During the combustion of 3.6 g of organic matter, 5.6 liters of carbon dioxide and 5.4 g of water were formed

During the combustion of 3.6 g of organic matter, 5.6 liters of carbon dioxide and 5.4 g of water were formed. The relative density of the substance in terms of hydrogen is 36. Derive the formula of the substance and name all its isomers.

Let’s write down given:

m (org. substance) = 3.6 g

v (CO2) = 5.6 l

m (H2O) = 5.4 g

DH2 (organic matter) = 36

Find the formula of a substance -?

Name isomers

Decision:

1) Calculate the mass of carbon in 3.6 g of the starting material:

V m = 22.4 l / mol

Let’s compose the ratio:

22.4 l (CO2) contains 12 g (C)

5.6 l (CO2) will contain x g (C), hence

m (C) = 5.6 l * 12 g / 22.4 l = 3 g

2) Calculate the mass of hydrogen in 3.6 g of the starting material.

M (H2O) = 1 * 2 + 16 * 1 = 18 g / mol

Let’s compose the ratio:

18 g (H2O) contains 2 g (H)

5.4 g (H2O) will contain y g (H), hence

m (H) = 5.4 g * 2 g / 18 g = 0.6 g

3) Find the mass of oxygen:

m (O) = 3.6 g – 3 g – 0.6 g = 0 g, this substance is a hydrocarbon.

4) Calculate the amount of substance of each element

x: y = 3/12: 0.6 / 1 = 0.25: 0.6 = 5: 12

So the simplest formula of the desired substance is C5H12

5) Find the true formula.

M (CxHy) = D H2 * M (H2), M (H2) = 2 g / mol

M (CxHy) = 36 * 2 g / mol = 72 g / mol

Let’s calculate the molar mass of the simplest formula and compare it with the true molar mass.

M (C5H12) = 12 * 5 + 1 * 12 = 72 g / mol

The molar mass of the simplest and true formulas is the same, therefore, the desired substance

C5H12.

Let’s write down the structural formulas of possible isomers:

CH3 – CH2 – CH2 – CH2 – CH3 n – pentane

CH3 – CH (CH3) – CH2 – CH3 2 – methylbutane

CH3 – CH (CH3) 2 – CH3 2, 2 – dimethylpropane

Answer: С5Н12