How to Calculate Atom Economy?

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The calculation of atom economy involves determining the total molar mass (Mr) of the products and dividing it by the total molar mass of the reactants, then multiplying by 100. However, to accurately assess the efficiency of producing a specific product, such as oxygen gas (O2), only the relevant product's molar mass should be considered. For example, in the reaction of sodium nitrite (NaNO2) and oxygen, the correct calculation for the efficiency of producing O2 is 18.8%, derived from the molar mass of O2 (16) divided by the molar mass of sodium nitrate (NaNO3, 85).

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  • Familiarity with the concept of conservation of mass
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atomecon.png


To work out the answer to this question, do I just find the total Mr of the products and divide it by the total Mr of the reactants (and then multiply by 100, and pick the biggest percentage?)

Cheers
 
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influx said:
atomecon.png


To work out the answer to this question, do I just find the total Mr of the products and divide it by the total Mr of the reactants (and then multiply by 100, and pick the biggest percentage?)

Cheers

If you did that, you would get %100 every time. Otherwise, you would be violating conservation of mass.

If you want to know the effency of producing Oxygen from your reactants, which product mass do you want to use in your equation?
 
flatmaster said:
If you did that, you would get %100 every time. Otherwise, you would be violating conservation of mass.

If you want to know the efficiency of producing Oxygen from your reactants, which product mass do you want to use in your equation?

If you do it for A, you get 118 %... (so I am doing something wrong?)

Mr of (NaNO2 + (1/2)O2) = 101
Mr of (NaNO3) = 85

(101/85)*100 = 118.8%

For A, wouldn't we use both products, since both products contain Oxygen?
 
Mr of (NaNO2 + (1/2)O2) = 101
You did this one wrong. You forgot to take 1/2 the mass of o2
For A, wouldn't we use both products, since both products contain Oxygen?No. We're only concerned with O2 as Oxygen gas. If the Oxygen is still in a compound, you haven't made oxygen.
 
flatmaster said:
If you did that, you would get %100 every time. Otherwise, you would be violating conservation of mass.

If you want to know the effency of producing Oxygen from your reactants, which product mass do you want to use in your equation?

flatmaster said:
Mr of (NaNO2 + (1/2)O2) = 101



You did this one wrong. You forgot to take 1/2 the mass of o2



For A, wouldn't we use both products, since both products contain Oxygen?


No. We're only concerned with O2 as Oxygen gas. If the Oxygen is still in a compound, you haven't made oxygen.


Oh, I always thought that when calculating the Mr, you do not consider the coefficients. Thanks for clearing that up..

So would this be a correct calculation (for A):

Mr of (1/2)O2 = 16
Mr of NaNO3 = 85

so 16/85 x 100 = 18.8%
 

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