EXPERIMENT 9 based on a mass fo 165 g sodium reaction the moles of alkoxi gi ven

Question

EXPERIMENT 9 based on a mass fo 165 g sodium reaction the moles of alkoxi gi ven o from the isomers For simpl reaction, but it is assumed that 100% yield of the alkox de is obtained heptene you will this is a two-step grams. Since a of the the stated your assigned alcohol wi h so heptene produced in your (Note: You must calc the theoretical yield of all products in order to obtain have the same molecular mass, use the total moles of necessary in you must calculate the be using the potassium number of moles.) Br alkoxide base 2-bromo heptane alkoxide heptene isomers base Name (You calculate only You calculate for (CH3 COK) Amount From PC Table Molecular Wt. From PC Table density g from PC Table Other Moles from 0.00717 You calculate limiting reagent Moles Maximum Moles of You calculate You calculate Designate the Limiting Reagent Limiting Reagent For this experiment, the lab will be divided into six sections by lab bench location, with each section preparing an alkoxide base using one of either: (1)methanol, (2) absolute ethanol, (3) isopropyl alcohol, (4) 2-butanol, or (5) isobutyl alcohol as the assigned alcohol, or using (6) purchased potassium tert butoxide, dissolved in t-butyl alcohol (sodium reacts too slowly with t-butyl alcohol to prepare this base in lab). See the lab map below for the alcohol assignments, identified by (1)-(6) as listed above. (1) (2) (4) (6) (3) (5)

Explanation / Answer

The first we need to do is to get the information of molecular weight of each specie except the alkoxyde which is given in the statement (.165 gr of sodium), the molecular weight as well as the 2bromoheptane density can be found on a book or a web. I got information from the internet.

In order to fin the ammount of alkoxyde, we take that moles are given on a basis of 0.165 gr of sodium so we make the following operation:

grams of sodium = 0.00717 moles * ( 0.165 gr of sodium / 1 mol ) = 0.00118305 gr of sodium.

With the density of 2 bromoheptane we make the next operation:

density = mass / volume

Making an arrangement:

mass = density * volume

mass = 1.136 g / ml * (0.22 ml ) = 0.24992 grams

moles = 0.24992 / 179 = 0.001396 moles.

We find that there are more moles of alkoxyde (0.007107) than 2 bromoheptane (0.00134), so we say that the 2 bromoheptane is the limiting reagent.

To find the excess of alkoxyde we just have to get the difference of moles availables - moles to react :

0.007107 - 0.00134= 0.00571 moles that will remain untouched

The chart will look like these.

2bromoheptane alkoxide heptane isomers

Ammount 0.24992 gr 0.00118 gr of sodium 0.1366512 gr

Molecular weight 179 g/mol 98 g/mol

Other density= 1.136g/ml

moles 0.0013944 0.007107 0.0013944

Maximum moles

of product: 0 0.00571moles

Limiting reagent : Since we have more alkoxyde we might say that 2 bromoheptane is the limiting reagent

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