2. For laboratory data related to carbon dioxide, Calculate the molecular weight

Question

2. For laboratory data related to carbon dioxide, Calculate the molecular weight of carbon dioxide. Then estimate the standard deviation of the result and discuss its reliability. a. Mass of evacuated container, me Mass of container + CO2, mr Ideal gas constant, R CO2 temperature, T Barometric pressure, P Barometer correction, Pc Volume of container, V -94.275 ± 0.003 g 94.675 0.003 g -0.08205 0.00001 L-atm-molK 297.3 0.3 K #748.4 0.2 torr 3.2 t 0.2 torr = 228.0 ± 0.5 mL Locate the measurement contributing most heavily to the uncertainty of the molecular weight value. What is the single most significant alteration that could be made to the experiment to decrease the uncertainty in MW? To what level could the single-value uncertainty be reduced by this alteration? b.

Explanation / Answer

a) Convert the pressures to atmospheres. We know that

Pb – Pc = (748.4 – 3.2) torr

= 745.2 torr

= (7452 torr)*(1 atm)/(760 torr) [1 atm = 760 torr]

= 0.9805 atm

0.981 atm

Also the volume of the container is 228.0 mL = (228.0 mL)*(1 L)/(1000 mL) = 0.228 L.

Use the numerical values without the uncertainty. Plug in values and obtain

MW = (mf – me)RT/(Pb – Pc)V

= (94.675 g – 94.275 g)*(0.08205 L-atm.mol-1.K-1)*(297.3 K)/(0.981 atm)(0.228 L)

= (0.400 g)*(0.08205 L-atm.mol-1.K-1)*(297.3 K)/(0.981 atm)(0.228 L)

= 43.6244 g.mol-1

43.624 g.mol-1 (correct to 3 decimal places).

This is not the correct MW of CO2. We need to determine the uncertainty associated with the result. We shall use the propagation of errors to calculate the uncertainty.

Uncertainty in me, me/me = (0.003 g)/(94.275 g) = 3.1822*10-5

Uncertainty in mf, mf/mf = (0.003 g)/(94.675 g) = 3.1687*10-5

Uncertainty in R, R/R = (0.00001 L-atm.mol-1.K-1)/(0.08205 L-atm.mol-1.K-1) = 1.2188*10-4

Uncertainty in T, T/T = (0.3 K)/297.3 K) = 1.0091*10-3

Uncertainty in Pb, Pb/Pb = (0.2 torr)/(748.4 torr) = 2.6724*10-4

Uncertainty in Pc, Pc/Pc = (0.2 torr)/(3.2 torr) = 0.0625

Uncertainty in V, V/V = (0.5 mL)/(228.0 mL) = 2.1930*10-3

The uncertainty in the MW of CO2 is given as

MW/MW = [(me/me)2 + (mf/mf)2 + (R/R)2 + (T/T)2 + (Pb/Pb)2 + (Pc/Pc)2 + (V/V)2]

= [(3.1822*10-5)2 + (3.1687*10-5)2 + (1.2188*10-4)2 + (1.0091*10-3)2 + (2.6724*10-4)2 + (0.0625)2 + (2.1930*10-3)2]

= 3.9122*10-3

= 0.0625

The absolute uncertainty in the MW of CO2 is (MW/MW)*(MW) = (0.0625)*(43.624 g.mol-1) = 2.7265 g.mol-1

2.726 g.mol-1.

The MW of CO2 is 43.6242.726 g.mol-1 (ans).

The accepted MW of CO2 is 44.01 g.mol-1; therefore, the absolute error in the measurement of the MW of CO2 = [(44.01 – 43.624)g.mol-1]/(44.01 g.mol-1)*100

= 0.877%.

Since the measured MW of CO2 varies from the accepted MW by less than 1%, hence, the calculations seems fairly accurate.

b) The parameter that contributes the most to the measurement of MW of CO2 is Pc with an uncertainty of 0.0625.

The experimental details are needed to answer the second part.

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