Question

black forest biologicals, a biotech startup, has a promising alzheimers drug candidate compound slt - 88 entering phase i trials this year. slt - 88 is the only product formed by the reaction of two precursor compounds a and b, both of which are quite expensive. the chief medicinal chemist of black forest is trying out different reaction conditions to minimize the cost of manufacturing slt - 88.
in the table below are listed the initial and final amounts of a and b used under two different trial conditions, and also the actual amount of slt - 88 recovered in each case. complete the table by calculating the theoretical yield of slt - 88 and the percent yield of slt - 88. round your amounts to the nearest milligram and your percentages to the nearest whole percent.

Explanation:

Trial 1:

Step1: Calculate the amount of A and B consumed

• Amount of A consumed: $750 - 0=750$ mg
• Amount of B consumed: $100 - 67 = 33$ mg

Step2: Determine the limiting reactant

Assume the reaction ratio of A and B is 1:1 (since no reaction equation is given, but based on stoichiometry concept). B is the limiting reactant (as 33 mg of B is consumed compared to 750 mg of A available, and if 1:1 ratio, B runs out first)

Step3: Calculate theoretical yield (assuming 1:1:1 ratio of A:B:SLT - 88 in reaction)

Theoretical yield (based on B) is 33 mg (if 1 mole B gives 1 mole SLT - 88 and molar masses are assumed same for simplicity as no molar mass data. If we assume they react in 1:1 mass ratio (since no other data), theoretical yield is 33 mg. But wait, maybe it's \(A + B
ightarrow SLT - 88\). If we assume that the mass of SLT - 88 is the sum of masses of reactants (in a simple addition reaction, no mass loss). But no, in a chemical reaction, mass is conserved. Wait, no, if it's \(A + B
ightarrow SLT - 88\), and assuming complete reaction of limiting reactant.
Wait, another approach: assume that the reaction is \(A + B
ightarrow SLT - 88\). The amount of B used is \(100 - 67=33\)mg. If 1 unit (by mass) of B reacts with 1 unit of A (but A is in excess) to form 1 unit of SLT - 88. So theoretical yield is 33mg. But that's not matching with the actual yield. Wait, no, maybe the reaction is \(nA + mB
ightarrow SLT - 88\). But since no reaction coefficients, assume \(A + B
ightarrow SLT - 88\). The mass of SLT - 88 should be mass of A reacted + mass of B reacted. Mass of A reacted: 750mg (but B is limiting. Wait no, if B is 33mg, and A is in excess. Wait, no, if \(A + B
ightarrow SLT - 88\), then moles of B: \(n_B=\frac{33}{M_B}\), moles of A: \(n_A=\frac{750}{M_A}\). If \(M_A = M_B\) (assume same molar mass for simplicity as no data). Then B is limiting. Moles of SLT - 88 = moles of B. Mass of SLT - 88 = mass of B (if \(M_{SLT - 88}=M_B\)). But actual yield is 595mg. This is wrong. Wait, no, maybe the problem has a typo. Wait, another approach: assume that the reaction is \(A + B
ightarrow SLT - 88\) and the mass of SLT - 88 is the sum of masses of A and B that react.
For trial 1: mass of A reacted = 750mg (since final A is 0), mass of B reacted=100 - 67 = 33mg. Theoretical yield \(=750+33=783\)mg
Percent yield \(=\frac{595}{783}\times 100\%\approx 76\%\)

Trial 2:

Step1: Calculate the amount of A and B consumed

• Amount of A consumed: \(150 - 0=150\)mg
• Amount of B consumed: \(550 - 374 = 176\)mg

Step2: Determine the limiting reactant

Assume \(A + B
ightarrow SLT - 88\). A is limiting (150mg of A vs 176mg of B. If 1:1 reaction, A runs out first)

Step3: Calculate theoretical yield

Theoretical yield \(=150 + 176=326\)mg (sum of masses of A and B that react, assuming \(A + B
ightarrow SLT - 88\) and no mass loss)
Percent yield \(=\frac{134}{326}\times 100\%\approx 41\%\)

Answer:

Trial	amount of A		amount of B		yield of SLT - 88
1	750 mg	0 mg	100. mg	67 mg	783 mg	595. mg	76%
2	150. mg	0 mg	550. mg	374. mg	326 mg	134. mg	41%