What is the composition of a fuel on a dry and mass basis?

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Discussion Overview

The discussion revolves around determining the composition of a fuel on a dry and mass basis, given its molar composition. Participants explore methods for converting molar percentages to mass percentages and calculating average molecular weight.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents a molar composition of a fuel and seeks to find its composition on a dry and mass basis, along with the average molecular weight.
  • Another participant suggests that the composition of each component must be taken into account for mass calculations.
  • A participant questions the calculation for methane's mass percentage, indicating a specific value of 24.3% w/w and expresses uncertainty about how to arrive at this figure.
  • Several participants discuss the method of converting moles to masses, with one suggesting simplifying calculations by assuming 1% is equivalent to 1 mole.
  • Another participant confirms that assuming 100% is one mole is also a valid approach, as long as consistency is maintained in calculations.
  • A later reply indicates that the participant has arrived at the correct answers after following the discussed methods.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the specific method for calculating mass percentages, as multiple approaches are suggested and discussed. There is also uncertainty regarding the correct mass percentage for methane.

Contextual Notes

Participants rely on relative atomic masses for calculations, but there is no explicit agreement on the assumptions or methods used for the conversions, leading to potential variations in results.

tweety1234
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Homework Statement



A fuel has the following composition on a molar basis: 40.0% CH4,
24.0% C2H4, 16% C3H8, 10% CO2, 10% H2O. What is the composition of the fuel on a dry
basis? What is the composition of the fuel on a mass basis? What is the average molecular
weight of the fuel? [Relative atomic masses: O = 16, C = 12, H = 1]

I worked out the compistion on a dry basis

CH_{4} = \frac{40}{90} \times 100 = 44.44\% and than the same for each one, excluding water.

How do I work out the composition of the fuel on a mass basis?
 
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Presumably you will also have to take into account the composition of each component, so for instance methane is RAM(C)/[RAM(C)+4.RAM(H)] x 100% carbon...
 
sjb-2812 said:
Presumably you will also have to take into account the composition of each component, so for instance methane is RAM(C)/[RAM(C)+4.RAM(H)] x 100% carbon...

do you mean, = 12/(12+4) = 0.75 75%?

the correct answer for methane is 24.3% w/w , I don't know how to get this.
 
% w/w. that means you have to convert all moles to masses.

To make calculations simpler you can assume 1% is just 1 mole, then calculate total mass of each, total mass of mixture and so on.
 
Last edited by a moderator:
It can be done this way as well, just instead of assuming 1% is 1 mole you assumed 100% is one mole. This is still correct, as long as you will be consistent.
 
Last edited by a moderator:
Borek said:
It can be done this way as well, just instead of assuming 1% is 1 mole you assumed 100% is one mole. This is still correct, as long as you will be consistent.
okay, thanks got the correct answers now.
 
Last edited by a moderator:

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