Methyl salicylate (oil of wintergreen) is prepared by heating salicylic acid (C7H6O3) with methanol (CH3OH)
C7H6O3 + CH3OH ® C8H8O3 + H2O
What is the percent yield if reaction of 8.021 g of salicylic acid with an excess of methanol produced a measured yield of 6.463 g of oil of wintergreen?
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How many grams of Ni(NO3)2 would you use to prepare 262.3 mL of 0.1614 M Ni(NO3)2?
Please show me how! Thank you!!! I will give you best answer!!
C7H6O3 + CH3OH ® C8H8O3 + H2O
What is the percent yield if reaction of 8.021 g of salicylic acid with an excess of methanol produced a measured yield of 6.463 g of oil of wintergreen?
--------------------------------------…
How many grams of Ni(NO3)2 would you use to prepare 262.3 mL of 0.1614 M Ni(NO3)2?
Please show me how! Thank you!!! I will give you best answer!!
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1) C7H6O3 + CH3OH ---> C8H8O3 + H2O
Methanol is in excess.
Theoretical yield:
Number of moles of salicylic acid reacted - n = m / MM = 8.021 g / 138.121 g/mol = 0.058 mol
C7H6O3 + CH3OH ---> C8H8O3 + H2O So from this equation we can see that salicylic acid and methyl salicylate are in a 1:1 molar ratio, that is to say that 1 mole of salicylic acid produces 1 mole of methyl salicylate ( as long as methanol is in excess, which it is).
So we can also say that 0.058 mol of salicylic acid produces 0.058 moles of methyl salicylate.
What is the mass of 0.058 mol methyl salicylate?
n = m / MM therefore m = n x MM = 0.058 mol x 152.1494 g/mol = 8.82 g
So theoretically we should produce 8.82 g , this is the maximum amount of product formed.
However we produce 6.463 g So our percentage yield = (theoretical / actual)*100 = (6.463 g / 8.82 g)*100 = 73.3%
2) Right how many moles of Ni(NO3)2 are present in 262.3 mL of a 0.1614M solution?
n = c x v/1000 = 0.1614 M x (262.3 mL / 1000) = 0.042 moles
So we need 0.042 moles of Ni(NO3)2 which as a mass is:
m = n x MM = 0.042 mol x 182.703 g/mol = 7.7g
Methanol is in excess.
Theoretical yield:
Number of moles of salicylic acid reacted - n = m / MM = 8.021 g / 138.121 g/mol = 0.058 mol
C7H6O3 + CH3OH ---> C8H8O3 + H2O So from this equation we can see that salicylic acid and methyl salicylate are in a 1:1 molar ratio, that is to say that 1 mole of salicylic acid produces 1 mole of methyl salicylate ( as long as methanol is in excess, which it is).
So we can also say that 0.058 mol of salicylic acid produces 0.058 moles of methyl salicylate.
What is the mass of 0.058 mol methyl salicylate?
n = m / MM therefore m = n x MM = 0.058 mol x 152.1494 g/mol = 8.82 g
So theoretically we should produce 8.82 g , this is the maximum amount of product formed.
However we produce 6.463 g So our percentage yield = (theoretical / actual)*100 = (6.463 g / 8.82 g)*100 = 73.3%
2) Right how many moles of Ni(NO3)2 are present in 262.3 mL of a 0.1614M solution?
n = c x v/1000 = 0.1614 M x (262.3 mL / 1000) = 0.042 moles
So we need 0.042 moles of Ni(NO3)2 which as a mass is:
m = n x MM = 0.042 mol x 182.703 g/mol = 7.7g