The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.21% O. What is the empirical formula for this gas?

Answer 1

To find the empirical formula for the gas produced during the bacterial fermentation of grain, we need to determine the ratio of carbon to oxygen in the compound.

First, we assume we have 100g of the gas. We are given that the percent composition is 27.29% C and 72.21% O.

So, in 100g of the gas, we have 27.29g of carbon and 72.21g of oxygen.

Next, we need to convert grams to moles using the molar mass of each element:

- The molar mass of carbon (C) is 12.01g/mol.
- The molar mass of oxygen (O) is 16.00g/mol.

Converting grams to moles:

For carbon (C): 27.29g × (1 mol/12.01g) ≈ 2.27 mol of carbon.
For oxygen (O): 72.21g × (1 mol/16.00g) ≈ 4.51 mol of oxygen.

To determine the empirical formula, we need to find the simplest whole-number ratio of carbon to oxygen. We'll divide the number of moles of each element by the smallest number of moles (2.27 mol):

C: 2.27 mol ÷ 2.27 mol = 1
O: 4.51 mol ÷ 2.27 mol ≈ 1.99 (approximately 2)

Therefore, the empirical formula for this gas is CO2.