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?
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.
To determine the empirical formula for the gas produced by bacterial fermentation, you can start by assuming that you have 100 grams of the gas.
Given that the gas is composed of 27.29% C and 72.21% O, we can calculate the number of grams for each element:
Mass of carbon (C) = 27.29 g
Mass of oxygen (O) = 72.21 g
Next, we need to convert the mass of each element to moles by dividing by their respective atomic masses:
Atomic mass of carbon (C) = 12.01 g/mol
Atomic mass of oxygen (O) = 16.00 g/mol
Moles of carbon (C) = 27.29 g / 12.01 g/mol ≈ 2.27 mol
Moles of oxygen (O) = 72.21 g / 16.00 g/mol ≈ 4.51 mol
Now, we need to find the simplest whole number ratio between the moles of each element. Divide each of the mole values by the smallest number of moles:
Moles of carbon (C) / 2.27 ≈ 1 mol
Moles of oxygen (O) / 2.27 ≈ 1.99 mol
The ratio of moles gives us approximately 1 mol of carbon to 2 moles of oxygen. Therefore, the empirical formula for the gas is CO2.
So, the empirical formula for the gas produced by bacterial fermentation is CO2.
To determine the empirical formula of the gas formed by bacterial fermentation, we need to find the ratio of different elements present in the gas.
Given:
Percent composition of carbon (C) = 27.29%
Percent composition of oxygen (O) = 72.21%
Step 1: Convert the percent composition to grams.
Assume we have 100g of the gas, this means:
- Mass of carbon (C) = 27.29g
- Mass of oxygen (O) = 72.21g
Step 2: Convert the grams to moles.
To convert grams to moles, we use the molar mass of each element.
The molar mass of carbon (C) is approximately 12 g/mol.
The molar mass of oxygen (O) is approximately 16 g/mol.
Number of moles of carbon (C) = Mass of carbon (C) / Molar mass of carbon (C)
= 27.29g / 12 g/mol
≈ 2.274 moles
Number of moles of oxygen (O) = Mass of oxygen (O) / Molar mass of oxygen (O)
= 72.21g / 16 g/mol
≈ 4.513 moles
Step 3: Find the simplest whole number ratio.
Divide the number of moles of each element by the smallest number of moles to get the simplest whole number ratio.
Carbon (C) = 2.274 moles / 2.274 moles = 1
Oxygen (O) = 4.513 moles / 2.274 moles ≈ 1.98 (approximately)
Since we need to find the empirical formula, we cannot have fractional subscripts. Therefore, we need to multiply the ratio by 2 to get whole numbers.
Carbon (C) = 1 * 2 = 2
Oxygen (O) = 1.98 * 2 ≈ 4
Step 4: Write the empirical formula.
The empirical formula for the gas formed by bacterial fermentation is CO₂.
So, the empirical formula for this gas is CO₂.