Stoichiometry Note #4: Molar Ratios

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What are molar ratios? Well, they are comparisons between the quantities of (some or all of) the products and reactants in a chemical reaction. To make this more understandable, let's look at the reaction between iron (III) and oxygen gas:

4Fe + 3O₂ → 2Fe₂O₃

In this reaction, we can say that every 4 atoms of Fe plus every 3 molecules of O₂ makes 2 molecules of Fe₂O₃. We can also say that every 4 moles of Fe plus every 3 moles of O₂ makes 2 moles of Fe₂O_3. Or, we can say that every 4 million atoms of Fe plus every 3 million molecules of O₂ makes 2 million molecules of Fe₂O_3.

See how the ratio between Fe, O₂ and Fe₂O₃ is always 4: 3: 2 no matter how we change the units/quantities? Yes, indeed, 4: 3: 2 is the molar ratio between Fe, O₂ and Fe₂O₃! It's the ratio between the quantities used or produced in a "perfect reaction."

And, as you can see, the molar ratio is very easy to obtain—we basically take the coefficients of the products and reactants in a balanced equation, and put them together!

Now, let's use the molar ratios in another problem (answer is in white, in the brackets):

• How many moles of O₂ is needed to react with 8 moles of Fe so that 4 moles of Fe₂O₃ can be produced? Answer: ( 6 ) moles

And a slightly more complicated problem:

• How much hydrogen and oxygen are needed to form 6 moles of water?
• Answer: ( 6 ) moles of hydrogen and ( 3 ) moles of oxygen gas are needed.
• Explanation:
  • First, you have to balance the equation and obtain the molar ratio between hydrogen, oxygen and water: 2H₂ + O₂ → 2H₂O (ratio: 2: 1: 2).
  • Then, using this ratio, you know that you need (6÷2×2=6) moles of hydrogen and (6÷2×1=3) moles of water to produce 6 moles of water (pay attention to the colors of the numbers, they show where the numbers come from). 

Problems involving mass
Very often, you'll see problems that involve the mass of the products and/or reactants. For example,

• How many grams of hydrogen and oxygen will be required to produce 36.0296 grams of water?

How should you approach this kind of problem? Well, the key is: When using the molar ratio, make sure that all values are in moles (that's why it's called molar ratio)! In other words, convert all values into moles before using molar ratio as part of your calculation. Then, after you are done with the molar calculations, convert the answers from moles to mass if you have to.

So for the above question,

1. Convert the mass of water (36.0296 g) to the number of moles: 
• n=(mass÷molar mass) = (36.0296÷18.0148) = 2 moles
2. Use molar ratio calculations:
   • 2H₂ + O₂ → 2H₂O (Ratio: 2: 1: 2)
   • Find number of moles of H₂ and O₂:
     • H₂: 2÷2×2= 2 moles
     • O₂: 2÷2×1= 1 mole
3. Convert answers to mass:
   • H₂: mass= (moles×molar mass)= (2×2.0158)= 4.03160 g
   • O₂: mass= (moles×molar mass)= (1×31.998)= 31.9980 g

That's it for this post.

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