Theoretical Yield Calculator with Steps | StatsUnlock

To calculate theoretical yield, convert each reactant to moles, find the limiting reagent (the one that runs out first), multiply its moles by the mole ratio from the balanced equation, and multiply by the product’s molar mass. For aspirin from 2.00 g of salicylic acid (138.122 g/mol) with acetic anhydride in excess:

Rule of thumb: theoretical yield = moles of limiting reagent × (product coefficient ÷ reactant coefficient) × molar mass of product. Percent yield = actual ÷ theoretical × 100, so 2.10 g of aspirin is an 80.5% yield, and a theoretical yield of 73 g with an actual yield of 62 g is 84.9%.

The theoretical yield is the amount of product, in grams or moles, that a reaction would make if it went to completion with no side reactions and no losses. It is set by the limiting reagent: the reactant that runs out first. The actual yield is what you weigh at the end; the percent yield compares the two.

A theoretical yield calculator does the stoichiometry for you. It balances the equation, converts each amount you enter into moles (masses with molar mass, liquids with density, solutions with molarity, gases with the molar volume), divides each by its coefficient to find the limiting reagent, then scales to the product with the mole ratio and its molar mass. It also reports the excess left over, the yield every reactant would give on its own, the atom economy and, if you enter it, the percent yield.

A worked one-liner: 10.0 g of CH₄ and 20.0 g of O₂ make CO₂ by CH₄ + 2O₂ → CO₂ + 2H₂O. O₂ gives 0.625 ÷ 2 = 0.3125, less than CH₄’s 0.623, so O₂ limits and the theoretical yield of carbon dioxide is 0.3125 × 44.009 = 13.8 g.

Who uses it: theoretical and percent yield are taught in NCERT Class 11 (“Some Basic Concepts of Chemistry”) and asked in JEE and NEET; GCSE, A level, AP Chemistry and IB courses set them every year, and every organic chemistry lab report (aspirin, cyclohexene, benzoic acid, Grignard reactions, esters) needs one. Process chemists use the same arithmetic to cost a batch.

What this free theoretical yield calculator solves

One tool covers every theoretical yield calculation in a typical course and lab: a theoretical yield calculator for a chemical reaction, a limiting reagent and theoretical yield calculator, a theoretical yield in grams calculator, a theoretical and percent yield calculator, a theoretical yield of aspirin calculator and a calculator for organic syntheses with liquids measured in mL. Press Try it to load an example, then press the button to see the working.

Free, with no sign-up. Everything runs in your browser: nothing you type is sent to a server or stored.

En español: ¿cómo se calcula el rendimiento teórico? Pasa cada reactivo a moles, encuentra el reactivo limitante, multiplica por la relación molar de la ecuación balanceada y por la masa molar del producto: 0,01448 mol × 180,159 g/mol = 2,61 g de aspirina.

How accurate is this theoretical yield calculator?

Molar masses use the CIAAW 2024 standard atomic weights (Prohaska et al., 2022), and equations are balanced exactly with integer coefficients. The answer is given to the number of significant figures of your least precise measurement. Liquid densities, when you leave the box empty, are typical handbook values at room temperature and are shown in the steps so you can replace them with your bottle’s value. Every answer is checked before it gets the Verified badge: no reactant may go below zero, and when every reactant is given, the mass of reactants used must equal the mass of products formed. The R and Python scripts in Sections 10 and 11 return the same 2.61 g of aspirin and 80.5% yield for the headline problem.

Studies of chemistry problem solving find that students often get a theoretical yield wrong because they skip the limiting reagent or apply the mole ratio upside down, not because of the arithmetic (Chandrasegaran, Treagust, Waldrip & Chandrasegaran, 2009). That is why the results show the product each reactant would give, side by side.

4.1 Theoretical, actual and percent yield

The theoretical yield is a calculated ceiling: the mass of product if the limiting reagent reacted completely and nothing was lost. The actual yield is the mass you weigh. Percent yield = actual ÷ theoretical × 100 tells you how close you came. Because the theoretical yield is the reference, every percent yield calculation starts with it.

