Finding Percent Yield

Question

Consider this reaction, which occurs in the atmosphere and contributes to photochemical smog:

SO₂(g) + H₂O(l) →H₂SO₃(g)

If there is 12.05 g SO₂ and excess H₂O present, the reaction yields 10.8 g H₂SO₃. Calculate the percent yield for the reaction.

Answer 1

This problem can be broken up into two parts:

1) Determining the theoretical yield

The first step is to make sure that the reaction is balanced. In this case, it is already balanced, but you can't assume that this will always be the case. Afterwards we will determine the limiting reagent. The problem tells you that there is excess H₂O present, which means that SO₂, our other reactant, is our limiting reagent. From here, we will determine the molar mass of SO₂ and H₂SO₃.

For SO₂, we add the molar mass of sulfur and oxygen as listed on the periodic table. SO₂ Has one sulfur atom and two oxygen atoms, so we will count sulfur once and oxygen twice:

(32.06 g/ 1 mol S) + 2(15.999 g/ 1 mol O) = 64.058 g/1 mol SO₂

We do the same for H₂SO_3:

(32.06 g/ 1 mol S) + 3(15.999 g/ 1 mol O) + 2(1.008 g/ 1 mol H) = 82.073 g/1 mol H₂SO₃

Next, we will determine the theoretical yield. First, we convert the starting amount (12.05 g) of our limiting reagent SO₂ into moles. We do this using the molar mass of SO₂ as a conversion factor:

12.05 g SO₂ x (1 mol SO₂)/(64.06 g SO₂) = 0.18811 mol SO₂

Then, using the coefficients of the chemical equation as a conversion, we convert from mol SO₂ to mol of H₂SO₃:

0.18811 mol SO₂ x (1 mol H₂SO₃)/(1 mol SO₂) = 0.18811 mol H₂SO₃

Lastly, we use the molar mass of H₂SO₃ to determine the theoretical yield of the reaction:

0.18811 mol H₂SO₃ x (82.073 g H₂SO₃)/(1 mol H₂SO₃) = 15.438816 g H₂SO₃ = theoretical yield

2) Calculating percent yield

The problem told us that the reaction yielded 10.8 g H₂SO₃. Using this and the theoretical yield we calculated earlier, we can determine the percent yield with the following formula:

Percent yield = 100% x (actual yield/theoretical yield)

Plugging in the corresponding values gives us the percent yield:

Percent yield = 100% x (10.8 g H₂SO₃ / 15.438816 g H₂SO₃) = 69.95355 %

Our actual yield had 3 significant figures, so our final answer should follow suit and also have 3 significant figures:

69.95355 % ≈ 70.0 %

Answer 2

To answer this question, we must first think about what "percent yield" is. Percent yield is essentially a comparison between the amount of a product collected during an experiment (in this case, the 10.8 g H₂SO₃) and the amount we would have expected to produce based on stoichiometry (theoretical yield).

Percent yield = (experimental yield/theoretical yield) *100

Steps:

1. Make sure the equation is balanced
2. Use stoichiometry to find the theoretical yield (convert from g SO₂ to g H₂SO₃) in the appropriate units (grams or moles). Since the experimental yield is given in grams, we will calculate theoretical yield in grams as well.
3. Use theoretical yield and experimental yield to determine percent yield.

Step 1: Balance the equation. In this case it was already balanced, so nothing to do here.

Step 2: Find theoretical yield. This is based on a few sub-steps.

1. Find molar mass and convert to moles of reactant
2. Use mole ratio to find moles of product produced
3. Use molar mass of product to find mass of product produced

We can use stoichiometry to determine the theoretical yield. Since we are given the mass of a reactant, we must use its molar mass to convert to moles.

1 mole SO₂ = 32.06 + 2(15.999)

= 64.058 g

So, there are 64.058 g/mol of SO₂

In this case, we are told that we have 12.05 g SO₂. So,

# moles SO₂ = 12.05 g SO₂ / 64.058 g/mol SO₂

= 0.1881 mol SO₂

Now, how many moles of the product would we expect to make from 0.1881 mol SO₂? We can use the coefficients in the balanced equation to predict this.

For every one mole of SO₂ we have, we should produce one mole of H₂SO₃. Their mole ratio is 1:1, so we expect to create 0.1881 mol H₂SO₃.

Now that we know how many moles of the product are expected, we can use the molar mass of the product to convert that to grams.

1 mol H₂SO₃ = 2(1.008) + 32.06 + 3(15.999)

= 82.073 g/mol

Now let's use that to convert from moles of H₂SO₃ to grams of H₂SO₃:

g H₂SO₃ = 0.1881 mol H₂SO₃ * (82.073 g/mol H₂SO₃)

= 15.4379 g H₂SO₃

So we would expect 15.4379 g of H₂SO₃ to be produced during this reaction - that's the theoretical yield.

Step 3 - Percent Yield

One last step - we need to make that comparison between the experimental and theoretical amounts.

% yield = (Experimental/Theoretical) *100

= (10.8/15.4379) * 100

= 69.96% (round to 70.0% for significant figures)