Car exhaust

Relevant wiki: Atomic and Molecular Weights

The fact that oxygen (atomic weight = 16 u) is significantly heavier than hydrogen (atomic weight = 1 u) makes it clear that the combustion of gasoline releases an amount of carbon dioxide, which weighs much more than the burned fuel.

But originally this task was about specifying the exact value for CO2 emissions in kilograms for a car with a consumption of 6 liters of gasoline per 100 kilometers. The density of gasoline with $\rho = 0.76 \,\text{kg}/\text{L}$ was given. Therefore, I ignore the overly simplified problem and make an exact quantitative calculation.

Gasoline burns with atmospheric oxygen according to the reaction equation $2 \, C_8 H_{18} + 25 \, O_2 \to 18 \, H_2O + 16\, CO_2$ The important thing here is that one part of gasoline is converted to eight parts of carbon dioxide. The molars mass of gasoline and carbon dioxide are $\begin{aligned} M_{ C_8 H_{18}} & \approx (8 \cdot 12 + 18 \cdot 1) \,\text{g}/\text{mol} = 114 \,\text{g}/\text{mol} \\ M_{ C O_2} & \approx (1 \cdot 12 + 2 \cdot 16) \,\text{g}/\text{mol} = 44 \,\text{g}/\text{mol} \end{aligned}$ Thus, the mass of the carbon dioxide results to $m_{ C O_2} = n_{ C O_2} M_{ C O_2} = 8 n_{ C_8 H_{18}} M_{ C O_2} = 8 m_{ C_8 H_{18}} \frac{M_{CO_2}}{M_{ C_8 H_{18}}} = 8 V \rho \frac{M_{CO_2}}{M_{ C_8 H_{18}}} = 8 \cdot (6 \cdot 0.76) \cdot \frac{44}{114} \,\text{kg} \approx 14 \,\text{kg}$ where $V = 6 \,\text{L}$ denotes the volume of the gasoline and $\rho = 0.76 \,\text{kg}/\text{L}$ the mass density.

Relevant wiki: Atomic and Molecular Weights

Carbon weighs 12 units and hydrogen weighs 1 unit. Since there are 8 carbon atoms they contribute 8x12 units to each gasoline molecule and the hydrogen contribute 18 units. That means carbon contributes most of the weight to the 4.5 Kg of gas. In fact, it contributes 96/114 of the 4.5 kg.

After burning, the carbon is paired with two oxygen atoms. Each oxygen atom is 14 times heavier than each displaced hydrogen atom so the end product must must weigh more than the original 4.5 kg.

Write the balanced chemical equation, calculate the molar masses of gasoline and carbon dioxide, multiply each by its respective stoichiometric coefficient, and see that for every 1 kg. of gasoline combusted, 3.0838... kg. of carbon dioxide are produced.

Full balanced equation:

$2 \, C_8 H_{18} + 25 \, O_2 \to 16\, CO_2 + 18 \, H_2O$

using the fact that the molar mass is: $C = 12u$ , $O = 16u$ and $H = 1u$ we can write the molar mass equation as:

$\begin{aligned} 2 \, C_8 H_{18} &+ 25 \, O_2 &\to 16\, CO_2 &+ 18 \, H_2O\\ 228u &+ 800 u &\to 704 u &+ 324 u \end{aligned}$

from the problem we know that $228u = 4.50kg$ so we just need to do a simple cross rule to see that $704 u \approx 13.89kg$ for the $CO_2$ .

$\begin{aligned} 2 \, C_8 H_{18} &+ 25 \, O_2 &\to 16\, CO_2 &+ 18 \, H_2O\\ 4.50kg &+ 15.79kg &\to 13.89kg &+ 6.39kg \end{aligned}$

Its amazing how gasoline can generate 3 times its weight as $CO_2$ ! Its not only more than 4.5kg... its way more!

I agree with Markus's solution for the actual weight of CO2. I'm wondering why the illustration of a car appears to be a Nissan Leaf which is an electric car and produces no CO2 directly.

The first step is to write and balance the equation:

2C8H18 + 25O2 = 16CO2 + 18H2O

Then, we can make a table and put in the values we know. The rest of the values can be found out using the formula mass = moles*molar mass until we get the mass of Carbon Dioxide produced

The last row is because of the 1:8 mole ratio, because there are 2 atoms of octane and 16 atoms of carbon dioxide.

