Question

Can someone please explain how to calculate this? I truely do not understand how to do theoretical yield...

calculate theoretical yield and percent yield for anthracene-9-methylmalemide and N-methylmalemide in water amounts used were 0.070 g of anthracene-9-methanol and 50 mL of water and 0.103 g of N-methylmaleimide. the weight of the product was 0.043 g

Answer 1

First you need to know the molecular weights of anthracene and n-methylmaleimide. (Which I will label as A and B)

MMa = 208.25 g/mol; MMb = 111.10 g/mol

Assuming this reaction has a 1:1 mole ratio then, calculate the number of mole of A and B:

moles A = 0.070 / 208.25 = 3.36x10^-4 moles

moles B = 0.103 / 111.10 = 9.27x10^-4 moles

These means that the limitant reactant is A and B is in excess so, the moles of A will react completely with a part of B and will produce moles of A, but with the molecular weight of the product (which you are not providing)

So, according to this reaction, and the reactants, this is likely to be a Diels - Alder reaction, so let's sum the molecular weight of both of the reactants to get the molecular weight of C (However you can provide the actual MM so I can fix this, but the procedure will be the same)

MMc = 208.24 + 111.10 = 319.34 g/mol

now the mass of C (product) = 3.36x10^.-4 * 319.34 = 0.1073 g. This would be the theorical yield.

The percent of yield : % = (0.043 / 0.1073) * 100 = 40.08%

Hope this helps