When the experimental values that I got are 1. Weight of inner air, separating f

Question

When the experimental values that I got are

1. Weight of inner air, separating funnel and rubber stooper (g) = 124.796g

2. Weight of sepearting funnel and the rubber stopper filled with CO2 (g) = 124.914

3. Temperature of CO2 gas (celsius) = 21.0

4. Volume of seperating funnel (mL) = 180mL

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Calculate the molar mass of CO2 based on the temperature and atmospheric pressure (1atm) of CO2 gas. (3 sig fig)

And compare the molar mass of CO2 calculated from the experiment with theoretical value. Calculate the error rate. (2 decimal places)

Explanation / Answer

mass of CO2=Weight of sepearting funnel and the rubber stopper filled with CO2 (g)- Weight of inner air, separating funnel and rubber stooper (g) =124.914g - 124.796g=0.118g

Volume of CO2 in the flask=180ml=V=0.180L

Temperature of CO2 gas (celsius) = 21.0=T=273+21=294K

P=atmospheric pressure=1atm

Now ,using ideal gas equation,

PV=nRT=(m/M)RT ,where n=moles of CO2=mass of CO2/molar mass of CO2

R=universal gas constant=0.0821L atm/K.mol

M=mRT/PV=0.118g*(0.0821 L atm K-1 mol-1)*294K/1atm*0.180L=15.823g/mol

M(experimental)=15.823g/mol

M(theoretical)=44.01g/mol

percentage error=|M(experimental)-M(theoretical)/M(theoretical)| *100=|15.823-44.01/44.01|*100=64.05% error

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