A 0.352 g sample of an unknown compound containing C, H, and O was burned to determine the carbon and hydrogen content. The analysis showed that 0.844 g carbon dioxide and 0.173 g water was formed. The molar mass of the compound is measured to be 110.1 g/mol. What is the molecular formula of this compound?

Explanation:

First, let's write the general reaction.

CₓYₐZₙ > CO₂ + H₂O

To get the molecular formula of this sample, we first need to know the empirical formula, and to do that, we need the mass of each element in the sample.

We have the mass of CO2 and water, so we can calculate the mass of C and H from there, using the molecular masses:

mol CO2 = 0.844 / 44 = 0.0192 moles

mol H2O = 0.173 / 18 = 0.0096 moles

Now that we have the moles, we can calculate the mass of C and H using the atomic mass of each element:

m C = 0.0192 * 12 = 0.2304 g

m H = 0.0096 * 2 = 0.0192 g

To get the mass of oxygen, we can get from difference of the total and the obtained masses:

m O = 0.352 - 0.2304 - 0.0192 = 0.1024 g

We have the mass of each element, we can calculate the empirical formula and then the molecular formula, using 6 steps:

Step 1: calculate the moles of each mass

In this case, we calculate the moles of each mass using the atomic weight:

mol C = 0.2304 / 12 = 0.0192 moles

mol H = 0.0192 / 1 = 0.0192 moles

mol O = 0.1024 / 16 = 0.0064 moles

Step 2: calculate the mole ratio of the elements

In this case, we just divide the previous moles by the smallest moles there. In this case, moles of oxygen

C = 0.0192 / 0.0064 = 3

H = 0.0192 / 0.0064 = 3

O = 0.0064 / 0.0064 = 1

Step 3: write the empirical formula with the previous results

The previous results show the number of atoms of each element in the empirical formula, so we can write now the empirical formula:

e.F = C₃H₃O

Step 4: calculate the molecular mass of the empirical formula

Now that we have the empirical formula, we can calculate the molecular mass of this, which is:

MM = (3*12) + (3*1) + (16*1) = 55 g/mol

Step 5: calculate the molecular mass ratio of both

With the molecular weight of the empirical formula, and the molecular formula which is 110.1, let's calculate the ratio between these 2. This will give us a factor number that will be used to get the molecular formula

f = 110.1 / 55 = 2.002

We can round this number to 2.

Step 6: multiply the f number with the empirical formula

With the previous obtained number, we can get the molecular formula:

mF = f * eF

mF = 2 * C₃H₃O

Molecular Formula = C₆H₆O₂

This is the molecular formula of the sample

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.