a hydrocarbon was burnt completely and the products were bubbled into excess of Ca(OH)2 solution. The resulting mixture was filtered, and the residue dried and weighed. The mass of the hydrocarbon was found to be 10.00g. In a separate experiment, the same amount of hydrocarbon was burnt an the mass of the water produced was 2.25g. Determine the empirical formula of the hydrocarbon

for the sake of my understanding could you explain what this resulting solution is?

Hydrocarbon+O2>>> CO2 + H2O

CO2+Ca(OH)2>>CaCO3 + H2O
mass CaCo3=10g
MassC=10/molmassCaCO3*atomicmassC
= 10/100*12=1.2 g of C=.1mol

Mass water=2.25
moles H = 2.25/18 * 2=.25

lowest ratio C:.1 H:.25 or
CH.25 or CH4 is the empirical formula.

C2H5

I should point out here that your problem makes no sense to me. True that the CO2 bubbled into Ca(OH)2 produces CaCO3 BUT you say the mass of the hydrocarbon is 10 g. So did you start with 10 g hydrocarbon or did the CaCO3 weigh 10 g. I don't call CaCO3 a hydrocarbon but chemistry may have changed since I was 15. I believe Bob Pursley made an "eye" typo, as he shows C as 0.1 and H as 0.25 but read that as C of 1 and H of 0.25. C of 0.1 and H of 0.25 gives C2H5.

Certainly! The resulting solution after the hydrocarbon was burnt completely and bubbled into excess of Ca(OH)2 solution is a mixture containing the products of the combustion reaction.

When a hydrocarbon is burned, it reacts with oxygen to produce carbon dioxide (CO2) and water (H2O) as the primary products. In this case, the products of the combustion reaction are captured in the solution.

The excess of Ca(OH)2 solution is added to the mixture to capture carbon dioxide. Calcium hydroxide (Ca(OH)2) reacts with carbon dioxide (CO2) to form calcium carbonate (CaCO3) and water (H2O). This reaction helps remove the carbon dioxide from the mixture and makes it easier to filter and separate the desired product later on.

After the solution is filtered, the residue left behind is dried and weighed, allowing us to determine the mass of the hydrocarbon that was originally burnt.

Now that we understand how the resulting solution is obtained, let's move on to determining the empirical formula of the hydrocarbon based on the given information.