please show how to calculate the percent yield, theoretical yield, and actual yi

Question

please show how to calculate the percent yield, theoretical yield, and actual yield of the nitration of methyl benzoate using nitric acid. I already have all the values but I am stuck because both starting materials are in volume and not grams

NITRATION REPORT TEMPLATE Read Only Compatibility Mode - Word able Tools Design Layout References Mailings Review View Help Design Layout Tell me what you want to do Calculations OCH3 MW 136.15 g/mol MW 63.01g/mol MW 181.15 gm Sta Reactant A methyl benzoate 5.4 mL Reactant B nitric acid 4.0 mL End Material methyl m-nitrobenzoate 2.17g Product Analyses and Yield Theoretical yield Actual ield Melting point Percent yield ID of major product: 64 C

Explanation / Answer

Density of methyl benzoate = 1.08 g/ ml

Given that 5.4 ml

Density = mass/ volume

Mass= density * volume

= 1.08*5.4

= 5.832 g

Number of mole of methyl benzoate = amount in g / molar mass

= 5.832 g/ 136.15 g/ mole

= 0.043 mole

According to reaction mole of m nitro benzoate = 0.043 mole

Theoretical amount = number of moles * molar mass

=0.043 * 181.15

= 7.79 g

Percentage yield = observed amount / Theoretical amount]*100

= 2.17/ 7.79*100

= 27.9 %

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