Significant Figures and Rounding in Yield Results

You are holding a calculated yield number and you do not know how many digits to write down. A rounding calculator is useful only after you have carried the unrounded values through the chemistry. Round intermediate moles too early and the final mass is wrong. Round the final mass to the wrong place and your percent yield fails the lab report. The rule for significant figures and rounding in yield results is this: intermediate moles carry at least one extra digit, called a guard digit, and the final mass is rounded to the fewest significant figures among the starting masses and molar masses used.

Those two numbers get rounded exactly when the final mass is determined. You get one worked example, one failure case, and the exact limit of the rule. The page owns precision through the full chemistry workflow, from the balance to the final percent yield, and no other page on this site covers this detail.

Take the reaction 2H₂ + O₂ → 2H₂O. You measure 4.0 g of hydrogen (two sig figs) and 32.0 g of oxygen (three sig figs). The limiting reactant is hydrogen because 4.0 g of H₂ at 2.016 g/mol gives 1.984 mol, while 32.0 g of O₂ at 32.00 g/mol gives 1.000 mol, and the mole ratio is 2:1, so H₂ produces less water.

Do not round that intermediate mole count. Keep 1.984 mol as 1.984, not 2.0. Use the mole ratio 2:1 to get 1.984 mol of H₂O. Multiply by the molar mass of water, 18.01528 g/mol, to get 35.74 g of H₂O. That is your intermediate final mass. Now round it to two sig figs, matching the least precise input (4.0 g has two sig figs). The final theoretical yield becomes 36 g.

The calculation walks through three steps: convert mass to moles, apply the mole ratio, and convert back to mass. Each step has a rounding rule.

Step 1: Convert Mass to Moles

Divide each measured mass by the molar mass from CIAAW standard atomic weights. For 4.0 g of H₂, molar mass is 2.016 g/mol. The division gives 1.984 mol. Keep four sig figs even though the input has two. This guard digit prevents error in later steps.

Step 2: Apply the Mole Ratio

From the balanced equation, 2 mol H₂ produce 2 mol H₂O, a 1:1 ratio. Multiply 1.984 mol by 1. No rounding here.

Step 3: Convert Moles to Mass

Multiply 1.984 mol by 18.01528 g/mol (water molar mass from CIAAW). The product is 35.74 g. Now round to the least number of sig figs among the inputs. The inputs are 4.0 g (two sig figs) and 32.0 g (three sig figs). The least is two, so 35.74 g rounds to 36 g. That is the final theoretical yield.

For percent yield, divide actual yield by theoretical yield and round to the same number of sig figs as the least precise of the two. If actual yield is 30.5 g (three sig figs) and theoretical yield is 36 g (two sig figs), the percent yield is 85%, not 84.7%.

The rule comes from IUPAC and NIST guidelines for measurement uncertainty. When you round an intermediate number, you discard information that later multiplication magnifies. The failure case below shows how a 0.7% error appears from rounding one intermediate value too early.

Molar masses from CIAAW have many sig figs, water is 18.01528 g/mol, hydrogen is 2.016 g/mol. If you use a truncated molar mass like 18.0 g/mol, you artificially limit the precision of the result. The 35.74 g from the proper calculation becomes 35.6 g with the truncated molar mass, an error of 0.14 g. That is a 0.4% error before you even round the final answer.

Failure Case: Rounding Intermediate Moles to Two Sig Figs

Take the same 4.0 g of H₂. Round 1.984 mol to 2.0 mol. Apply the mole ratio: 2.0 mol H₂O. Multiply by 18.01528 g/mol: 36.03 g. Round to two sig figs: 36 g. The correct calculation gave 35.74 g, rounded to 36 g. The error is 0.26 g, or 0.7%. That 36 g is the same as the correct answer by coincidence in this example, but the intermediate error is real and can shift a percent yield by a full percentage point in other reactions.

When the Rule Breaks Down

The rule assumes all starting measurements have uniform precision. If one reactant mass has two sig figs and the other has four, the rule still applies: the final answer rounds to the least precise input. But if the molar mass itself is the limiting factor, such as using a value from an old periodic table with three sig figs, the result may be less precise than the measurements warrant. Always use CIAAW atomic weights from the IUPAC biennial update.

Value	When to Round	Number of Sig Figs	Example
Measured mass	Before conversion to moles	As measured	4.0 g (2 sig figs)
Intermediate moles	Never	At least 1 extra digit	1.984 mol (4 sig figs)
Final theoretical yield	After mass calculation	Least sig figs among inputs	36 g (2 sig figs)
Percent yield	After division	Least sig figs of actual and theoretical	85% (2 sig figs)

Before you write the theoretical yield in your lab notebook, check the sig figs on every input mass and every molar mass you used. Identify the one with the fewest sig figs. That number determines where the final answer stops. If you are using the site's rounding tool, feed it the unrounded intermediate values from your own calculation; do not let the tool round for you until the final step.

The single thing that most often goes wrong: students round intermediate moles to match the sig figs of the starting mass, which introduces a cumulative error that shifts the final yield by up to one percent. Keep that guard digit. The final rounding step is the only one that matters.