Question

I need to figure out my percent yield of luminol. My actual yield was 0.097 g, but I can't figure out the theoretical yield to find the percent yield. Also, the molar ratio for each.

	3-Nitrophthalic Acid	Hydrazine	Sodium dithionate	Luminol
MW	211.1284 g/mol	32.0452 g/mol	174.11 g/mol	177.16 g/mol
Amount g or ml	0.202 g	0.8 ml	0.6 g	0.097
Moles	9.57 x10-4	0.025 mol	0.0034 mol	5.48 x 10-4

Answer 1

first lets calculate the reritical yield

from your reaction it is clear that one mole of 3-Nitrophthalic Acid will give one mole of Luminol

accordingly you have 9.57 x10-4 moles of 3-Nitrophthalic Acid will give 9.57 x10-4 moles of Luminol

now you know the Luminol moles you can actually calculate the weight of Luminol theritically

= moles x MW

= 9.57 x10-4 x 177.16

= 0.1659 grams this is theritical yield

now actual yield

= product mmoles / rectant millimoles x 100 these all are experimentally

= 5.48 x 10-4 / 9.57 x10-4 x 100

= 57.26%

here reactent is 3-Nitrophthalic Acid remaining all are reactant