Salicyclic acid reacts with methanol to produce methylsalicyclate or winter green and water according to the following reaction:
C7H6O3 (s) + CH3OH (l) --> C8H8O3 (s) + H2O (l)
What mass methylsalicyclate can be produced from 15.00 grams of salcyclic acid? What mass of methanol is required? If the methanol has a density of 0.79 g cm-3 What volume of methanol is required?
To find the mass of methylsalicyclate produced from 15.00 grams of salicylic acid, we need to use stoichiometry.
Step 1:
Write down the balanced chemical equation:
C7H6O3 (s) + CH3OH (l) → C8H8O3 (s) + H2O (l)
Step 2:
Calculate the molar mass of salicylic acid (C7H6O3):
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of salicylic acid = (7 * C) + (6 * H) + (3 * O) = (7 * 12.01) + (6 * 1.01) + (3 * 16.00) = 138.12 g/mol
Step 3:
Convert the given mass of salicylic acid to moles:
Moles of salicylic acid = Mass / Molar mass = 15.00 g / 138.12 g/mol = 0.1084 mol (rounded to four decimal places)
Step 4:
Using the balanced equation, the stoichiometry tells us that 1 mole of salicylic acid reacts with 1 mole of methanol to produce 1 mole of methylsalicyclate.
So, the moles of methylsalicyclate produced is also 0.1084 mol.
Step 5:
Calculate the molar mass of methylsalicyclate (C8H8O3):
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of methylsalicyclate = (8 * C) + (8 * H) + (3 * O) = (8 * 12.01) + (8 * 1.01) + (3 * 16.00) = 152.15 g/mol
Step 6:
Calculate the mass of methylsalicyclate produced:
Mass of methylsalicyclate = Moles of methylsalicyclate × Molar mass of methylsalicyclate = 0.1084 mol × 152.15 g/mol = 16.49 g (rounded to two decimal places)
Therefore, the mass of methylsalicyclate produced from 15.00 grams of salicylic acid is 16.49 grams.
To find the mass of methanol required, we look at the stoichiometry of the balanced equation.
According to the equation:
1 mole of salicylic acid reacts with 1 mole of methanol.
Since we have already determined that 0.1084 moles of salicylic acid reacted, we know that we need an equal amount of moles of methanol. Thus, the moles of methanol required is also 0.1084 moles.
However, we need to convert moles to grams using the molar mass of methanol.
Step 1:
Calculate the molar mass of methanol (CH3OH):
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of methanol = (1 * C) + (4 * H) + (1 * O) = (12.01) + (4 * 1.01) + (16.00) = 32.04 g/mol
Step 2:
Calculate the mass of methanol required:
Mass of methanol = Moles of methanol × Molar mass of methanol = 0.1084 mol × 32.04 g/mol = 3.48 g (rounded to two decimal places)
Therefore, the mass of methanol required is 3.48 grams.
To find the volume of methanol required, we'll use the density of methanol to convert the mass to volume.
Given: Density of methanol = 0.79 g cm^-3
Step 1:
Calculate the volume of methanol using the mass and density:
Volume = Mass / Density = 3.48 g / 0.79 g cm^-3 = 4.41 cm^3 (rounded to two decimal places)
Therefore, the volume of methanol required is 4.41 cm^3.
Here is a link to an example stoichiometry problem I've posted. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html