4.2 Why the limiting reagent matters

Reactants are rarely mixed in exact proportions. The one that runs out first stops the reaction, so only it determines how much product forms. It is not always the reactant with the smaller mass: 20.0 g of O₂ limits the burning of 10.0 g of CH₄ because each CH₄ needs two O₂. Dividing moles by coefficients, as the calculator does, removes that trap (Chandrasegaran et al., 2009).

4.3 Excess reagents

Chemists often add one reactant in excess on purpose: acetic anhydride in the aspirin synthesis, so that all the more valuable salicylic acid reacts. The calculator reports how much excess is left (0.0384 mol, or 3.92 g, of acetic anhydride for the headline problem), which you will need to remove or destroy in the work-up.

4.4 Percent yields above 100%

A real percent yield cannot exceed 100%. A value above it means the product still holds solvent or water, contains an impurity or unreacted starting material, or a mass was misread. Dry the product to constant mass and check its purity, for example by melting point (aspirin melts at about 135 °C).

4.5 Typical percent yields

4.6 Atom economy

Atom economy asks a different question: what fraction of the mass of all the reactants ends up in the desired product, even at 100% yield (Trost, 1991)? The aspirin synthesis has an atom economy of 75.0% because acetic acid is made as a by-product; the Haber process (N₂ + 3H₂ → 2NH₃) has 100%. Green chemistry tries to raise both yield and atom economy (Constable, Curzons & Cunningham, 2002).

4.7 Multi-step syntheses

When a product is made in several steps, the overall yield is the product of the step yields as fractions. Three steps of 85%, 90% and 75% give 0.85 × 0.90 × 0.75 = 57.4% overall, which is why long syntheses need every step to be efficient.

4.8 A field note on where these problems go wrong

In marking yield calculations, the same slips come up: using grams straight in the mole ratio, choosing the limiting reagent by the smaller mass, inverting the ratio (multiplying by 1/2 instead of 2), forgetting the density for a liquid measured in mL, and weighing a wet product. Amounts tables like the calculator’s results table make these errors easy to see (Olmsted, 1999).

▶ Solve a problem above to fill all five cards with your own answer.

The theoretical yield formula is m(product) = n(limiting) × (b ÷ a) × M(product), where a and b are the coefficients of the limiting reagent and the product. Every other equation below feeds into it or uses it.

This flowchart shows how to calculate the theoretical yield of aspirin from salicylic acid, on the same problem used in the Excel, R, Python and by-hand sections: 2.00 g of salicylic acid (C₇H₆O₃) reacts with 5.00 mL of acetic anhydride (C₄H₆O₃, density 1.08 g/mL). What is the theoretical yield of aspirin (C₉H₈O₄), and the percent yield if 2.10 g is obtained?

Check any box yourself: the Excel sheet in Section 9, the R script in Section 10 and the Python script in Section 11 all return 2.61 g and 80.5% for this problem.

1. Choose what you have. Pick an equation with amounts, starting material and product names without an equation, or two of the three yields.
2. Type the equation. Write C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2; it is balanced for you, and one row appears for each reactant.
3. Enter the reactant amounts. Use g, mg, mol or mmol, mL of a liquid (with density), mL of a solution (with molarity) or litres of a gas at STP.
4. Leave excess reactants empty. If only one amount is given, it is treated as the limiting reagent.
5. Pick the product. Choose which product you want the theoretical yield of, for example aspirin rather than acetic acid.
6. Without an equation, type names and the ratio. Enter salicylic acid and aspirin with coefficients 1 and 1, or methyl benzoate, bromobenzene (2) and triphenylmethanol.
7. Add the actual yield. Type the mass of dry product to get the percent yield.
8. Press the button. Nothing is calculated before you press it, and changing any input clears the old answer.
9. Read the limiting reagent, the steps and the charts. The Verified badge means no reactant goes below zero and mass is conserved.
10. Copy, share or batch. Copy the answer, share a link, download a report, or solve a whole worksheet on the Batch tab.

Excel handles the whole calculation in one small table: a row per reactant with its grams, molar mass and coefficient, a column for moles and one for moles ÷ coefficient. MIN picks the limiting reagent and two more cells give the theoretical and percent yield. The sheet uses the headline problem, so it must return 2.61 g of aspirin and 80.5%, the same as the theoretical yield calculator above.

