Determine the mass of the ester formed by the reaction of 90 g of formic acid and 70 g of methanol

Determine the mass of the ester formed by the reaction of 90 g of formic acid and 70 g of methanol, if the yield of the reaction product is 80% of the theoretically possible.

Let’s find the amount of formic acid substance НСООН according to the formula:

n = m: M.

M (HCOOH) = 46 g / mol.

n = 90 g: 46 g / mol = 1.96 mol.

Let’s find the amount of alcohol substance CH3OH:

n = m: M.

M (CH3OH) = 32 g / mol.

n = 70 g: 32 g / mol = 2.18 mol.

NSOOH is given in excess, we solve for methanol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

CH3OH + HCOOH → HCOOCH3 + H2O.

According to the reaction equation, 1 mol of CH3OH accounts for 1 mol of HCOOCH3 ether. The substances are in quantitative ratios of 1: 1.

The amount of substance will be the same.

n (HCOOH) = n (HCOO CH3) = 1.96 mol.

Let’s find the mass of the ether by the formula:

m = n × M,

M (HCOOCH3) = 60 g / mol.

m = 1.96 mol × 60 g / mol = 117.6 g.

117.6 g was calculated (theoretical yield).

According to the condition of the problem, the ether output was 80%.

Let’s find a practical way out of the product.

117.6 g – 100%

x g – 80%,

x = (117.6 g × 80%): 100 g = 94.08 g.

Answer: 94.08 g.