[Borek]

100 g of 23% w/w solution of an aldehyde dissolved 19 g of the next aldehyde of the homologous series. 2 g of the solution reduces 4.35 g of silver from the solution of the ammoniacal silver nitrate.

Give molecular formulas of both aldehydes.

[Schrödinger]

sbeznyqrulqr rgunany?

[Borek]

sbeznyqrulqr rgunany?

Huh?

[Schrödinger]

rot13. To avoid publicly displaying the answer.

This happens to be my 1000th post! \m/

[123456789]

100 g of 23% w/w solution of an aldehyde dissolved 19 g of the next aldehyde of the homologous series. 2 g of the solution reduces 4.35 g of silver from the solution of the ammoniacal silver nitrate.

Give molecular formulas of both aldehydes.

by 2g of the solution, do you mean the original 23% solution, or the solution with the  further dissolved alderhyde?

[Borek]

Solution with both aldehydes.

[123456789]

(CHO)2 and CH3COCHO?

[Borek]

(CHO)2 and CH3COCHO?

I don't see how these are homologous aldehydes.

sbeznyqrulqr rgunany?

Decoded to "formaldehyde ethanal" - and that's the correct answer.

[Rutherford]

I hope that it's not a problem that I write my calculations which are wrong, so someone can correct them.

Molar mass of the first aldehyde is x.
Molar mass of the second aldehyde is x+14 (because of -CH₂-).
It is obvious that the first aldehyde is in excess, so his number of moles after the aldol reaction will be: n=23/x-19/(x+14).
The number of moles of the aldol is n_a=19/(x+14).
With silver, both the aldol and the remaining quantity of the first aldehyde react, when summed that is: 23/x.
In 2g of the solution, 4.35/108 mol of Ag reacted, in 119g of the solution 517.65/216 mol of Ag react. As two moles of Ag react per 1 mole of the aldehyde and aldol, 23/x=517.65/432 and x=19g/mol. What am I doing wrong?

[Schrödinger]

There isn't any aldol reaction, really. Just oxidation by Tollen's reagent.

[Rutherford]

Oh, silly me ! The term "dissolved" confused me and I thought that it is a non-neutral solution. Without the aldol reaction, it is easier and I got that x=29.96g/mol which is HCHO. Thanks for the indication.

[Schrödinger]

Even if the conditions were somehow hypothetically aldol-conducive, there is still one flaw in your numericals : Aldol condensation needs an alpha-H right? The question doesn't state anywhere that the aldehydes contain alpha hydrogen. As a matter of fact, one of the answers methanal doesn't have one.

[Rutherford]

If it was aldol, then it should be obvious that at least one of the aldehydes must have an α-H.
If HCHO is the first aldehyde (doesn't have an α-H), the second aldehyde must have an α-H.

[Schrödinger]

What if the first aldehyde is 2,2-dimethylpropanal? And the next one contains just one CH₂ extra. It could be anywhere. Not necessarily alpha; like 2,2-dimethylbutanal.

[Rutherford]

But only the molecular formula is needed: "Give molecular formulas of both aldehydes." so if I e.g. determine a molecular formula C₄H₉CHO, it won't mean the aldehyde you proposed, but any other that has α-H.