On complete oxidation, one mole of an organic compound produced four moles of water. Which one of the following substances could it be?
CH4
C2H6
C3H7OH
C4H10
So I added O2 to each of these compounds, but I must be doing something wrong because I can't get 4 moles of water. Help? Thanks.
products: 4H2O + XCO2
so you know H is at least 8 in the water, and to get CO2 (required, it has to be greater).
so start with C4H10+ ?O2>>4H2O+ X CO2 Which has CxH8 in it?
only one fits.
C3H7OH right?
To determine which organic compound could produce four moles of water upon complete oxidation, let's analyze the structures of each compound and their oxidation products.
1. CH4 (Methane):
CH4 + 2O2 → CO2 + 2H2O
For every mole of methane oxidized, only two moles of water are produced, not four. Therefore, CH4 is not the answer.
2. C2H6 (Ethane):
C2H6 + 3.5O2 → 2CO2 + 3H2O
For every mole of ethane oxidized, three moles of water are produced. Again, not four moles. So C2H6 is not the answer.
3. C3H7OH (Isopropanol):
C3H7OH + 4.5O2 → 3CO2 + 4H2O
For every mole of isopropanol oxidized, four moles of water are produced. However, the given information states that only one mole of organic compound produces four moles of water. So C3H7OH does not fit the criteria.
4. C4H10 (Butane):
C4H10 + 6.5O2 → 4CO2 + 5H2O
For every mole of butane oxidized, five moles of water are produced, which satisfies the given information. Therefore, C4H10 could be the organic compound that produces four moles of water upon complete oxidation.
In summary, the organic compound that could produce four moles of water upon complete oxidation is C4H10 (Butane).
To find out which organic compound produces four moles of water upon complete oxidation, we need to balance the equation for the combustion of each compound. Let's go through each compound and balance the equation to figure out the correct one.
1. CH4 (methane):
The balanced equation for the combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O
In this reaction, one mole of methane produces only two moles of water, not four.
2. C2H6 (ethane):
The balanced equation for the combustion of ethane is:
2C2H6 + 7O2 -> 4CO2 + 6H2O
In this reaction, two moles of ethane produce six moles of water, which is also not four.
3. C3H7OH (isopropyl alcohol):
The balanced equation for the combustion of isopropyl alcohol is:
C3H7OH + 9/2O2 -> 3CO2 + 4H2O
In this reaction, one mole of isopropyl alcohol (C3H7OH) produces four moles of water. Therefore, C3H7OH is the correct compound that produces four moles of water upon complete oxidation.
4. C4H10 (butane):
The balanced equation for the combustion of butane is:
2C4H10 + 13O2 -> 8CO2 + 10H2O
In this reaction, two moles of butane produce ten moles of water, which is also not four.
So, out of the given options, the compound that can produce four moles of water upon complete oxidation is C3H7OH, which is isopropyl alcohol.