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