ok thank you for elaborating and you are correct in a neutralization reaction the reactants are the acid and base. the salt and water would be the product and the two elements could be anything meaning that it is most likely a mixture
Hope this helps
Answer:
THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.
Explanation:
Mass of the unknown substance = 0.50 g
Freezing point of the solution = 3.9 °C
Freezing point of pure benzene = 5.5 °C
Freezing point dissociation constant Kf = 5.12°C/m
First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.
Change in temperature = 5.5 -3.9 = 1.6 °C
Next is to calculate the number of moles or molarity of the compound that dissolved.
Using the formula:
Δt = i Kf m
Assume i = 1
So,
1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene
x = 1.6 * 0.008 / 5.12
x = 0.0128 / 5.12
x = 0.0025 moles.
Next is to calculate the molar mass using the formula, molarity = mass / molar mass
Molar mass = mass / molarity
Molar mass = 0.50 g /0.0025
Molar mass = 200 g/mol
Hence, the molar mass of the unknown compound is 200 g/mol
The empirical formula of the compound is calculated as follows
first calculate the mass of oxygen= 12-(4.09 +3.71)= 5.02g
then calculate the moles of each element, moles = mass/ molar mass
moles of K = 4.09g/39 g/mol(molar mass of K) = 0.105 moles
moles of Cl = 3.71g/35.5 g/mol(molar mass of Cl) = 0.105 moles
moles of O = 5.02g/ 16g/mol(molar mass of O) = 0.314 moles
then calculate e mole ratio by dividing each mole by the smallest number of moles ( 0.105 moles)
K=0.105/0.105= 1
Cl=0.105 /0.105=1
O= 0.314/0.105=3
therefore the empirical formula = KClO3
I am pretty sure it is 4.002602