9.1 The functions you need

9.2 Step by step

9.3 Awkward cases

• Three or more reactants. Extend the table; MIN still finds the limiting reagent.
• Solutions. Put =molarity*volume_mL/1000 in the moles column.
• Multi-step syntheses. =PRODUCT(B2:B4/100)*100 (entered as an array formula in older Excel) gives the overall yield.

9.4 Why your Excel answer might differ

9.5 Charting in Excel

1. Next to each reactant, compute the grams of product it could make: =E2*$B$6*$C$6.
2. Insert → Clustered Column; the shortest bar is the theoretical yield, as in chart 1 above.

9.6 Excel errors you will probably hit

This base R script keeps the reactants in a data frame, converts grams to moles, finds the limiting reagent with which.min, calculates the theoretical yield, the excess left over and the percent yield, checks five textbook answers and saves a 300 dpi chart. For the headline problem it prints 2.61 g and 80.5%, matching the theoretical yield calculator on this page.

10.1 The complete script

R, base only, saves theoretical_yield.png

# ============================================================
# Theoretical yield calculator in R (base R only):
# moles -> limiting reagent -> theoretical yield -> percent yield
# StatsUnlock - https://statsunlock.com/theoretical-yield-calculator/
# ============================================================

# ---- 1. Your reaction ---------------------------------------------------
# C7H6O3 (salicylic acid) + C4H6O3 (acetic anhydride) -> C9H8O4 (aspirin) + CH3COOH
reactants <- data.frame(
  name  = c("salicylic acid", "acetic anhydride"),
  coef  = c(1, 1),
  M     = c(138.122, 102.089),          # g/mol
  grams = c(2.00, 5.00 * 1.08)          # 5.00 mL x 1.08 g/mL for the liquid
)
product <- list(name = "aspirin", coef = 1, M = 180.159)
actual_yield <- 2.10                    # grams of dry product (NA if unknown)

# ---- 2. Moles and the limiting reagent ----------------------------------
reactants$moles  <- reactants$grams / reactants$M
reactants$extent <- reactants$moles / reactants$coef
lim <- which.min(reactants$extent)
print(reactants, digits = 5)
cat("Limiting reagent:", reactants$name[lim], "\n")

# ---- 3. Theoretical yield ----------------------------------------------
n_product   <- reactants$extent[lim] * product$coef
theoretical <- n_product * product$M
cat(sprintf("Theoretical yield = %.5f mol x %.3f g/mol = %.2f g %s\n", n_product, product$M, theoretical, product$name))
left <- reactants$moles - reactants$extent[lim] * reactants$coef
for (i in seq_len(nrow(reactants))[-lim])
  cat(sprintf("Excess %s: %.4f mol = %.2f g left over\n", reactants$name[i], left[i], left[i] * reactants$M[i]))
pct <- actual_yield / theoretical * 100
cat(sprintf("Percent yield = %.2f / %.4f x 100 = %.1f%%\n", actual_yield, theoretical, pct))

# ---- 4. Checks ----------------------------------------------------------
stopifnot(reactants$name[lim] == "salicylic acid",
          round(theoretical, 2) == 2.61,
          round(pct, 1) == 80.5,
          round(62 / 73 * 100, 1) == 84.9,             # theoretical 73 g, actual 62 g
          round(0.85 * 0.90 * 0.75 * 100, 1) == 57.4)   # three-step overall yield
cat("Verdict: Verified (all checks pass)\n")

# ---- 5. Figure: product each reactant could make, and the actual yield ----
vals <- c(reactants$extent * product$coef * product$M, actual_yield)
png("theoretical_yield.png", width = 2100, height = 1350, res = 300)
par(mar = c(4, 4.5, 3, 1))
barplot(vals, names.arg = c(reactants$name, "actual"), col = c(ifelse(seq_along(reactants$name) == lim, "#ea580c", "#93c5fd"), "#2563eb"),
        ylab = paste("g of", product$name), main = "Product from each reactant: the smallest is the theoretical yield")
abline(h = theoretical, lty = 2, col = "#16a34a", lwd = 1.5)
invisible(dev.off())
cat("Saved theoretical_yield.png\n")

Output: Limiting reagent: salicylic acid, Theoretical yield = 0.01448 mol x 180.159 g/mol = 2.61 g aspirin, Excess acetic anhydride: 0.0384 mol = 3.92 g left over, Percent yield = 80.5%, Verdict: Verified.

