Combustion analysis of a 13.42-g sample of an unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 10.06 g H2O. The molar mass of the compound is 288.38 g/mol . Part A Find the molecular formula of the unknown compound. Express your answer as a chemical formula.

Explanation:

Hello,

In this case, combustion analyses help us to determine the empirical formula of a compound via the quantification of the released carbon dioxide and water since the law of conservation of mass is leveraged to attain it. In such a way, as 36.86 g of carbon dioxide were obtained, this directly represents the mass of carbon present in the sample, thus, we first compute the moles of carbon:

Then, into the water one could find the moles of hydrogen:

Now, we compute the moles of oxygen by firstly computing the mass of oxygen:

Then, we have the mole ratio:

Whose molar mass is 12x6+1x8+16=96 g/mol, but the whole compound molar mass is 288.38, the factor is 288.38/96 =3, Therefore the formula:

Regards.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).