Complete combustion of 2.30 g of a hydrocarbon produced 7.39 g of CO2 and 2.52 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary.

C3H5

Explanation:

Since it’s a hydrocarbon, it contains only carbon and hydrogen. We proceed to deduce it’s empirical formula as follows:

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 7.39/44 = 0.168

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

From the mass of water produced, we can obtain the number of moles of hydrogen. The number of moles of water produced is 2.52/18 = 0.14

Since there are two moles of hydrogen in one mole of water, the number of moles of hydrogen is thus 0.14 * 2 = 0.28 moles

We now proceed to get the actual number of atoms of each of the elements. The ratio of C to H is 0.168:0.28

= 0.6:1

Let’s multiply through by 5 to get a whole number = 3:5

The empirical formula is thus C3H5

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol