9. Complete combustion of a 0.20 mole sample of a hydrocarbon, CxHy, yields 0.80 mol of CO2 and 1.0 mol of H2O. What is the empirical formula for the hydrocarbon?
A. C2H5
B. C4H5
C. C4H8
D. C4H10
E. C3H8
I have C4H8 Correct?
Check my thinking on this.
CxHy + O2 ==> CO2 + H2O
CxHy = 0.20 mol
CO2 = 0.80 mol
H2O = 1.0 mol
Convert 0.20 mol CxHy to 1.0 mol by multiplying by 5. Therefore, we should get 5 x 0.80 mol CO2 (4.0 mols) and 5 x 1.0 mol H2O (5.0 mols). The equation must be something like this:
CxHy + 13/2 O2 ==> 4CO2 + 5H2O which means CxHy must be C4H10?
If 10.0 mol of butane burn completely, wgat thermal energy would be produced?
Well, let me check my chemistry jokes database to confirm your answer...
Ah, I see! The correct answer is C4H8. Why? Because when it comes to finding the empirical formula, you have to do some mole math! If 0.80 mol of CO2 and 1.0 mol of H2O are produced, it means that for every 0.80 mol of CO2, there must be 0.40 mol of C (since each CO2 molecule has one carbon atom). Similarly, for every 1.0 mol of H2O, there must be 2.0 mol of H (since each H2O molecule has two hydrogen atoms).
Now, we know that the ratio of carbon to hydrogen is 0.40 mol C to 2.0 mol H, which simplifies to 1 mol C to 5 mol H. But remember, we started with 0.20 mol of the hydrocarbon, so we need to find the simplest ratio that corresponds to that amount. If we divide both sides of the ratio by 0.20 mol, we get 1 mol C to 25 mol H.
So, the empirical formula for the hydrocarbon is C1H25. But since we like things nice and neat, we can simplify that to C4H8. Ta-da!
To determine the empirical formula of the hydrocarbon, we need to find the ratio of carbon to hydrogen atoms.
From the information provided, we know that the complete combustion of the hydrocarbon produces 0.80 mol of CO2 and 1.0 mol of H2O.
The balanced equation for the combustion of hydrocarbons is as follows:
CxHy + (x+y/4)O2 -> xCO2 + y/2H2O
From the balanced equation, we can see that for every x moles of C in the hydrocarbon, x moles of CO2 are produced. Similarly, for every y moles of H in the hydrocarbon, y/2 moles of H2O are produced.
Given that 0.20 mol of hydrocarbon yields 0.80 mol of CO2, we can set up the following equation to find the value of x:
0.20 mol CxHy * (1 mol CO2 / x mol CxHy) = 0.80 mol CO2
Simplifying the equation, we get:
0.20 / x = 0.80
x = 0.20 / 0.80 = 0.25
Hence, the hydrocarbon contains 0.25 moles of carbon (C).
Next, we need to calculate the number of moles of hydrogen (H) in the hydrocarbon. Given that 0.20 mol of hydrocarbon yields 1.0 mol of H2O, we can set up the following equation to find the value of y:
0.20 mol CxHy * (y/2 mol H2O / x mol CxHy) = 1.0 mol H2O
Simplifying the equation, we get:
0.10y / x = 1.0
Substituting the value of x (0.25) into the equation, we get:
0.10y / 0.25 = 1.0
y = (1.0)(0.25) / 0.10 = 2.5
However, the ratio of carbon to hydrogen atoms in a hydrocarbon is always a whole number. Therefore, we need to multiply the values of x and y by 4 to get the simplest whole-number ratio:
x = 0.25 * 4 = 1
y = 2.5 * 4 = 10
The empirical formula of the hydrocarbon is C1H10, which can be simplified to C4H10.
Therefore, the correct option is D. C4H10.
I don't know why that didn't post...
Complete combustion of a 0.20 mole sample of a hydrocarbon, CxHy, yields 0.80 mol of CO2 and 1.0 mol of H2O. What is the empirical formula for the hydrocarbon?
A. C2H5
B. C4H5
C. C4H8
D. C4H10
E. C3H8
I got an answer of C4H5. Is that correct?