Chemistry, 23.06.2019 23:20 Kaitneedshelps

An unknown organic compound (0.315 g) containing only c, h, and o produces 0.771 of co2 and 0.105 g of h2o when it undergoes complete combustion. the approximate molar mass is 108 g/mol. which of the following compounds could be the identification of the unknown?

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Answer from: shahedraed82

C₆H₄O₂

Explanation:

Given parameters:

Mass of the unknown compound = 0.315g

Atoms contained in the compound C, H and O

Mass of CO₂ produced = 0.771g

Mass of H₂O = 0.105g

Molar mass of compound = 108gmol⁻¹

The complete combustion of most hydrocarbon compounds like this would produce carbon dioxide and water only.

Solution

We first find the mass of the Carbon, hydrogen and Oxygen in the compounds given. We then continue to derive the empirical formula of the compound. From the empirical formula, we can find the compound from the given molar mass.

We first find the mass of the Carbon, hydrogen and Oxygen in the compounds given.

Mass of carbon in compound = $\frac{Molar mass of Carbon}{Molar mass of CO_{2} }$ x mass of CO₂

Molar mass of C = 12

Molar mass of CO₂ = 12 + (16x2) = 44

Mass of CO₂ produced = 0.771g

Mass of carbon in compound = $\frac{12}{44 }$ x 0.771=0.2103g

Mass of H in compound = $\frac{Molar mass of H}{Molar mass of H_{2}O }$ x mass of H₂O

Molar mass of H in H₂O = 2

Molar mass of H₂O = 2 + 16 = 18

Mass of H₂O = 0.105g

Mass of H in compound = $\frac{2}{18}$ x 0.105 = 0.012g

Now, to find the mass of oxygen in the compound, we sum the mass H and C and subtract from the mass of the compound given:

Mass of oxygen = 0.315 - (0.2103 + 0.012)= 0.0927g

We then continue to derive the empirical formula of the compound.

C H O

mass of

atoms 0.2103 0.012 0.0927

Moles 0.2103/12 0.012/1 0.0927/16

0.018 0.012 0.006

Dividing by

the smallest 0.018/0.006 0.012/0.006 0.006/0.006

3 2 1

The empirical formula of the compound is C₃H₂O

From the empirical formula, we can find the compound from the given molar mass:

Molecular formula = (Empirical formula)n

Where n is the number of repeating times of the empirical formula present in one mole of the molecule.

Therefore n = $\frac{molar mass of the compound}{molar mass of the empirical formula of the compound}$

Molar mass of the empirical formula C₃H₂O = (12x3) + (2x1) + (16) = 54g/mol

n = $\frac{108}{54}$ = 2

The molecular formula of the compound is = 2(C₃H₂O) = C₆H₄O₂

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An unknown organic compound (0.315 g) containing only c, h, and o produces 0.771 of co2 and 0.105 g...