5.14-5.16 - Yield, Atom Economy And Practical Evidence

Comparing Yield And Atom Economy

A reaction can disappoint a chemist in two different ways. It might make less of the desired product than the balanced equation predicts, or it might make the desired product while also sending many atoms into unwanted products.

[DIAGRAM: asset_name: Edexcel A Level Chemistry L028 Yield Vs Atom Economy; asset_slug: edexcel_a_level_chemistry_l028_yield_vs_atom_economy; recommended_method: deterministic_code; description: Actual and theoretical calcium oxide masses give yield; coefficient-weighted equation masses give atom economy. Pale 1600 by 900 diagram rendered with Pillow; technical geometry and Unicode chemical labels inspected.]

Percentage yield and atom economy answer different questions.

Percentage yield is practical. It changes if the reaction is incomplete, the product is lost during transfer, the product is impure, or the product is not fully dry.

Atom economy is theoretical for a stated equation. It does not depend on whether the student collected the product neatly. It depends on the balanced equation and which product is considered the desired product.

The two quantities can disagree. A reaction can have excellent atom economy but poor percentage yield if the practical method is inefficient. A reaction can have high percentage yield but poor atom economy if it reliably makes a lot of unwanted by-product.

Common-error contrast

Percentage Yield From Experimental Results

Percentage yield begins with the normal amount-of-substance route. The balanced equation predicts a theoretical yield; the experiment gives an actual yield.

Use this route:

balanced equation → moles of known substance → mole ratio to desired product → theoretical mass of desired product → percentage yield

If the question says a reactant is in excess, the other reactant controls the theoretical yield. If no excess is stated and there is more than one reactant amount, decide which reactant is limiting before calculating the theoretical product.

Worked example: calculate percentage yield

Calcium carbonate decomposes on heating:

A student heats $10.0\,\mathrm{g}$ of calcium carbonate and obtains $4.20\,\mathrm{g}$ of calcium oxide. Calculate the percentage yield of calcium oxide. Use $M_{\mathrm r}(\ce{CaCO3}) = 100.1$ and $M_{\mathrm r}(\ce{CaO}) = 56.1$.

Step 1: convert the known mass to moles.

Step 2: use the equation ratio.

1 mol $\ce{CaCO3}$ : 1 mol $\ce{CaO}$, so theoretical $n(\ce{CaO}) = 0.0999001\,\mathrm{mol}$

Step 3: convert theoretical moles of product to mass.

theoretical mass of $\ce{CaO} = n \times M_{\mathrm r} = 0.0999001 \times 56.1 = 5.60\,\mathrm{g}$

Step 4: compare the actual yield with the theoretical yield.

$\%\ \text{yield} = 4.20 / 5.6044 \times 100 = 74.9\%$ (using the unrounded theoretical mass)

Chemical interpretation: the balanced equation predicts up to about $5.60\,\mathrm{g}$ of calcium oxide from $10.0\,\mathrm{g}$ of calcium carbonate. The student collected less than this, so possible explanations for a low mass of isolated pure oxide include loss during transfer or product left in the apparatus. Incomplete decomposition reduces the amount of oxide made, but unreacted carbonate remaining in the weighed residue contaminates it and can increase its apparent mass. Do not identify the mass of a mixed residue as pure oxide yield.

Guided practice

Magnesium burns in oxygen:

A student burns $0.600\,\mathrm{g}$ of magnesium and obtains $0.940\,\mathrm{g}$ of magnesium oxide. Use $A_{\mathrm r}(\ce{Mg}) = 24.3$ and $A_{\mathrm r}(\ce{O}) = 16.0$ to calculate the percentage yield.

Route:

2 mol $\ce{Mg}$ : 2 mol $\ce{MgO}$, so theoretical $n(\ce{MgO}) = 0.0247\,\mathrm{mol}$

$M_{\mathrm r}(\ce{MgO}) = 40.3$

theoretical mass of $\ce{MgO} = 0.0247 \times 40.3 = 0.995\,\mathrm{g}$

The result is below $100\%$, so it is chemically plausible. A lower yield could come from some magnesium oxide escaping as smoke or not all magnesium reacting.

Atom Economy From Equations

Atom economy asks how efficiently atoms are incorporated into the desired product. It does not ask whether the product was collected successfully.

The denominator is built from the products in the balanced equation. If a product has a coefficient, multiply its molar mass by that coefficient. Atom economy is a percentage and has no unit.

Worked example: atom economy for thermal decomposition

Calcium carbonate decomposes:

If calcium oxide is the desired product:

$M_{\mathrm r}(\ce{CaO}) = 56.1$ and $M_{\mathrm r}(\ce{CO2}) = 44.0$

Chemical interpretation: only about $56\%$ of the product mass is in the desired calcium oxide. The rest is carbon dioxide, which is not the desired product in this calculation.

If carbon dioxide were the desired product instead:

The balanced equation has not changed. The "desired product" chosen by the chemical process has changed.

Worked example: coefficient care

Hydrogen reacts with nitrogen to form ammonia:

If ammonia is the desired product, there are no by-products in this equation.

The coefficient $2$ appears in both numerator and denominator because the equation produces $2\,\mathrm{mol}$ of ammonia. The atom economy is still $100\%$ because every product atom is in the desired product.

Common-error contrast

For:

if sodium carbonate is desired, the numerator is $M_{\mathrm r}(\ce{Na2CO3})$, and the denominator is:

The usual specification formula builds the denominator from coefficient-weighted products. An equivalent denominator $2\times M_{\mathrm r}(\ce{NaHCO3})$ is also chemically valid: mass conservation gives the same total for coefficient-weighted reactants and products in a balanced equation. Do not reject a correct equivalent route merely because it uses the reactant total.

Observations To Full And Ionic Equations

A simple test-tube reaction gives observations first: colour change, fizzing, warming, a solid disappearing, a metal coating, or a precipitate forming. The equation is the chemical claim that explains the observation.