(a) What percentage of methane by weight does a gas of 0.65 specific gravity contain if it is composed only of methane and ethane? What percentage by volume?
(b) Explain why the percentage by volume is greater than the percentage by weight.
The Correct Answer and Explanation is :
To solve the problem, we’ll break it into parts. The specific gravity (SG) is a ratio of the density of a gas to the density of air. Let’s assume that the air’s molecular weight is approximately 29 g/mol.
Part (a): Percentage by weight and volume of methane
Given:
- Specific gravity of the gas = 0.65
- The gas is composed only of methane (CH₄) and ethane (C₂H₆).
We need to find the weight and volume percentages of methane in the mixture.
Step 1: Relate specific gravity to molecular weight
The specific gravity (SG) of a gas is the ratio of its molecular weight (M) to the molecular weight of air (Mₐir = 29 g/mol):
[
SG = \frac{M}{M_{\text{air}}}
]
Rearranging to solve for the molecular weight of the gas mixture:
[
M = SG \times M_{\text{air}} = 0.65 \times 29 = 18.85 \, \text{g/mol}
]
Step 2: Set up equations for weight fractions
Let’s assume the gas is composed of ( x ) moles of methane and ( (1 – x) ) moles of ethane.
- Molecular weight of methane (CH₄) = 16 g/mol
- Molecular weight of ethane (C₂H₆) = 30 g/mol
The total molecular weight of the mixture is:
[
M_{\text{mixture}} = x(16) + (1 – x)(30)
]
We know that ( M_{\text{mixture}} = 18.85 ), so:
[
x(16) + (1 – x)(30) = 18.85
]
Solving for ( x ):
[
16x + 30 – 30x = 18.85
]
[
-14x = -11.15
]
[
x = 0.796
]
This means 79.6% of the moles of the gas are methane.
Step 3: Calculate the percentage by weight
The weight fraction of methane is:
[
\text{Weight fraction of methane} = \frac{0.796 \times 16}{(0.796 \times 16) + (0.204 \times 30)} = 0.426
]
So, the percentage by weight of methane is:
[
\text{Percentage by weight} = 42.6\%
]
Step 4: Calculate the percentage by volume
Since gases behave ideally, the volume fraction is the same as the mole fraction. Therefore, the percentage by volume of methane is:
[
\text{Percentage by volume} = 79.6\%
]
Part (b): Explanation
The key reason the percentage by volume is greater than the percentage by weight lies in the difference between the molecular weights of methane and ethane. Methane is lighter (16 g/mol) than ethane (30 g/mol), so a mole of methane occupies more volume than a mole of ethane under the same conditions of temperature and pressure.
- In terms of volume, methane makes up a larger proportion of the mixture because gases are measured by the number of moles, not by weight.
- In terms of weight, methane contributes less than its volume percentage because it has a lower molecular weight. Thus, fewer grams of methane are needed to make up a given volume compared to ethane.
In summary, the percentage by volume is greater because volume is directly related to the number of molecules (or moles), and methane, being lighter, contributes more moles per unit volume than ethane.
I will now generate an image to visually explain this.
Here is the educational diagram explaining the relationship between weight and volume percentages of methane and ethane. It illustrates how methane, despite having a lower molecular weight, contributes more to the volume due to its smaller molecular size. Let me know if you’d like further clarification!
