Question

1.50 g sample of hydrocarbon undergoes complete combustion to produce

4.40 g of CO2 and 2.70 g of H2O. What is the empirical formula of this compound? In

addition, its molecular weight has been determined to be about 78. What is the

molecular formula?​

Answer 1

Explanation:

The first step here is to determine the mass of C in CO2 and the mass of H in H2O. This is done by dividing the atomic mass by the molecular mass and then multiplying by the mass of compound produced.

For C:

(12.011 g / 44.009 g) x 4.40 g = 1.1999 g

For H:

(1.0079 x 2 / 18.0148 g) x 2.70 g = 0.3021 g

The next step is to convert these masses into moles. This is done by dividing the mass by the relative atomic mass of the element:

C: 1.1999 g / 12.011 g mol-1 = 0.0999 mol

H: 0.3021 g / 1.0079 g mol-1 = 0.2997 mol

The final step is to divide each of these two values by the smallest number, in this case this is the number of moles of carbon:

C: 0.0999 mol / 0.0999 mol = 1

H: 0.2997 mol / 0.0999 mol = 3

We therefore have a ratio of 1 carbon atom to 3 hydrogen atoms, thus the empirical formula for this hydrocarbon is CH3.

Answer 2

Answer:

see the attachment ✌️❣️