Chemistry Labs

Problem 3

A 10 cm3^3 sample of an unknown gaseous hydrocarbon was mixed with 70 cm3^3 of oxygen and the mixture was set on fire by means of an electric spark. When the reaction was over and water vapour was liquefied, the final volume of gases decreased to 65 cm3^3. This mixture then reacted with a potassium hydroxide solution and the volume of gases decreased to 45 cm3^3. What is the molecular formula of the unknown hydrocarbon, all volumes being measured at standard temperature and pressure (STP)?
Step 1 of 3: Read off the gas volumes
← PreviousNext →
V(COX2)=65−45=20 cm3;V(OX2)left=45 cm3V(\ce{CO2}) = 65 - 45 = 20\ \text{cm}^3; \quad V(\ce{O2})_{\text{left}} = 45\ \text{cm}^3
Analysis

KOH removes CO2 quantitatively, so the drop from 65 to 45 cm3 is the CO2 formed; the residual 45 cm3 is unreacted O2. Hence O2 consumed = 70 − 45 = 25 cm3.