Question

A combustion analysis of a 0.44g sample of an unknown compound yields 0.88 g CO​_2​ and 0.36 g H​_2​O. If the sample a molar mass of 132 g/mol, what is the molecular formula of the sample? Give your answer in the form C#H#O# where the number following the element’s symbol corresponds to the subscript in the formula. (Don’t include a 1 subscript explicitly). For example, the formula CH​_2​O would be entered as CH2O.

Answer 1

C₆H₁₂O₃

Step-by-step explanation:

The combustion reaction occurs between a fuel and oxygen. How the products only have carbon, oxygen, and hydrogen, so the compound must be CxHyOz. All the moles of carbon will form CO₂, and all moles of hydrogen will form H₂O. So, let's calculate the number of moles of each element.

The molar masses are C = 12 g/mol, O= 16 g/mol, H = 1 g/mol

CO₂ = 12x1 + 16x2 = 44 g/mol

H₂O = 1x2 + 16x1 = 18 g/mol

So, the number of moles of CO₂ is the mass divided by the molar mass:

n = 0.88/44 = 0.02 mol

In 1 mol of CO₂, there is 1 mol of C, so the number of moles of C is 0.02 mol.

For water,

n = 0.36/18 = 0.02 mol

In 1 mol of H₂O, there are 2 moles of H, so the number of moles of H is 0.04 mol.

So, the number of atoms in the compound will be the number of moles multiplied by the molar mass and divided by the total mass of the compound:

C = 0.02x132/0.44 = 6

H = 0.04x132/0.44 = 12

So, to know the total of oxygen, we must calculate the molar mass of the compound:

6x12 + 1x12 + zx16 = 132

16z = 48

z = 3

The compound is C₆H₁₂O₃