Asked by Anthony
A reaction requires 0.5 g of pure methyl oleate (density = 0.87g/mL) (MW = 310g/mol). How many mL of 70% pure methyl oleate are needed for the reaction.
I don't exactly know how to start this off. I think what I'm supposed to do is multiply the 70% by the molecular weight and then go on from there but I tried it and it doesn't give me the right answer. Any tips?
I don't exactly know how to start this off. I think what I'm supposed to do is multiply the 70% by the molecular weight and then go on from there but I tried it and it doesn't give me the right answer. Any tips?
Answers
Answered by
DrBob222
0.5/0.70 = approx 0.714 g of the 70% stuff you need.
Then convert grams to mL using density. Remember the %yield problems you worked early in chemistry?
70% stuff x what number = 0.5 g real stuff
what number = 0.5/0.70 = about 0.7 g. Doesn't that make sense. You must take 0.714 g of 70% stuff to give you 0.5 of the real stuff. Sure it does.
Then convert grams to mL using density. Remember the %yield problems you worked early in chemistry?
70% stuff x what number = 0.5 g real stuff
what number = 0.5/0.70 = about 0.7 g. Doesn't that make sense. You must take 0.714 g of 70% stuff to give you 0.5 of the real stuff. Sure it does.
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