Dr. Ray's Learning Center

Mass-to-Mass Stoichiometry

General Chemistry · Reaction Stoichiometry

Dr. Ray’s Learning Center · covers: Mass-to-Mass Stoichiometry

  • 20–25 min
  • 7 figures
  • 3 worked examples
  • 4 practice sets

Before this lesson:

Balances weigh grams, not moles. A lime-kiln operator knows how many tonnes of limestone go in; a combustion engineer knows the mass flow of fuel. To predict the mass of a product, or the mass of another reactant that is needed, you have to cross from grams to moles and back.

Mass-to-mass stoichiometry is that crossing: convert grams to moles, apply the mole ratio from the balanced equation, and convert moles back to grams (Figure 1).

  1. start (g)Mass of Agrams (g)
  2. ×1MA\times \dfrac{1}{M_{\mathrm{A}}}molar mass of A
  3. start (mol)Moles of Amol
  4. ×ca\times \dfrac{c}{a}mole ratio
  5. Moles of Cmol
  6. ×MC\times M_{\mathrm{C}}molar mass of C
  7. answerMass of Cgrams (g)

The middle arrow is the mole ratio from the previous lesson.All three worked examples in this lesson use all three arrows.

Figure 1. The stoichiometry roadmap. Every problem in this lesson follows this path for a reaction written as aA→cCa\,\mathrm{A} \rightarrow c\,\mathrm{C}. You never jump straight from grams of A to grams of C: the balanced equation only relates moles.

Learning Objectives

After this lesson, you should be able to:

  1. Convert a given mass of one species to the mass of another (mass to moles, mole ratio, moles to mass).
  2. Calculate the mass of a reactant required, including the O2 needed for combustion.
  3. Report the answer to the correct number of significant figures.

Key Ideas

1. Converting Between Mass and Moles

n=mMn = \dfrac{m}{M}
  • nn = amount of substance (mol)
  • mm = mass (g)
  • MM = molar mass (g/mol)

One mole of every substance contains the same number of particles, but a mole of a heavier formula weighs more (Figure 2). The molar mass is the conversion factor: divide grams by it to get moles, multiply moles by it to get grams.

Mass of one mole of four substancesHorizontal bars for the mass of one mole: H2 2.02 g, H2O 18.02 g, CO2 44.01 g, CaCO3 100.09 g. Each bar is one mole, the same number of particles, 6.022 times 10 to the 23rd.One mole of each substancesame number of particles: 6.022 × 1023​eachH2​2.02 gH2​O18.02 gCO2​44.01 gCaCO3​100.09 g
Figure 2. Same amount, different mass. Every bar is exactly 1 mol, yet one mole of CaCO3 weighs about 50 times as much as one mole of H2. The molar mass n=mMn = \frac{m}{M} is the conversion factor between grams and moles.

Key idea: only moles go into a mole ratio. Molar masses take you into moles and back out to grams.

2. Coefficients Relate Moles, Not Grams

In 2H2+O2→2H2O2\,\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\,\mathrm{H_2\!O} the amounts are in the ratio 2 : 1 : 2, but the masses are about 1 : 8 : 9 (Figure 3). Applying the coefficients to grams gives a wrong answer.

To get a product mass from a reactant mass, always route through moles:

mass of reactant → moles of reactant → moles of product → mass of product (Figure 1)

Amount (mol)Bar chart of amounts: H2 2 mol, O2 1 mol, H2O 2 mol.Amount (mol)H2​2 molO2​1 molH2​O2 molMass (g)Bar chart of masses: H2 4.04 g, O2 32.00 g, H2O 36.04 g.Mass (g)H2​4.04 gO2​32.00 gH2​O36.04 g

Not the same ratio: 2 : 1 : 2 in mol, about 1 : 8 : 9 in g.

Figure 3. Coefficients are mole ratios, not mass ratios. For the formation of water the amounts are 2 : 1 : 2, but the masses are about 1 : 8 : 9. Mass is still conserved: 4.04g+32.00g=36.04g4.04\,\text{g} + 32.00\,\text{g} = 36.04\,\text{g}.

