The ester below can be separated into phenol and an acetate salt, via saponification: heating the ester with a strong base, such
as sodium hydroxide, will produce phenol and sodium acetate. the two chemicals can then be separated by heat and filtration.
a. calculate the unsaturation number of the ester above.
b. calculate the molecular weight of the ester above
a. calculate the unsaturation number of the ester above.
b. calculate the molecular weight of the ester above
1 answer:
I found a similar question online which will help me answer your incomplete question. To make it easier, show all the elements of the compound given. It is shown in the second picture attached.
a.) The formula for unsaturation number is shown in the 3rd picture attached. Following this,
n = 8
m = 2(8) + 2 + 0 + 0 - 0 = 18
Thus,
x = Unsaturation number = (18 - 8)/2 = 5
<em>The unsaturation number is 5.</em>
b.) The molar mass of C is 12.01 g/mol; H is 1 g/mol; O is 16 g/mol. So, the molecular weight is:
Molecular weight = 12.01(8) + 8(1) + 2(16) = 136.08 g/mol
<em>The molecular weight of the ester is 136.08 g/mol.</em>
a.) The formula for unsaturation number is shown in the 3rd picture attached. Following this,
n = 8
m = 2(8) + 2 + 0 + 0 - 0 = 18
Thus,
x = Unsaturation number = (18 - 8)/2 = 5
<em>The unsaturation number is 5.</em>
b.) The molar mass of C is 12.01 g/mol; H is 1 g/mol; O is 16 g/mol. So, the molecular weight is:
Molecular weight = 12.01(8) + 8(1) + 2(16) = 136.08 g/mol
<em>The molecular weight of the ester is 136.08 g/mol.</em>
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This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.