A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 44./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 24.01g water 13.10g Use this information to find the molecular formula of X.

C3H6.

Explanation:

Data obtained from the question:

Mass of the compound = 8g

Mass of CO2 = 24.01g

Mass of H2O = 13.10g

Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01

=> 12/44 x 24.01 = 6.5g

Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1

=> 2x1/18 x 13.1 = 1.5g

Mass of O in the compound = Mass of compound – (mass of C + Mass of H)

=> 8 – (6.5 + 1.5) = 0

Next, we shall determine the empirical formula of the compound. This is illustrated below:

C = 6.5g

H = 1.

Divide by their molar mass

C = 6.5/12 = 0.54

H = 1.4/1 = 1.

Divide by the smallest

C = 0.54/0.54 = 1

H = 1/0.54 = 2

Therefore, the empirical formula is CH2

Finally, we shall determine the molecular formula as follow:

The molecular formula of a compound is a multiple of the empirical formula.

Molecular formula = [CH2]n

[CH2]n = 44

[12 + (2x1)]n = 44

14n = 44

Divide both side by 14

n = 44/14

n = 3

Molecular formula = [CH2]n = [CH2]3 = C3H6

Therefore, the molecular formula of the compound is C3H6

i just got this problem. it's 90.3

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.