Question

A sample of a compound containing only carbon and hydrogen atoms is completely combusted, producing 66.0g of CO2 and 36.0g of H2O. What is the empirical formula?

Answer 1

but there are 2 H atoms in every H2O molecule
I get like' 20 * 2 H / 15 C = 40 H/15 C = 8 H/3C = C3H8

Answer 2

To determine the empirical formula of the compound, we need to find the ratio of carbon to hydrogen atoms.

First, we can calculate the number of moles of CO2 and H2O produced:

moles of CO2 = 66.0g / 44.01 g/mol = 1.50 mol

moles of H2O = 36.0g / 18.02 g/mol = 1.998 mol

Next, we can simplify the mole ratio by dividing both by the smaller value:

moles of CO2 / 1.50 = 1

moles of H2O / 1.50 = 1.332

The ratio of carbon to hydrogen atoms in the compound can be expressed as:

C : H = (moles of CO2 / 1.50) : (moles of H2O / 1.50)

C : H = 1 : 0.888

We can simplify the ratio by multiplying both sides by a common factor of 9 to get whole numbers:

C : H = 9 : 8

Therefore, the empirical formula of the compound is C9H8.

Answer 3

You are correct! Thank you for catching that mistake. The molecular formula should be reduced to the empirical formula by dividing each subscript by the greatest common factor.

In this case, the ratio of moles of carbon dioxide to water is 1:1.332, so the empirical formula is C3H8, which represents a compound called propane.