Since approximately 13.89 kg of CO2 is produced, the answer is More than 4.5 kg

Complete combustion of C8H18:

$C_{8}H_{18} + \frac{25}{2} O2 \longrightarrow 8CO_2 + 9H_2O$

$gmol^{-1}$ of $C_8H_{18}$ = $114$ and $gmol^{-1}$ of $CO_2$ = $44$

Moles of $C_8H_{18}$ = $\frac{4500}{114}$ $mol$

$C_8H_{18} : CO_2$

$1 : 8$

$8\cdot\frac{4500}{114}\cdot44 =13894.7g = 13.9kg$ which is greater than $4.5kg$ .

Writing the balanced equation for this reaction, $25O_2+2C_8H_18 \rightarrow 16CO_2 + 18H_2O$ . This leads to a comparison between $2C_8H_18$ and $16CO_2$ . Since the answers are approximate, we can also approximate; there's the same amount of carbon in each, so that won't make a difference. We know that hydrogen is much lighter than oxygen, and there's about the same amount of each, so the carbon dioxide must be heavier.

Variables A= chemical B = chemical being solved for g = gram L = liter mol = mole All other letters are numbers that are to be plugged in $\frac{mg/LAr}{1}$ | $\frac{1molA}{xg/LA}$ | $\frac{ymolB}{zmolAr}$ | $\frac{ugB}{omolB}$ = ngB

I used basic Highschool chemistry to solve this problem 1.balance equation C8H18 + 25O2 $-->$ 9H2O+8CO2 2.Now covert kilograms to grams 4.5kgCO2 *10^3 = 4500gCO2 Then plug everything in

$\frac{4500gC8H18}{1}$ | $\frac{1molC8H18}{114gC8H18r}$ | $\frac{8molCO2}{1molC8H18}$ | $\frac{28gCO2}{1molCO2}$ = 9000gCO2 = 9kg so amount of CO2 emitted is greater than 4.2g Notes: I am doing what I was taught in highschool, which I know is not the best way to solve this equation still works well for this problem and other problems like this though it is not good in the real world .

In each molecule of the fuel all the 18 hydrogen atoms are replaced by 16 oxygen atoms. Each oxygen atom is 16 times heavier than each hydrogen atom. 1 x 18 < 16 x 16

The solution requires information not presented in the problem, i.e., the relative atomic weights of H & O. Therefore, strictly speaking, the problem is not solvable.

C8O16 and H18O9 seems to be the ratio in what comes out of the reaction. Just the C8O16 alone is bigger than C8H18, so more than 4.5kg

You can get into stoichiometry and molar masses, but I think the quicker back-of-the-envelope answer is to notice that there are 2.25 H atoms for every C atom in gasoline. Then each of those C atoms gets 2 O atoms in carbon dioxide. The mass of 2 O atoms is more than 2.25 H atoms, so more mass is emitted than is consumed.

This is the partially balanced equation for the complete combustion of octane (I skipped out the water bit because we don't need to look at it and I only balanced the octane with carbon dioxide).

$\ce{C8H18}$ + 8 $\ce{O2}$ ---> 8 $\ce{CO2}$ + ...

As you can see, 1 mole of octane produces 8 moles of carbon dioxide. So next we need to find out how many moles we are dealing with.

The relative molecular mass of octane is 114g (8 $\times$ 12 + 18 $\times$ 1) and carbon is 44g (1 $\times$ 12 + 2 $\times$ 16)

$\frac{4500}{114}$ = 39.47 moles of octane

39.47 $\times$ 8 = 315.79 moles of carbon dioxide

315.79 $\times$ 44= 13.894kg of carbon dioxide produced.

As 13.9 is greater than 4.5, the answer is more than 4.5kg

This is exactly a routine problem I give my freshman chemistry students when we get to stoichiometry. Most generally I expect them to solve it using the complete balanced equation (à la Mr Michelmann, below). One can more quickly sum up the problem by remarking that each C in octane is bonded to 2.25 H's on average, which are replaced by 2 O atoms - a substitution of 32 mass units for 2.25 or, including the C, 16.25 units now weighs 44 units; thus 4.5 kg of octane will generate 13.9 kg of CO2. Nice when discussing climate change and the like.

Molare Masse(C8H18)=8 12+1 18= 114g/mol Masse m (C8H18)= 4,5kg=4500g Stoffmenge n=m/M=4500g/114g/mol=39,5mol Aus einem mol C8H18 werden 8 mol CO2 n(CO2)=8 39,5=315,789mol M(CO2)=2 16+12=44g/mol m(CO2)=n M=44g/mol 315,789mol=13894g ~14kg

My SOLUTION is somewhat a guess work as in gasoline C8H18 there are 8 atoms of carbon. So when 1 molecule of gasoline dissociates it releases 8 atoms of carbon. Thus more amount of carbon is released per molecule and hence the amount of carbon dioxide is also more 😂😂