A compound of carbon, hydrogen and oxygen was burned inoxygen, and 1.000g of the compound produced 1.418g CO2 and 0.871gH2O. In another experiment, 0.1103g of the compound was dissolvedin 45.0g of water. This solution had a freezing point of -0.0734C. What is the molecular formula of the compound

C₂H₆O₂

Explanation:

Using the amount of CO2 and H2O we can determine the empirical formula. With the other experiment, we can determine the molar mass. With molar mass and empirical formula we can determine the empirical formula:

Empirical formula:

1g of the compound react producing CO₂ and H₂O. All Carbon of the compound produce CO₂ and all H produce H₂O:

Moles CO₂ = Moles C:

1.418g CO2 * (1mol / 44g) = 0.03223 moles C

Moles H₂O = 1/2 Moles H:

0.871g H₂O * (1mol / 18g) = 0.04839 moles H₂O * 2 = 0.09678 moles H

Mass of C and H:

Mass C: 0.03223 moles C * (12g / mol) = 0.387g C

Mass H: 0.09678 moles H * (1g / mol) = 0.097g H

Mass O: 1g - 0.387g C - 0.097g H = 0.516g O

Moles O:

0.516g O * (1mol / 16g) = 0.03225 moles O

Empirical formula:

C: 0.03223moles C / 0.03223 = 1

Dividing in the lower number of moles

H: 0.09678 moles H / 0.03223 = 3

O: 0.03225 moles O / 0.03223 = 1

Empirical formula:

CH₃O

And weighs: 12g/mol + 3*1g/mol + 16g/mol = 31g/mol

Molar mass:

Following freezing point depression:

ΔT = Kf*m*i

Where ΔT is change in freezing point (0.0734°C).

Kf is freezing point depression constant (1.86°C/m)

m is molality (Moles substance / kg water)

i is Van't Hoff factor (1 for organic compounds with C, H and O)

ΔT = Kf*m*i

0.0734°C = 1.86°C/m*m*1

m = 0.03946m = Moles compound / kg of water

As mass of water is 45.0g = 0.045kg:

Moles = 0.045kg * 0.03946 = 1.776x10⁻³ moles of compound}

In 0.1103g:

Molar mass:

0.1103g / 1.776x10⁻³ moles = 62g/mol

As molar mass of the compound is twice the molar mass of the empirical formula, the molecular formula is twice the empirical formula, that is:

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