Aldehydes A and B are consecutive aldehydes in the homological row. They have masses 0,39 g and 0.32 g , respectively, and they are mixed together. Tollens' reagent is added to the system and it has been calculated that for complete oxidization of both aldehydes 0,0202 moles of Ag+ ions have been used. Calculate the molar mass of A and B. The problem is that the molar masses I get by calculation are way too random and they dont fit for any aldehyde out there. Starting from the equation of the reaction between an aldehyde and Tollens' reagent:
RCHO+ 2[Ag(NH3)2]OH = RCOOH + 2Ag + 4NH3 + H2O
For every molecule of aldehyde, 2 ions of Ag+ are used, thus n(A)/n(Ag+)=1/2 and n(B)/n(Ag+)=1/2
Now if the amount of Ag+ ions is the sum of the ions that reacted with A and the ions that reacted with B
n(Ag+)total=n(Ag+)forA+n(Ag+)forB=2n(A)+2n(B)=0.0202 mol
m(A)/M(A)+m(B)/M(B)=0.0101 mol
m(A)/M(A)+m(B)/[M(A)+M(methilene group)]=0.0101 From here I get to a quadratic equation where x is M(A) and I get 64.65 g/mol which doesnt fit at all