Mass-to-Mass Stoichiometry (Grams to Grams)

You have 25.0 g of sodium chloride and need to know how many grams of chlorine gas can be made from it. The answer is 15.2 g of Cl₂. This is a grams to grams stoichiometry problem, and you solve it in exactly three steps: grams to moles, mole ratio, then moles back to grams. Every mass-to-mass stoichiometry calculation on a homework set or lab report follows that same map.

Stoichiometry is the arithmetic of chemical reactions. The coefficients in a balanced reaction tell you the mole ratio, the exact proportion of reactants that combine and products that form. Because you weigh solids in grams, not in moles, you need a bridge between the two units. The bridge is the conversion formula n = m / M, where n is amount in moles, m is mass in grams, and M is molar mass in g mol⁻¹. You use it twice: once to leave grams, once to return.

Every grams-to-grams stoichiometry problem uses the same three-step path. Step one: convert the given mass (grams) to moles by dividing by the substance’s molar mass. Step two: use the mole ratio from the balanced reaction to find the moles of the target substance. Step three: convert those moles back to grams by multiplying by the target substance’s molar mass.

• Step 1: Grams of given → Moles of given (÷ molar mass)
• Step 2: Moles of given → Moles of target (× mole ratio)
• Step 3: Moles of target → Grams of target (× molar mass)

Memorise the sequence. A common mistake is to skip Step 2 and try to convert directly from grams of one substance to grams of another. That does not work because the mole ratio is the only link between different substances in a reaction. Without it, you are comparing apples to chlorine atoms by weight, which has no chemical meaning.

Before you start any stoichiometry calculation, the chemical reaction must be balanced. The coefficients are the only source of the mole ratio you will use in Step 2. If the reaction is wrong, every number that follows is wrong.

For example, consider the formation of water from hydrogen and oxygen. The unbalanced reaction H₂ + O₂ → H₂O says nothing about how many molecules react. The balanced version, 2 H₂ + O₂ → 2 H₂O, tells you the mole ratio of hydrogen to water is 2:2, which simplifies to 1:1. A student who used the unbalanced reaction would assume a 1:1 ratio and produce an answer off by a factor of two. Always balance first, then extract the mole ratio.

The process of balancing, adjusting coefficients until the same number of each atom appears on both sides, is a separate skill, but it is the necessary first step for any stoichiometry calculation. OpenStax Chemistry 2e sections 3.1 and 4.3 cover the mole concept and reaction stoichiometry in detail, including how to derive mole ratios from balanced reactions.

Problem

How many grams of water (H₂O) can be produced from 10.0 g of hydrogen gas (H₂)? The balanced reaction is 2 H₂ + O₂ → 2 H₂O.

Step 1: Grams to Moles of Given

Molar mass of H₂ = 2.016 g mol⁻¹ (from the periodic table, each hydrogen atom is 1.008 g mol⁻¹, and there are two atoms per molecule). Moles of H₂ = 10.0 g / 2.016 g mol⁻¹ = 4.96 mol (rounded to three significant figures, matching the 10.0 g input).

Step 2: Mole Ratio

From the balanced reaction, the mole ratio of H₂ to H₂O is 2:2, which simplifies to 1:1. So moles of H₂O = 4.96 mol × (2 mol H₂O / 2 mol H₂) = 4.96 mol.

Step 3: Moles to Grams of Target

Molar mass of H₂O = 18.015 g mol⁻¹ (IUPAC CIAAW). Grams of H₂O = 4.96 mol × 18.015 g mol⁻¹ = 89.4 g (rounded to three significant figures).

The answer: 10.0 g of hydrogen gas produces 89.4 g of water.

Problem

How many grams of carbon dioxide (CO₂) are produced when 50.0 g of methane (CH₄) burns completely in oxygen? The balanced reaction is CH₄ + 2 O₂ → CO₂ + 2 H₂O.

Step 1: Grams to Moles of Given

Molar mass of CH₄ = 12.011 g mol⁻¹ (carbon) + 4 × 1.008 g mol⁻¹ (hydrogen) = 16.043 g mol⁻¹. Moles of CH₄ = 50.0 g / 16.043 g mol⁻¹ = 3.12 mol (three significant figures).

Step 2: Mole Ratio

The balanced reaction shows a 1:1 mole ratio between CH₄ and CO₂. So moles of CO₂ = 3.12 mol × (1 mol CO₂ / 1 mol CH₄) = 3.12 mol.

Step 3: Moles to Grams of Target

Molar mass of CO₂ = 44.01 g mol⁻¹ (from IUPAC CIAAW). Grams of CO₂ = 3.12 mol × 44.01 g mol⁻¹ = 137 g (rounded to three significant figures).

The answer: 50.0 g of methane produces 137 g of carbon dioxide.

The three-step map above assumes the given reactant is the one that runs out first, the limiting reactant. In real reactions, you often have amounts of two or more reactants. The limiting reactant is the one that produces the least product when you run the calculation on each. The other reactant is in excess and some of it remains unreacted.

For example, if you have 10.0 g of H₂ and 80.0 g of O₂ in the water-formation reaction, you would run the three-step calculation twice: once starting from H₂ (giving 89.4 g H₂O) and once from O₂ (giving, after similar steps, about 90.0 g H₂O). The smaller product amount (89.4 g) is the theoretical yield, and H₂ is the limiting reactant. You cannot get more than that.

After the reaction, you might only collect 72.0 g of water. That is the actual yield. The percent yield = (actual / theoretical) × 100% = (72.0 / 89.4) × 100% = 80.5%. Percent yield is always less than 100% in practice due to incomplete reactions, side reactions, or losses during handling.

The three-step map still works for both the limiting reactant and percent yield contexts. You just run it more than once for limiting reactant, and you compare the result to the actual yield for percent yield.

The most frequent error is using the wrong molar mass. Students often grab the atomic mass of an element from the periodic table without multiplying by the number of atoms in the molecule. For oxygen gas (O₂), the molar mass is 31.998 g mol⁻¹, not 15.999 g mol⁻¹. For water, it is 18.015 g mol⁻¹, not 18.0 g mol⁻¹. The difference of 0.015 g mol⁻¹ matters in lab work.

Another mistake is forgetting the mole ratio entirely. Some students try to convert grams of reactant directly to grams of product using a single factor. That bypasses the mole ratio and gives a number that is chemically meaningless. The mole ratio is the only way to move between different substances.

Unit mismatch is another failure mode. The mass must be in grams, not kilograms, for the formula n = m / M to work. If a problem gives mass in kilograms, convert to grams first by multiplying by 1000.

Significant figures also trip people up. The answer cannot have more significant figures than the least precise measurement in the problem. In the water example, 10.0 g has three significant figures, so the answer 89.4 g has three. Reporting 89.352 g would be over-precise and wrong.

Finally, rounding intermediate values too aggressively introduces error. Keep at least one extra digit in each step, then round only the final answer. For example, in Step 1 of the methane problem, storing 3.1165 mol instead of 3.12 mol in your calculator prevents the final answer from drifting.

Substance	Formula	Molar Mass (g mol⁻¹)	Source
Hydrogen gas	H₂	2.016	IUPAC CIAAW (atomic weights)
Oxygen gas	O₂	31.998	IUPAC CIAAW
Water	H₂O	18.015	IUPAC CIAAW
Carbon dioxide	CO₂	44.01	IUPAC CIAAW
Methane	CH₄	16.043	IUPAC CIAAW
Sodium chloride	NaCl	58.44	IUPAC CIAAW
Glucose	C₆H₁₂O₆	180.156	IUPAC CIAAW