A 0.67671 g (± 0.2 mg) sample of sodium carbonate (FM= 105.99 g) was transferred
ID: 488617 • Letter: A
Question
A 0.67671 g (± 0.2 mg) sample of sodium carbonate (FM= 105.99 g) was transferred to a calibrated 250 mL volumetric flask and diluted to the full mark with DI water. After mixing, three 10 mL aliquots of this sample were transferred to another beaker to be used in a separate chemical reaction. The volumetric flask and the volumetric pipet were calibrated, and found to contain 250.12 ± 0.08 mL and 9.996 ± 0.005 mL.
a)Calculate the moles of sodium carbonate that were transferred for use in the separate chemical reaction.
b)Which step contributed the largest relative percentage error to the overall preparation? 1. sample weighing 2. dilution of sample in the volumetric flask 3. pipetting of two 10 mL volumes of sample into beaker
c) Using the error values provided, estimate the absolute error in the Molar concentration.
Explanation / Answer
Mass of sodium carbonate taken = 0.67671 g.
Gram FM of sodium carbonate = 105.99 g/mol.
a) Moles of sodium carbonate transferred = (mass of sodium carbonate transferred)/(gram FM of sodium carbonate) = (0.67671 g)/(105.99 g/mol) = 0.0063846 mole 0.00638 mole (ans).
b) Calculate the relative error in measurement of mass u(m)/m = (uncertainty in mass measured)/(mass measured) = (0.2 mg)/(0.67671 g) = (0.2 mg)*(1 g/1000 mg)/(0.67671 g) = 0.00029
Relative uncertainty in the volume of the volumetric flask u(V’)/V’ = (uncertainty in volume calibrated)/(volume of volumetric flask) = ((0.08 mL)/(250.12 mL) = 0.003198 0.00032.
Relative uncertainty in the volume of the pipette u (V”)/V” = (uncertainty in volume calibrated)/(volume of pipette) = (0.005 mL)/(9.996 mL) = 0.0005
All the three uncertainties are of the same order (10-4); however the relative uncertainty in volume calibration of the pipette is the highest and will contribute the maximum to the uncertainty in the measured molar concentration (ans).
c) Molar concentration of sodium carbonate = (moles of sodium carbonate taken)/(volume of sodium carbonate taken in L) = (0.00638 mole)/[(250.12 mL)*(1 L/1000 mL)] = 0.0255 mol/L = 0.0255 M.
Relative uncertainty in the measured molar concentration u(M)/M = [u(m)/m]2 + [u(V’)/V’]2 + u(V”)/V”]2 = [(0.00029)2 + (0.00032)2 + (0.0005)2] = 0.00066
Uncertainty in measured molar concentration u(M) = 0.00066*M = 0.00066*(0.0255 mol/L) = 0.00001683 M 0.000017 M = 1.7*10-5 M.
The measured molar concentration of sodium carbonate is 0.0255 1.7*10-5 M (ans).
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