10.2 R traps

• which.min returns the first minimum; when two reactants tie, both are limiting.
• Inside sprintf, write a percent sign as %%.
• Keep molar masses to three decimals; rounding them to whole numbers changes the third significant figure.

10.3 What each part does

10.4 The figure

The script saves theoretical_yield.png at 2100 × 1350 pixels and 300 dpi: the grams of aspirin each reactant could make, the actual yield, and a dashed line at the theoretical yield.

This script does the same job in plain Python with a dictionary of reactants: moles, the limiting reagent, the theoretical yield, the excess, the percent yield, five checks and a 300 dpi chart. It is the same method the theoretical yield calculator runs in your browser.

Install once: pip install matplotlib

11.1 The complete script

Python, saves theoretical_yield.png

# ============================================================
# Theoretical yield calculator in Python:
# moles -> limiting reagent -> theoretical yield -> percent yield
# StatsUnlock - https://statsunlock.com/theoretical-yield-calculator/
# pip install matplotlib
# ============================================================
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# ---- 1. Your reaction ---------------------------------------------------
# C7H6O3 (salicylic acid) + C4H6O3 (acetic anhydride) -> C9H8O4 (aspirin) + CH3COOH
reactants = {
    #  name               coefficient  molar mass (g/mol)  grams
    "salicylic acid":   (1, 138.122, 2.00),
    "acetic anhydride": (1, 102.089, 5.00 * 1.08),   # 5.00 mL x 1.08 g/mL
}
product_name, product_coef, product_M = "aspirin", 1, 180.159
actual_yield = 2.10                                   # grams of dry product (or None)

# ---- 2. Moles and the limiting reagent ----------------------------------
moles = {r: g / M for r, (a, M, g) in reactants.items()}
extent = {r: moles[r] / reactants[r][0] for r in reactants}     # n / coefficient
limiting = min(extent, key=extent.get)
for r in reactants:
    print(f"{r:17s} {moles[r]:.5f} mol  / {reactants[r][0]} = {extent[r]:.5f}"
          f"{'   <- limiting' if r == limiting else ''}")

# ---- 3. Theoretical yield ----------------------------------------------
n_product = extent[limiting] * product_coef
theoretical = n_product * product_M
print(f"Theoretical yield = {n_product:.5f} mol x {product_M} g/mol = {theoretical:.2f} g {product_name}")
for r in reactants:
    if r != limiting:
        left = moles[r] - extent[limiting] * reactants[r][0]
        print(f"Excess {r}: {left:.4f} mol = {left * reactants[r][1]:.2f} g left over")
if actual_yield is not None:
    pct = actual_yield / theoretical * 100
    print(f"Percent yield = {actual_yield} / {theoretical:.4f} x 100 = {pct:.1f}%")

# ---- 4. Checks ----------------------------------------------------------
assert limiting == "salicylic acid"
assert round(theoretical, 2) == 2.61
assert round(pct, 1) == 80.5
assert round(62 / 73 * 100, 1) == 84.9                   # theoretical 73 g, actual 62 g
assert round(0.85 * 0.90 * 0.75 * 100, 1) == 57.4         # three-step overall yield
print("Verdict: Verified (all checks pass)")

# ---- 5. Figure: product each reactant could make, and the actual yield ----
names = list(reactants) + ["actual"]
values = [extent[r] * product_coef * product_M for r in reactants] + [actual_yield or 0]
colors = ["#ea580c" if r == limiting else "#93c5fd" for r in reactants] + ["#2563eb"]
fig, ax = plt.subplots(figsize=(7, 4.5))
ax.bar(names, values, color=colors)
ax.axhline(theoretical, color="#16a34a", ls="--", lw=1.5, label=f"theoretical yield {theoretical:.2f} g")
ax.set_ylabel(f"g of {product_name}")
ax.set_title("Product from each reactant: the smallest is the theoretical yield")
ax.legend(); fig.tight_layout()
fig.savefig("theoretical_yield.png", dpi=300)
print("Saved theoretical_yield.png")

Output: salicylic acid 0.01448 mol / 1 = 0.01448 <- limiting, Theoretical yield = 0.01448 mol x 180.159 g/mol = 2.61 g aspirin, Percent yield = 80.5%, Verdict: Verified.

