fill in reaction table below. Make sure you correctly calculate the molar amount

Question

fill in reaction table below. Make sure you correctly calculate the molar amounts on your reactive materials. Just need Q2-4 answered please

CH3 catalytic HO OH + 2x CH3OH HO3S Figure 1. Reaction Scheme Prelab Questions 1) The Material Safety Data Sheets (MSDS) for all the chemicals involved in this lab are on iLearn. Read these and answer the following questions a) Which chemical is the most dangerous in this lab? b) Explain why you chose your answer for part a), and the safety precautions you will take when handling this material 2) Fill in the reaction table below. Make sure you correctly calculate the molar amounts of your reactive materials formula mol.-eq mmol amount 0.50 mL 0.75 mL name Benzaldehyde Dimethyl 1.00 Acetal 2-Methylpropane-1,3-diol Camphorsulfonic acid Dichloromethane product 3) Based on your answers to Q2, which is the limiting reagent in this reaction? 4) Calculate the Theoretical Yield of your product, i.e. the mass you would expect to recover, assuming 100% conversion to product.

Explanation / Answer

Name

Formula

Mol. Eq.

Mw (g/mol)

Mmol

Amount

Benzaldehyde Dimethyl Acetal

C9H12O2

1.00

152.19

3.30 mmoles

0.50 mL

2-Methylpropane-1,3-diol

C4H10O2

2.56

90.121

8.447 mmoles

0.75 mL

Camphorsulfonic acid

--

--

--

--

5 mg

Dichlorometane

--

--

--

--

3 mL

Product

C11H14O2

1.0

178.228

3.30 mmoles

Product = 5-Methyl-2-phenyl-1,3-dioxane.

To know millimoles of these compounds we need to know the density, since with density we can calculate the mass, with mass and molar mass we can calculate moles and then convert into millimoles.

Density of Benzaldehyde Dimethyl Acetal = 1.014 g/mL

Density of 2-Methylpropane-1,3-diol = 1.015 g/mL

Calculating millimoles of Benzaldehyde Dimethyl Acetal:

Density= mass / volume

We know density (1.014 g/mL) and we know volume (0.50 mL)

We rearrange and solve for mass:

Mass = density x volume

Mass = 1.014 g/mL x 0.50 mL

Mass = 0.507 g

Calculating moles:

Moles = mass / molar mass

Moles = 0.507 g / 152.19 g/mol

Moles = 0.0033 moles of Benzaldehyde Dimethyl Acetal

Millimoles of Benzaldehyde Dimethyl Acetal:

1 mol = 1000 mmoles

0.0033 moles (1000 mmoles / 1 mol) = 3.30 mmoles

Calculating millimoles of 2-Methylpropane-1,3-diol:

Density= mass / volume

We know density (1.015 g/mL) and we know volume (0.75 mL)

We rearrange and solve for mass:

Mass = density x volume

Mass = 1.015 g/mL x 0.75 mL

Mass = 0.7613 g

Calculating moles:

Moles = mass / molar mass

Moles = 0.7613 g / 90.121 g/mol

Moles = 0.0084 moles of 2-Methylpropane-1,3-diol

Millimoles of 2-Methylpropane-1,3-diol:

1 mol = 1000 mmoles

0.0084 moles (1000 mmoles / 1 mol) = 8.447 mmoles

We calculate mol equivalence dividing each mol of compound between the moles of the limiting reagent, since it is the one that determines the reaction.

Mol equivalent of 2-Methylpropane-1,3-diol:

0.008447 moles / 0.0033 moles = 2.56 mol equivalents

Mol equivalent of product:

0.0033 moles / 0.0033 moles = 1.0 mol equivalents

We can know which is the limiting reagent because the limiting reagent is the compound that will have the less amount of moles or millimoles, it will produce the less amount of product and it is not in excess. This compound will determine the reaction, since from this compound we obtain the minimum amount of product (product yield).

The limiting reagent is Benzaldehyde Dimethyl Acetal.

Comparing, we can see that Benzaldehyde Dimethyl Acetal has less number of millimoles than 2-Methylpropane-1,3-diol and it will produce less amount of product.

Benzaldehyde Dimethyl Acetal = 3.30 mmoles > 2-Methylpropane-1,3-diol = 8.447 mmoles

We do this from millimoles of Benzaldehyde Dimethyl Acetal. Since Benzaldehyde Dimethyl Acetal is the limiting reagent.

As we can see in the reaction, 1 mol of Benzaldehyde Dimethyl Acetal produce 1 mol of product, so we have 1:1 ratio.

1 mol Benzaldehyde Dimethyl Acetal = 1 mol of product

3.30 mmoles Benzaldehyde Dimethyl Acetal = 3.30 mmoles of product

0.0033 moles Benzaldehyde Dimethyl Acetal = 0.0033 moles of product

If 1 mol of product has 178.228 g/mol, how many grams will be in 0.0033 moles?

0.0033 moles of product (178.228 g / 1 mol of product) = 0.5822 g of product

In aperfect experiment, 0.5822 g of product will be produced.

Name

Formula

Mol. Eq.

Mw (g/mol)

Mmol

Amount

Benzaldehyde Dimethyl Acetal

C9H12O2

1.00

152.19

3.30 mmoles

0.50 mL

2-Methylpropane-1,3-diol

C4H10O2

2.56

90.121

8.447 mmoles

0.75 mL

Camphorsulfonic acid

--

--

--

--

5 mg

Dichlorometane

--

--

--

--

3 mL

Product

C11H14O2

1.0

178.228

3.30 mmoles

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