Question

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

Answer 1

Combustion analysis is a technique used to determine the composition of a compound that contains carbon and hydrogen. In this method, the compound is burned with excess oxygen, and the resulting gases (CO2 and H2O) are measured. From the measurements, we can calculate the amounts of carbon, hydrogen, and sulfur in the compound.

Step-by-step explanation:

Combustion analysis is a technique used to determine the composition of a compound that contains carbon and hydrogen. In this method, the compound is burned with excess oxygen, and the resulting gases (CO2 and H2O) are measured. By comparing the masses of CO2 and H2O produced, we can calculate the amount of carbon and hydrogen in the sample.

In the given question, a 32.149 mg sample of the compound is combusted, yielding 57.271 mg of CO2 and 23.444 mg of H2O. From these measurements, we can determine the amounts of carbon and hydrogen in the original sample. Additionally, in another experiment, 22.345 mg of the compound is reacted with excess oxygen to produce 9.656 mg of SO2, which allows us to determine the amount of sulfur in the compound.

Answer 2

C₅H₁₀SO₂

Step-by-step 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 = (molar \ mass \ of \ C )/(molar \ mass \ of \ CO_2)* 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= (molar \ mass \ of \ H )/(molar \ mass \ of \ H_2O)* mass \ of \ H_2O`

`Mass \ of \ Hydrogen= (2* 1.008 )/(18)* 23.444 \ mg`

Mass of hydrogen = 2.6257 mg

The mass of sulphur is:

`Mass \ of \ Sulphur = (molar \ mass \ of \ S )/(molar \ mass \ of \ SO_2)* mass \ of \ SO_2`

Mass of Sulphur =
`(32.07)/(64.06) * (9.656)/(1) * (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:

`= (15.6178)/(12) : (2.6257)/(1.008) : (6.9550)/(32.07) : (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