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

[tweety1234]

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?

 

[sjb-2812]

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...

 

[tweety1234]

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.

 

[Borek]

% 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.

 

[tweety1234]

% 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.

--
chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods

okay, mass = mr x moles so for $$CH_{4} = 0.40 \times 16 = 6.4$$

Is this what you mean?

 

[Borek]

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.

 

[tweety1234]

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.