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marissa [1.9K]
2 years ago
8

g Calculate the theoretical yield (in grams) of your product if you start with 0.50 grams of E-stilbene. The molecular weight of

E-stilbene is 180.25 g/mol, and the molecular weight of the product is 340.058 g/mol. 0.5109 grams 0.9433 g 0.2342 g 0.6312 g
Chemistry
1 answer:
sattari [20]2 years ago
4 0

Answer:

0.9433g

Explanation:

Theoretical yield is defined as the mass produced assuming all reactant reacts producing the product.

Assuming the reaction is 1:1, we need to find the moles of E-stilbene (Reactant). If all reactant reacts, the moles of E-stilbene = Moles of product.

Using the molar mass of the product we can find the theoretical yield as follows:

<em>Moles E-stilbene:</em>

0.50g * (1mol/180.25g) = 0.00277 moles = Moles Product

<em>Mass Product = Theoretical yield:</em>

0.00277 moles * (340.058g/mol) = 0.9433g

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Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

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We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

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The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


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