The middle step is the mole ratio from the previous lesson, Mole Ratios and Mole-to-Mole Conversions:

nC=nA×can_{\mathrm{C}} = n_{\mathrm{A}} \times \dfrac{c}{a}

and the last step turns moles back into grams:

m=n×Mm = n \times M

Worked Example 1: Mass-to-Mass Stoichiometry

Problem: Calcium carbonate decomposes on heating to calcium oxide and carbon dioxide. Calculate the theoretical mass of CO2 produced from 10.0 g of pure CaCO3, assuming complete decomposition.

During strong heatingA crucible with a loosely fitting lid sits on a pipe-clay triangle on a tripod over a burner flame. It holds 10.0 g of calcium carbonate, and carbon dioxide gas escapes.During strong heatingCaCO3​(s)10.0 g at startCO2​(g)escapespipe-claytriangletripodburner flame,strong heatAfter coolingAfter cooling, 5.60 g of calcium oxide remains in the crucible.After coolingCaO(s)5.60 g leftcrucibleand lid
Figure 4. What Worked Example 1 describes in the lab. Strong heating (about 900 °C, as in a lime kiln) drives off CO2 gas. Weighing the crucible before and after shows the mass lost: 10.0g−4.40g=5.60g10.0\,\text{g}-4.40\,\text{g} = 5.60\,\text{g} of CaO remains.

Step 1. Write the balanced equation.

CaCO3→CaO+CO2\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}

The mole ratio of CaCO3\mathrm{CaCO_3} to CO2\mathrm{CO_2} is 1 : 1.

Step 2. Determine the molar masses.

  • MCaCO3=100.09g/molM_{\mathrm{CaCO_3}} = 100.09\,\text{g/mol}
  • MCO2=44.01g/molM_{\mathrm{CO_2}} = 44.01\,\text{g/mol}

Step 3. Convert the mass of CaCO3 to moles.

nCaCO3=10.0g100.09g/mol=0.09991moln_{\mathrm{CaCO_3}} = \dfrac{10.0\,\text{g}}{100.09\,\text{g/mol}} = 0.09991\,\text{mol}

Step 4. Apply the mole ratio.

nCO2=0.09991mol CaCO3×1mol CO21mol CaCO3=0.09991mol CO2\begin{aligned}n_{\mathrm{CO_2}} &= 0.09991\,\mathrm{mol\ CaCO_3} \times \dfrac{1\,\mathrm{mol\ CO_2}}{1\,\mathrm{mol\ CaCO_3}} \\ &= 0.09991\,\mathrm{mol\ CO_2}\end{aligned}

Step 5. Convert moles of CO2 to mass.

mCO2=0.09991mol×44.01g/mol=4.397g≈4.40g\begin{aligned}m_{\mathrm{CO_2}} &= 0.09991\,\text{mol} \times 44.01\,\text{g/mol} \\ &= 4.397\,\text{g} \approx 4.40\,\text{g}\end{aligned}

Final Answer: 4.40 g CO2

  1. CaCO3 given10.0g10.0\,\text{g}
  2. Step 3×1mol100.09g\times \dfrac{1\,\text{mol}}{100.09\,\text{g}}grams to moles
  3. CaCO30.09991mol0.09991\,\text{mol}
  4. Step 4×1mol CO21mol CaCO3\times \dfrac{1\,\mathrm{mol\ CO_2}}{1\,\mathrm{mol\ CaCO_3}}mole ratio 1 : 1
  5. CO20.09991mol0.09991\,\text{mol}
  6. Step 5×44.01g/mol\times 44.01\,\text{g/mol}moles to grams
  7. CO2 answer4.397g≈4.40g4.397\,\text{g} \approx 4.40\,\text{g}Final answer to 3 significant figures, set by 10.0 g
Figure 5. Worked Example 1 on the roadmap (Figure 1). Each arrow is one step of the solution; carry the 4-significant-figure values and round only the final answer.

This is the theoretical amount, assuming pure CaCO3 and complete conversion. Keep the extra digit in intermediate values and round only the final answer (three significant figures, set by 10.0 g).

Exam tip: the given value sets the significant figures. 10.0 g has three, so the answer is reported as 4.40 g, not 4.397 g or 4.4 g.