11.2 Python traps

• min(extent, key=extent.get) returns the name of the limiting reagent, not its value.
• Store the density conversion in the grams column (5.00 * 1.08) so liquids and solids are handled the same way.
• Round only when printing; keep full precision for the percent yield.

11.3 Packages

pip install chempy balances equations with balance_stoichiometry, and molmass gives the molar mass of any formula, so the script can start from formulas instead of typed molar masses.

11.4 What each part does

11.5 The figure

The script saves theoretical_yield.png at 300 dpi: the product each reactant could make, with the theoretical yield marked.

12.1 Molar masses and densities of common lab reagents

Molar masses from CIAAW 2024 atomic weights; densities are typical handbook values near 20 to 25 °C. These are the values the calculator uses when you type a name or leave the density box empty.

12.2 Theoretical yields of common reactions

Every row was computed by the same engine as the theoretical yield calculator.

12.3 Theoretical yield calculations worksheet with answers

Twenty-five questions in three groups: theoretical yield from one reactant, with a limiting reagent, and percent yield. Every answer was checked by the calculator engine. Cover the last column and try them first.

Eight examples run through the calculator, from aspirin to a Grignard reaction. Each figure is drawn from the computed numbers: one bar per reactant for the product it could make, with the limiting reagent in orange, and the actual yield in blue where there is one.

This is the written method examiners expect, on the same problem as Sections 7, 9, 10 and 11: 2.00 g of salicylic acid reacts with 5.00 mL of acetic anhydride (1.08 g/mL); 2.10 g of aspirin is obtained.

14.1 How to calculate theoretical yield step by step

1. Write the balanced equation. C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + CH₃COOH (1 : 1 : 1 : 1).
2. Moles of salicylic acid. 2.00 g ÷ 138.122 g/mol = 0.01448 mol.
3. Moles of acetic anhydride. 5.00 mL × 1.08 g/mL = 5.40 g; 5.40 ÷ 102.089 = 0.0529 mol.
4. Limiting reagent. Divide by the coefficients (both 1): 0.01448 < 0.0529, so salicylic acid limits.
5. Moles of aspirin. 0.01448 mol × 1/1 = 0.01448 mol.
6. Theoretical yield in grams. 0.01448 mol × 180.159 g/mol = 2.61 g.
7. Percent yield. 2.10 ÷ 2.61 × 100 = 80.5%.

14.2 How to calculate theoretical yield from the limiting reagent

For 10.0 g CH₄ and 20.0 g O₂ in CH₄ + 2O₂ → CO₂ + 2H₂O: CH₄ 10.0 ÷ 16.043 = 0.623 mol (÷ 1 = 0.623); O₂ 20.0 ÷ 31.998 = 0.625 mol (÷ 2 = 0.3125). O₂ is limiting even though there is more of it by mass. CO₂ = 0.3125 mol × 44.009 g/mol = 13.8 g.

14.3 How to calculate theoretical yield from mL

Turn the volume of a liquid into grams with its density first. 10.0 mL of cyclohexanol (0.962 g/mL) is 9.62 g, or 9.62 ÷ 100.161 = 0.0960 mol. Dehydration gives cyclohexene 1 : 1, so the theoretical yield is 0.0960 × 82.146 = 7.89 g.

14.4 How to calculate theoretical yield in organic chemistry

Organic procedures rarely give a full equation; you need only the mole ratio of the limiting starting material to the product. In the Grignard synthesis of triphenylmethanol, two PhMgBr add to one methyl benzoate: 1.50 g methyl benzoate (136.150 g/mol) is 0.01102 mol, and 4.00 g bromobenzene (157.010 g/mol) is 0.02548 mol, or 0.01274 per 2. Methyl benzoate limits, so the theoretical yield is 0.01102 × 260.336 = 2.87 g.

14.5 How to calculate actual yield from theoretical yield

Rearrange the percent yield equation: actual = theoretical × % ÷ 100. A theoretical yield of 2.61 g at 75% gives 1.96 g. Going the other way, theoretical = actual ÷ (% ÷ 100): 2.10 g at 80.5% means a theoretical yield of 2.61 g.

