Chemistry: A 32.149 mg sample of a chemical known to contain only carbon hydrogen sulfur and oxygen is put into a combustion analysis apparatus yielding 57.271 mg of co2 and 23.444 mg of h2o. In another experiment 22.345 mg of the compound is reacted with excess oxygen to produce 9.656 mg of sulfur dioxide

A 32.149 mg sample of a chemical known to contain

gonzalesrosalinda66

05.02.2022

C₅H₁₀SO₂

Explanation:

From the given information:

The objective is to identify the empirical formula of the compound.

To start with determining the mass of carbon from carbon dioxide; we have:

$Mass \ of \ Carbon = \dfrac{molar \ mass \ of \ C }{molar \ mass \ of \ CO_2}\times mass \ of \ CO_2$

Mass of carbon = 12/44 × 57.271 mg

Mass of carbon = 0.2727 × 57.271 mg

Mass of carbon = 15.6178 mg

The mass of hydrogen is:

$Mass \ of \ Hydrogen= \dfrac{molar \ mass \ of \ H }{molar \ mass \ of \ H_2O}\times mass \ of \ H_2O$

$Mass \ of \ Hydrogen= \dfrac{2\times 1.008 }{18}\times 23.444 \ mg$

Mass of hydrogen = 2.6257 mg

The mass of sulphur is:

$Mass \ of \ Sulphur = \dfrac{molar \ mass \ of \ S }{molar \ mass \ of \ SO_2}\times mass \ of \ SO_2$

Mass of Sulphur = $\dfrac{32.07}{64.06} \times \dfrac{9.656}{1} \times \dfrac{32.149}{ 22.345}$

Mass of Sulphur = 6.9550 mg

The Mass of oxygen can now be = mass of (Sample - Carbon - Hydrogen - Sulphur)

= (32.149 - 15.6178 - 2.6257 - 6.9550)g

= 6.9505 g

Recall that:

number of moles = mass/molar mass

Thus:

The moles of C : H : S : O are:

$= \dfrac{15.6178}{12} : \dfrac{2.6257}{1.008} : \dfrac{6.9550}{32.07} : \dfrac{6.9505}{16}$

= 1 : 2 : 0.2 : 0.4

Divide by the smallest; we have:

= 1/0.2 : 2/0.2 : 0.2/0.2 : 0.4/0.2

= 5 : 10 : 1 : 2

Thus, the empirical formula is = C₅H₁₀SO₂

0,0(0 оценок)