Answer:
The number before any molecular formula applies to the entire formula. So here you have five molecules of water with two hydrogen atoms and one oxygen atom per molecule. Thus you have ten hydrogen atoms and five oxygen atoms in total.
CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).
Answer:
Enthalpy of vaporization = 30.8 kj/mol
Explanation:
Given data:
Mass of benzene = 95.0 g
Heat evolved = 37.5 KJ
Enthalpy of vaporization = ?
Solution:
Molar mass of benzene = 78 g/mol
Number of moles = mass/ molar mass
Number of moles = 95 g/ 78 g/mol
Number of moles = 1.218 mol
Enthalpy of vaporization = 37.5 KJ/1.218 mol
Enthalpy of vaporization = 30.8 kj/mol
