Atom Economy: How to Calculate It?

In summary, when calculating the efficiency of producing Oxygen from reactants, you would use the mass of Oxygen gas (O2) and not the mass of Oxygen present in a compound. This means taking into account the coefficients of the products in the equation.
  • #1
influx
164
2
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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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%
 

Related to Atom Economy: How to Calculate It?

1. What is atom economy?

Atom economy is a measure of the efficiency of a chemical reaction. It calculates the percentage of atoms in the reactants that end up in the desired product. A higher atom economy means less waste and a more sustainable process.

2. How is atom economy calculated?

Atom economy is calculated by dividing the molecular weight of the desired product by the sum of the molecular weights of all the reactants, and then multiplying by 100. This gives a percentage of the total reactants that end up in the desired product.

3. Why is atom economy important?

Atom economy is important because it reflects the efficiency and sustainability of a chemical reaction. A higher atom economy means less waste and a more environmentally friendly process. It also helps to conserve natural resources and reduce the cost of production.

4. What is the ideal atom economy?

The ideal atom economy is 100%, which means all the atoms in the reactants are used to form the desired product and there is no waste. However, in most cases, achieving 100% atom economy is not possible due to practical limitations.

5. How can atom economy be improved?

Atom economy can be improved by designing more efficient and selective reactions, using catalytic processes, and recycling or reusing reactants. It is also important to consider atom economy in the early stages of chemical process development to maximize efficiency and sustainability.

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