Your turn: mass-to-mass

Same five steps as Worked Example 1: molar masses, grams to moles, mole ratio, moles to grams.

Loading the practice problem…

Worked Example 2: Mass of Reactant Required

Problem: Nitrogen reacts with hydrogen to produce ammonia. What mass of hydrogen (H2\mathrm{H_2}) must react to produce 25.0 g of ammonia (NH3\mathrm{NH_3}) with sufficient nitrogen?

Step 1. Write the balanced equation.

N2+3H2→2NH3\mathrm{N_2} + 3\,\mathrm{H_2} \rightarrow 2\,\mathrm{NH_3}

Step 2. List the given information.

  • mNH3=25.0gm_{\mathrm{NH_3}} = 25.0\,\text{g}
  • MNH3=17.03g/molM_{\mathrm{NH_3}} = 17.03\,\text{g/mol}
  • MH2=2.02g/molM_{\mathrm{H_2}} = 2.02\,\text{g/mol}

Step 3. Convert the mass of NH3 to moles.

nNH3=25.0g17.03g/mol=1.468moln_{\mathrm{NH_3}} = \dfrac{25.0\,\text{g}}{17.03\,\text{g/mol}} = 1.468\,\text{mol}

Step 4. Apply the mole ratio. 2 mol of NH3\mathrm{NH_3} are formed from 3 mol of H2\mathrm{H_2}; hydrogen is wanted, so it goes on top.

nH2=1.468mol NH3×3mol H22mol NH3=2.202mol H2\begin{aligned}n_{\mathrm{H_2}} &= 1.468\,\mathrm{mol\ NH_3} \times \dfrac{3\,\mathrm{mol\ H_2}}{2\,\mathrm{mol\ NH_3}} \\ &= 2.202\,\mathrm{mol\ H_2}\end{aligned}

Step 5. Calculate the mass of H2.

mH2=2.202mol×2.02g/mol=4.448g≈4.45gm_{\mathrm{H_2}} = 2.202\,\text{mol} \times 2.02\,\text{g/mol} = 4.448\,\text{g} \approx 4.45\,\text{g}

Final Answer: 4.45 g H2

  1. NH3 wanted (given)25.0g25.0\,\text{g}
  2. Step 3×1mol17.03g\times \dfrac{1\,\text{mol}}{17.03\,\text{g}}grams to moles
  3. NH31.468mol1.468\,\text{mol}
  4. Step 4×3mol H22mol NH3\times \dfrac{3\,\mathrm{mol\ H_2}}{2\,\mathrm{mol\ NH_3}}mole ratio 3 : 2, wanted over given
  5. H22.202mol2.202\,\text{mol}
  6. Step 5×2.02g/mol\times 2.02\,\text{g/mol}moles to grams (H2 is 2.02, not 1.01)
  7. H2 required4.448g≈4.45g4.448\,\text{g} \approx 4.45\,\text{g}Final answer to 3 significant figures, set by 25.0 g
Figure 6. Worked Example 2 on the roadmap (Figure 1). The path runs from the product back to a reactant, but it is the same three arrows.

Only 4.45 g of hydrogen is needed for 25.0 g of ammonia, because most of the mass of NH3 comes from nitrogen.

Exam tip: check with conservation of mass. The nitrogen needed is 1.468×12=0.7340mol1.468 \times \frac{1}{2} = 0.7340\,\text{mol}, which has a mass of 0.7340mol×28.01g/mol=20.56g0.7340\,\text{mol} \times 28.01\,\text{g/mol} = 20.56\,\text{g}. Nitrogen plus hydrogen: 20.56g+4.448g=25.01g20.56\,\text{g} + 4.448\,\text{g} = 25.01\,\text{g}, which matches 25.0 g of ammonia.

Your turn: mass of reactant required

Same method as Worked Example 2: the mass of one substance is given and the mass of a reactant is wanted.

Loading the practice problem…

Worked Example 3: Combustion Stoichiometry

Problem: Methane is burned completely. Determine the theoretical mass of oxygen required to burn 16.0 g of methane.

Step 1. Write the balanced combustion reaction.

