5 calculations

Organic Chemistry Laboratory Calculations

Balancing Chemical Reactions

How to balance a chemical reaction that yields a mixture of isomers, two or more products with the same molecular formula but different structures?

Dehydration of 2-methylcyclohexanol in the presence of phosphoric acid as a catalyst yields a mixture of 1-methylcyclohexene and 3-methylcyclohexene.

The following reaction is correctly balanced because only one mole of the mixture of the two products forms per each mole of the starting material, 2-methylcyclohexanol. Concentration Units

The concentration of solutions can be expressed in molarity, mass per volume, and percentage.

The concentration expressed in molarity gives the number of moles of solute per liter of solution in which it is contained:

Another example is the concentration expressed as the mass of solute (g or mg) per volume of solution (1 mL, 100 mL, or 1 L). For instance, 50 g of NaCl is dissolved in water, with the final volume of the solution being 1000 mL.

[NaCl] = 50 g/1000 mL = 50 g/L = 0.050 g/mL = 50 mg/mL

One can also express the concentration as a weight or volume percentage. The weight percentage expresses the mass of solute as a percentage of the mass of the solution, while the volume percentage expresses the volume of solute as a percentage of the volume of the solution. The weight percentage is represented as w/w, weight in weight, while the volume percentage is represented as v/v, volume in volume.

If 100 g of NaCl is dissolved in water to make 1000 g of solution, then [NaCl] is:

[NaCl] = 100 g/1000 g x 100 = 10% (w/w)

If 25 mL of MeOH is mixed with water, with the final volume of the solution being 100 mL, then [MeOH] is:

[MeOH] = 25 mL/100 mL x 100 = 25% (v/v)

molarity = # of moles solute

volume of solution (L)

CH3

OH

CH3 CH3

H3PO4

heat

water

1-methylcyclohexene 3-methylcyclohexene


The concentration can be expressed as weight in volume, w/v,

percentage, the mass of solute per 100 mL of solution. If 12 g of NaCl is dissolved in water to make 200 mL solution, then [NaCl]

is:

[NaCl] = 12 g/200 mL x 100 = 6% (w/v)

The final concentration notation is commonly useful for mixtures of liquids. It describes a solution of liquids by the volume of each individual component in the mixture. A mixture of 30 mL water, 20 mL MeOH, and 10 mL acetone can be expressed as 30:20:10 or 3:2:1. Moles and Millimoles

For your synthesis reactions, calculating the correct molar amounts is crucial. To calculate the number of moles of a given compound:

If the molecular mass of aspirin is 180.15 g/mol, then the mol equivalence of 0.25 g of aspirin is

# of moles = 0.25 g/180.15 g/mol = 0.00139 mol

In cases where milligram amounts are used, a smaller unit, the millimole (mmol), is preferred.

1 mmol = 10-3 mol; 1000 mmol = 1 mol

To convert moles into mmoles the number is multiplied by 1000.

0.00139 mol x 1000 mmol/mol = 1.39 mmol

To calculate the number of moles in a volume of pure liquid, the volume is first converted into mass using the density (d):

To calculate the number of moles or mmoles for 2.5 mL of MeOH, d = 0.7915 g/mL and molecular mass = 32.04 g/mol, first calculate the amount in grams:

0.7915 g/mL x 2.5 mL = 1.98 g # of moles = 1.98 g/32.04 g/mol = 0.0618 mol = 61.8 mmol

# of moles = mass (g)

molecular mass (g/mol)

mass (g) = d (g/mL) x volume (mL)


To calculate the number of moles of a chemical solution, the

concentration, g/mL, and volume, mL, are needed. If the concentration is given, the mass of solute can be obtained:

How many grams of NaCl is in 1.3 mL solution of NaCl in water with a concentration of 0.5 g/mL?

mass = 0.5 g/mL x 1.3 mL = 0.65 g Mineral Acids

The concentration of HCl is 37% w/w, which means there are 37 g of HCl per 100 g of solution.

To calculate the number of moles of acid, figure out the mass of the solution:

mass solution (g) = d (g/mL) x volume (mL) mass solution = 1.2 g/mL x 1000 mL = 1200 g

Calculate how much of this mass is pure HCl:

mass HCl (g) = mass solution x %HCl (w/w) mass HCl = 1200 g x 37 g HCl/100 g sol. = 444 g HCl

Calculate the number of moles:

# moles = mass (g)/ MW (g/mol) # moles = 444 g/36.46 g/mol = 12.2 mol

Yield Calculation

In the isolation of a final product from a mixture, the percentage yield (% yield) is calculated:

To calculate the % yield of a reaction, first calculate the number of moles of reactants and products. Use the balanced equation to determine the maximum amount of product the reaction can yield, the theoretical yield.

In the chemical reaction mixture there is often a deficit of one of the reactants and an excess of others. The starting materials in excess will be present at the end of the reaction, while the chemical present in deficit would be consumed. This compound is referred to as the limiting reagent and is used to determine the theoretical yield of the reaction.

mass (g) = concentration (g/mL) x volume (mL)

% yield = mass of product x 100

mass of mixture


In the above shown reaction, acetic acid reacts with MeOH in the presence of sulfuric acid, a catalyst, to yield an ester, methyl acetate, and water. Given that you have mixed 1.0 mL of acetic acid with 5 mL MeOH in the presence of 100 µL of the catalyst, you obtain 770 mg of the final product. What is the % yield of the reaction?

Start by calculating the number of mmoles from mass (g) by using the following equations:

mass (g) = concentration (g/mL) x volume (mL)

# of moles = mass (g)


Acetic acid MeOH H2SO4 Methyl acetate volume 1.0 mL 5.0 mL 100 µL* density (g/mL)

1.049 0.7915 1.84 for 98% w/w

0.9342


60.05 32.04 98.08 74.08

mass (g) 1.05 3.96 0.180 0.770 mmol 17.5 123.6 1.83 10.4

*The conversion between µL and mL - 1 µL = 10-6 L = 10-3 mL, 100 µL = 0.100 mL.

Based on the reaction equation, the reaction between the acetic acid and

MeOH is 1 to 1, meaning, 1 mol of MeOH is required for each mole of acetic acid. Since the number of mmoles for MeOH is larger than the number of mmoles for acetic acid, acetic acid is the limiting reagent. Sulfuric acid is not the limiting reagent because it participates in the reaction as a catalyst. It is consumed and regenerated during the reaction to speed-up the reaction rate.

If acetic acid is fully transformed into the ester, the maximum number of mmol of methyl acetate that can be produced is equal to the mmol amount of acetic acid; therefore the theoretical yield for the reaction is equal to 17.5 mmol.

The estertification reaction is in equilibrium, meaning, at the end of the reaction there is always some reactant left; however, for the theoretical yield calculation it is assumed that the reaction is irreversible and goes to completion.

O

OHCH3OH

H2SO4

heat

O

OH2O


The actual yield of the reaction is the amount obtained experimentally. In

our case that amount is 10.4 mmol. The percent yield can be calculated by the following equation:

% yield = actual yield x 100

theoretical yield

% yield = 10.4 mmol x 100 = 59.4 %

17.5 mmol

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