Combustion of a $20g$ mass of compound gives $44g$ $CO_2$, and $7.99g$ of water. Given that the molecular mass is $180g*mol^-1$ what are the empirical and molecular formulae?

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Combustion of a $20g$ mass of compound gives $44g$ $CO_2$, and $7.99g$ of water. Given that the molecular mass is $180g*mol^-1$ what are the empirical and molecular formulae?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

We can burn organic in a furnace to produce (i) carbon dioxide; and (ii) water. The carbon and hydrogen content of these products COME directly from the hydrocarbon. Sometimes these are expressed as a percentage by mass and SOMETIMES, as here, the measured percentages do not add up to 100%. The missing percentage is presumed to be oxygen, which is NOT measured, but obtained by difference, i.e. $O%=100%-%C-%H$.

Here, the carbon content is measured by $CO_2$, and the hydrogen content is measured by WATER (I think you made a mistake when you posed the question!).

So let's see if we can get something useful.

In $20*g$ compound, there are $(44*g)/(44*g*mol^-1)C=1*mol*C$.

In $20*g$ compound, there are $(7.99*g)/(18.01*g*mol^-1)xx2H=0.89*mol*H$.

In $20*g$ compound, there are presumed to be the following molar quantity of oxygen,

$(20.0*g-12.011*g-0.89*g)/(16*g*mol^-1)O=0.44*mol*O$. (That is the mass balance was due to oxygen!).

We divide thru by the SMALLEST molar quantity, that of oxygen to get a trial formula of $C_(2.27)H_2O$, which we must multiply by $4 $ to get whole numbers, i.e. $C_9H_8O_4$, which is the $"empirical formula"$.

But we know that the $"molecular formula"$ is a simple whole number multiple of the $"empirical formula"$

And thus............................,

$nxx(9xx12.011+8xx1.00794+4xx15.999)*g*mol^-1$

$=180*g*mol^-1$

Clearly, $n=1$, so here $"molecular formula"$ $=$ $"empirical formula"$ $=$ $C_9H_8O_4$

I am sorry for making such a meal of this answer, but if you are an undergrad, you should be aware of the background to this question. If you are at A level, your teacher has no business asking you these sorts of questions.

Combustion of a $20*g$ mass of compound gives $44*g$ $CO_2$, and $7.99*g$ of water. Given that the molecular mass is $180*g*mol^-1$ what are the empirical and molecular formulae?

1 Answers

Combustion of a $20g$ mass of compound gives $44g$ $CO_2$, and $7.99g$ of water. Given that the molecular mass is $180g*mol^-1$ what are the empirical and molecular formulae?