Chemistry Labs

Problem 2

In a gaseous mixture of CO and COX2\ce{CO2}, a mass ratio of carbon : oxygen = 1 : 2 was determined. 2.1 Calculate the mass percent composition of the mixture. 2.2 Calculate the volume percent composition. 2.3 Indicate the values of the carbon : oxygen mass ratio for which both gases cannot be present simultaneously in the mixture.
Step 2 of 4: Solve and convert to mass percent
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x=y=1.389 mol;%COX2=1.389×44100=61.11 %,%CO=38.89 %x = y = 1.389\ \text{mol};\qquad \%\ce{CO2} = \tfrac{1.389 \times 44}{100} = 61.11\,\%,\quad \%\ce{CO} = 38.89\,\%
Analysis

The ratio equation simplifies to x=yx = y; substitution into 28x+44x=10028x + 44x = 100 gives x=y=1.389x = y = 1.389 mol, i.e. 61.11 % COX261.11\,\%\ \ce{CO2} and 38.89 % CO38.89\,\%\ \ce{CO} by mass.