite, CaCO3 (MM 100.085) , had a total mass of 12.3 ± 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.) (Input the answer in terms of absolute uncertainty.)

B. A volume of 28.47 ± 0.06 mL of HNO3 solution was required for complete reaction with 0.9721 ± 0.0008 g of Na2CO3, (FM 105.988 ± 0.001). Find the molarity of the HNO3 and its absolute uncertainty.

Answer 1

(A). Five mineral samples of equal mass of calcite, CaCO3 (MM 100.085) , had a total mass of 12.3 ± 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.) (Input the answer in terms of absolute uncertainty.)

In CaCO₃ Fw=100.085g/mol

So:

%Ca=(40.078g/100.085g)x100%=40.04%

%C=(12.0107g/100.085g)x100%=12.00%

%O=(47.9982g/100.085g)x100%=47.96%

Total Sample Mass: (12.3±0.1)g

Total Calcium in Sample Mass: (12.3±0.1)g x 40.04% = (12.3±0.1)g x 0.4004= (4.92±0.04)g

Average Calcium in Sample Mass: (4.92±0.04)g/5 samples= (0.98±0.01)g of Ca for sample

(B). A volume of 28.47 ± 0.06 mL of HNO₃ solution was required for complete reaction with 0.9721 ± 0.0008 g of Na₂CO₃, (FM 105.988 ± 0.001). Find the molarity of the HNO₃ and its absolute uncertainty.