CH4+2O2→CO2+2H2O\mathrm{CH_4} + 2\,\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\,\mathrm{H_2\!O}
Particle diagram of CH4 + 2 O2 forming CO2 + 2 H2OOne methane molecule and two oxygen molecules on the reactant side; one carbon dioxide molecule and two water molecules on the product side. Carbon 1 in, 1 out; hydrogen 4 in, 4 out; oxygen 4 in, 4 out.+→+REACTANTSPRODUCTSHHHHCCH4​OOOO2 O2​OOCCO2​HHOHHO2 H2​O✓ Atoms balance: C 1 in, 1 out · H 4 in, 4 out · O 4 in, 4 out
Figure 7. Why methane needs two O2. The four oxygen atoms end up in one CO2 (two O) and two H2O (one O each). This 1 : 2 ratio is the mole ratio used in Step 4.

Step 2. List the given information.

  • mCH4=16.0gm_{\mathrm{CH_4}} = 16.0\,\text{g}
  • MCH4=16.04g/molM_{\mathrm{CH_4}} = 16.04\,\text{g/mol}
  • MO2=32.00g/molM_{\mathrm{O_2}} = 32.00\,\text{g/mol}

Step 3. Convert the methane mass to moles.

nCH4=16.0g16.04g/mol=0.9975moln_{\mathrm{CH_4}} = \dfrac{16.0\,\text{g}}{16.04\,\text{g/mol}} = 0.9975\,\text{mol}

Step 4. Apply the mole ratio. Each mole of CH4\mathrm{CH_4} requires 2 mol of O2\mathrm{O_2}.

nO2=0.9975mol CH4×2mol O21mol CH4=1.995mol O2\begin{aligned}n_{\mathrm{O_2}} &= 0.9975\,\mathrm{mol\ CH_4} \times \dfrac{2\,\mathrm{mol\ O_2}}{1\,\mathrm{mol\ CH_4}} \\ &= 1.995\,\mathrm{mol\ O_2}\end{aligned}

Step 5. Calculate the oxygen mass.

mO2=1.995mol×32.00g/mol=63.84g≈63.8gm_{\mathrm{O_2}} = 1.995\,\text{mol} \times 32.00\,\text{g/mol} = 63.84\,\text{g} \approx 63.8\,\text{g}

Final Answer: 63.8 g O2

Engineering note: This is the stoichiometric (theoretical) oxygen requirement. Real burners are fed excess air, but the stoichiometric amount is always the starting point of that calculation.

Exam tip: write the species next to every unit (mol CH4, mol O2). If the ratio is the right way up, the given species cancels and only the wanted one is left.

Your turn: combustion stoichiometry

Same method as Worked Example 3, with a different fuel and mass each time.

Loading the practice problem…

Common Mistakes

Common mistake: Treating coefficients as mass ratios. Coefficients are mole ratios. Convert grams to moles before you use them (Key Idea 2).

Common mistake: Stopping at moles. When the question asks for grams, finish with m=n×Mm = n \times M for the substance asked about.

Common mistake: Using the wrong molar mass for a diatomic element. O2 is 32.00 g/mol, not 16.00, and H2 is 2.02 g/mol, not 1.01. Using 1.01 in Worked Example 2 gives 2.22 g instead of 4.45 g.

Common mistake: Writing the mole ratio upside down, or leaving it out. Put the wanted species on top and the given species on the bottom, and never assume a 1 : 1 ratio. See the previous lesson, Wanted Over Given.

Common mistake: Rounding too early. Keep 4 significant figures in intermediate steps and round only the final answer to 3.

Common mistake: Expecting the theoretical amount in practice. Real yields are lower because of incomplete conversion, side reactions and losses, and when two reactant amounts are given one of them may run out first. Those are the Percent Yield and Limiting Reactant lessons.

Key Takeaway

Nearly every mass-to-mass problem is the same three moves: convert the given mass to moles, apply the mole ratio, and convert to the mass asked for. Round only at the end.

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

Loading the practice problem…

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Loading your results…

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. The practice problems are generated and checked by an audited engine; see the lesson audit log.

Leave a Reply