14.6 How to calculate theoretical yield with two products

Each product has its own theoretical yield from the same limiting reagent. Burning 22.0 g of propane in 64.0 g of oxygen (C₃H₈ + 5O₂ → 3CO₂ + 4H₂O): O₂ limits (2.000 ÷ 5 = 0.400), giving 3 × 0.400 = 1.20 mol CO₂ (52.8 g) and 4 × 0.400 = 1.60 mol H₂O (28.8 g).

14.7 On a scientific calculator

On a Casio fx-991EX or fx-82MS, type the whole chain at once: 2.00 ÷ 138.121 × 180.159 = gives 2.6087. Then 2.10 ÷ Ans × 100 = gives 80.50 for the percent yield.

Keep the unrounded theoretical yield (Ans) for the percent yield. Key names differ between models; check your manual.

14.8 Do it in the lab: synthesis of aspirin

Example data. A melting point a little below and broader than 135 °C points to a trace of salicylic acid or water; recrystallising would sharpen it but lower the yield further.

14.9 How to calculate theoretical yield: practice problems with answers

Ten practice problems, easy to hard. Open each one to see the answer.

14.10 Mistakes that lose marks

Research on stoichiometry finds that most errors come from the mole ratio and the limiting reagent, not the arithmetic (Gulacar et al., 2013; Mulford & Robinson, 2002).

• Using grams in the mole ratio instead of moles.
• Choosing the limiting reagent by the smaller mass.
• Inverting the ratio: multiplying by a ÷ b instead of b ÷ a.
• Forgetting the density for a liquid measured in mL.
• Calculating the yield from the excess reagent.
• Weighing a wet product and reporting more than 100%.
• Rounding the theoretical yield before working out the percent yield.

Four real examples

Decision tree

1. Do you have a balanced (or unbalanced) equation and amounts? Yes: equation mode. No: go on.
2. Do you know only the starting material, the product and their ratio? Yes: mole-ratio mode, with names such as salicylic acid.
3. Do you already have two of theoretical, actual and percent yield? Yes: percent yield mode.
4. Several steps? Add their yields to the multi-step box.

Limitations. A theoretical yield calculator gives the stoichiometric maximum, not a prediction of what you will isolate. It assumes the reaction goes to completion as written; for equilibria, stereoisomer mixtures or competing pathways the practical ceiling is lower. Atom economy and percent yield measure different things: a reaction can have a high yield and a poor atom economy (Sheldon, 2017).

What the answer is

Theoretical yield = moles of limiting reagent × mole ratio × molar mass of the product. For 2.00 g of salicylic acid it is 2.61 g of aspirin.

What it means chemically

It is the ceiling set by stoichiometry. Comparing it with what you weigh gives the percent yield, a measure of how well the reaction and work-up went.

Watch out for

Work in moles, find the limiting reagent by moles ÷ coefficient, use density for liquids, put the ratio the right way up, and dry the product before weighing.

Next steps

Practise with the worksheet in Section 12.3 and the ten problems in Section 14.9, then use the calculator for your next lab report.

Ratios are algebra, and comparing yields across repeat experiments is statistics; these tools cover both.

Peer-reviewed sources for the methods and data used in this theoretical yield calculator: standard atomic weights, atom economy and green chemistry metrics, and research on how students learn stoichiometry and limiting reagents.

1. Chandrasegaran, A. L., Treagust, D. F., Waldrip, B. G., & Chandrasegaran, A. (2009). Students’ dilemmas in reaction stoichiometry problem solving: Deducing the limiting reagent in chemical reactions. Chemistry Education Research and Practice, 10(1), 14–23. https://doi.org/10.1039/B901456J
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4. Furió, C., Azcona, R., & Guisasola, J. (2002). The learning and teaching of the concepts ‘amount of substance’ and ‘mole’: A review of the literature. Chemistry Education Research and Practice, 3(3), 277–292. https://doi.org/10.1039/B2RP90023H
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8. Mulford, D. R., & Robinson, W. R. (2002). An inventory for alternate conceptions among first-semester general chemistry students. Journal of Chemical Education, 79(6), 739. https://doi.org/10.1021/ed079p739
9. Nakhleh, M. B. (1992). Why some students don’t learn chemistry: Chemical misconceptions. Journal of Chemical Education, 69(3), 191–196. https://doi.org/10.1021/ed069p191
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