The bacterial fermentation of grain to produce ethanol forms a gas with a
percent composition of 27.29% C and 72.71% O, What is the empirical formula
for this gas? (3)
(At.wt: C=12, O=16)
Answers
Answer:
Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages). (.7546) (100 g) = 75.46 g C. ...
Convert the mass of each element to moles. ...
Find the ratio of the moles of each element. ...
Use the mole ratio to write the empirical fomula.
The empirical formula of the gas, which is CO2, carbon dioxide.
To determine the empirical formula of the gas produced by bacterial fermentation of grain to produce ethanol, we need to use the percent composition of carbon and oxygen provided. We assume that the gas is made up of only carbon and oxygen.
To begin, we need to convert the percent composition to grams. Assuming we have 100 grams of the gas, we can calculate that it contains 27.29 grams of carbon and 72.71 grams of oxygen.
Next, we need to find the ratio of carbon to oxygen in the gas. We can do this by dividing the number of grams of each element by its atomic weight and then dividing by the smallest result. In this case, the smallest result is 1.707 for oxygen.
So, the ratio of carbon to oxygen is:
Carbon: 27.29 g / 12 g/mol / 1.707 = 1.000 (to three significant figures)
Oxygen: 72.71 g / 16 g/mol / 1.707 = 2.000 (to three significant figures)
This gives us the empirical formula of the gas, which is CO2, carbon dioxide.
In conclusion, the empirical formula for the gas produced by bacterial fermentation of grain to produce ethanol is CO2, meaning that the gas is composed of one carbon atom and two oxygen atoms.
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