Stoichiometry

Stoichiometry is the arithmetic of "how much". You know what reacts with what; you want to know how many grams of one thing you need, or how many grams of another you can expect.

Nearly every failed attempt at these problems shares one root cause, and it is worth stating before anything else.

Coefficients are moles, never grams

In the equation

N₂ + 3H₂ → 2NH₃

the 3 does not mean three grams of hydrogen, or three times the mass of nitrogen. It means three molecules of hydrogen per molecule of nitrogen, and therefore three moles per mole. The masses in that ratio are wildly unequal: one mole of nitrogen is 28.014 g and three moles of hydrogen only 6.048 g.

Grams cannot cross the arrow. Moles can. That single restriction dictates the shape of every calculation in this topic.

The bridge and the hops around it

The mole ratio is the bridge. Everything else is getting to it and getting back:

1. Convert whatever you were given into moles of the substance you were given.
2. Cross the equation using the ratio of coefficients.
3. Convert the resulting moles into whatever unit the question asked for.

Step 2 is stoichiometry proper and it is one multiplication. Steps 1 and 3 are conversions you already know — by mass through the molar mass, by particle count through Avogadro's constant, by solution volume through concentration, by gas volume through molar volume. The unit changes; the middle step never does.

Everything downstream assumes the equation is correct. An unbalanced equation gives a wrong ratio and therefore a wrong answer that survives every subsequent check, which is why balancing comes first and is not negotiable.

Worked: nitrogen into ammonia

How much ammonia can 28.0 g of N₂ produce, given plenty of hydrogen?

Given to moles. 28.0 ÷ 28.014 = 0.9995 mol N₂.

Across the equation. The ratio of NH₃ to N₂ is 2 to 1, so multiply by 2/1:

0.9995 × 2 = 1.999 mol NH₃

Moles to the requested unit. Ammonia's molar mass is 17.031 g/mol:

1.999 × 17.031 = 34.0 g

Write the ratio as a fraction with the wanted substance on top and the given substance underneath, and the moles of the given substance cancel visibly. Getting the fraction upside down is by far the most common error at this step, and it is undetectable in the final answer unless you have written the units out.

A quick sanity check that costs nothing: two moles of ammonia weigh 34.06 g while one mole of nitrogen weighs 28.01 g, so the product mass should be a bit larger than the reactant mass. It is. The extra mass came from the hydrogen.

Which reactant runs out first

The ammonia problem said "plenty of hydrogen". When both quantities are specified, one will run out before the other, and the one that runs out — the limiting reagent — decides everything. The other is in excess and some of it simply survives the reaction.

Take the reduction of iron ore, as it happens in a blast furnace:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

with 100.0 g of Fe₂O₃ (molar mass 159.687) and 50.0 g of CO (molar mass 28.010).

Fe₂O₃: 100.0 ÷ 159.687 = 0.6262 mol
CO:     50.0 ÷ 28.010  = 1.7851 mol

There is far more CO by mole count, which tempts people into declaring it the excess reagent. It is not, because the equation demands three of it per Fe₂O₃.

Method one — divide by the coefficient. This is the fastest and the least error-prone:

Fe₂O₃: 0.6262 ÷ 1 = 0.6262
CO:    1.7851 ÷ 3 = 0.5950

The smaller result identifies the limiting reagent: CO. These numbers are not moles of anything; they are how many times over each reactant could supply the reaction as written, which is why they are comparable when raw mole counts are not.

Method two — compare demand with supply. Reacting all 0.6262 mol of Fe₂O₃ would require 3 × 0.6262 = 1.879 mol of CO, and only 1.785 mol is available. Same conclusion, more arithmetic.

Now finish from the limiting reagent and ignore the other entirely:

Fe: 1.7851 × (2/3) = 1.1901 mol
    1.1901 × 55.845 = 66.5 g

Using the excess reagent here would have given 69.9 g of iron, an answer that is wrong by five per cent and looks entirely plausible.

What is left over, and the check that proves it

Excess means leftovers, and questions often ask for them.

The CO consumed 1.7851 ÷ 3 = 0.5950 mol of Fe₂O₃, out of the 0.6262 mol present. The remainder is 0.0312 mol, or 0.0312 × 159.687 = 4.98 g of unreacted iron oxide.

That opens up the best check available in the whole topic. Mass is conserved, so everything that went in must be accounted for:

• iron produced: 66.45 g
• carbon dioxide produced: 1.7851 × 44.009 = 78.56 g
• iron oxide unreacted: 4.98 g
• total: 149.99 g

Against 100.0 + 50.0 = 150.0 g in. The tenth of a gram is rounding. If your three figures do not sum to the starting mass, you have made an error somewhere and you now know it before an examiner does — and this check works on every limiting-reagent problem ever set.

Theoretical, actual, and the gap between them

The 66.5 g above is the theoretical yield: what the equation permits. Real processes deliver less, and the comparison is the percent yield:

percent yield = actual ÷ theoretical × 100

If 61.2 g of iron were recovered, that is 61.2 ÷ 66.45 × 100 = 92.1%.

The shortfall is not sloppiness. It comes from reactions that reach equilibrium rather than completion, side reactions consuming reactants into unwanted products, product left behind during separation, and reactants that were not perfectly pure to begin with. A yield above 100% is therefore always a signal of an error — most often a product that has not been fully dried, so the mass includes solvent.

Percent yield is measured. It cannot be predicted from the equation, and any question asking for it must give you an actual mass.

Atom economy asks a different question

A reaction can have a superb yield and still waste most of its input mass, because the equation itself sends atoms into by-products. Atom economy measures that, and unlike yield it is computable from the equation alone:

atom economy = mass of desired product ÷ total mass of all products × 100

For the iron reaction, the desired product is 2Fe at 111.69, and the total product mass is 111.69 + (3 × 44.009) = 243.72. That gives 45.8% — even a flawless run sends more than half the mass of the inputs out as carbon dioxide.

The two metrics answer different complaints. A low yield means the process is inefficient. A low atom economy means the reaction was inefficient before anyone ran it, and improving it requires a different reaction rather than better technique.

Two molar volumes, both called STP

Gas-volume problems replace the molar mass with a molar volume, and here textbooks genuinely disagree with each other.

IUPAC changed the definition of standard temperature and pressure in 1982 from 0 °C and 1 atm to 0 °C and 100 kPa. The older definition gives a molar volume of 22.4 L/mol; the current one gives 22.7 L/mol. Many chemistry courses still teach 22.4, and many use "room temperature and pressure", 25 °C and 1 atm, where the figure is 24.0 L/mol instead.

None of these is wrong. They are answers under different stated conditions, and a marked answer that uses 22.4 where the paper assumed 24.0 is out by seven per cent through no fault of the arithmetic. Find out which convention your course uses, and state the conditions you assumed whenever a question does not.