Fractions, Ratios, and Stoichiometry in Chemistry
TL;DR: Stoichiometry is proportional reasoning with chemical formulas. Turn coefficients into conversion fractions, keep everything in moles, and the rest is arithmetic. If you’d rather skip the grind, we do chemistry homework—private, fast, A/B Guarantee.
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Table of Contents
- What Is Stoichiometry?
- Fractions and Ratios Refresher
- Stoichiometry with Mole Ratios
- Stoichiometry Steps (Template)
- Limiting Reactant (Worked Example)
- Gas Stoichiometry (Mini Example)
- Percent Yield (Worked Example)
- Common Mistakes (and Fixes)
- Practice Set (with Answers)
- Platform Notes: ALEKS • WebAssign • MyLab
- How FMMC Helps
- FAQs
- Next Reads (Internal Links)
1) What Is Stoichiometry?
Stoichiometry uses balanced chemical equations to relate amounts of reactants and products via mole ratios. It powers lab prep, limiting-reactant calls, and yield calculations.
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2) Fractions and Ratios Refresher
Coefficients are ratios. For 2 H2 + O2 → 2 H2O, the proportion is 2:1:2 in moles. Use fractions that equal 1 to convert between species:
| From | To | Conversion Fraction |
|---|---|---|
| mol H2 | mol H2O | (2 mol H2O) / (2 mol H2) |
| mol O2 | mol H2O | (2 mol H2O) / (1 mol O2) |
| g → mol | mol → g | (1 mol / M) ↔ (M g / 1 mol), where M is molar mass |
NA=6.022×1023 mol−1 • 1 L = 1000 mL •
Molar volume (STP)=22.414 L·mol−1 • 1 atm = 760 mmHg = 101.325 kPa
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3) Stoichiometry with Mole Ratios
Workflow: mass → moles → stoichiometric conversion (mole ratio) → moles → mass. Keep units visible; cancel as you go.
Need platform-specific help? See Dimensional Analysis in Chemistry and our dedicated ALEKS Stoichiometry Answers page.
- Balance the equation.
- Convert given amounts to moles.
- Use the mole ratio (coefficients) to convert between species.
- Convert moles to the requested unit (grams, liters of gas, volume from molarity, particles via NA).
- Check units & sig figs; round once at the end.
4) Limiting Reactant (Worked Example)
Problem: For 2 Al + 3 Cl2 → 2 AlCl3, if you react 10.0 g Al with 20.0 g Cl2, which reactant limits and how many grams of AlCl3 can you form?
- Convert to moles: M(Al)=26.98 g/mol; M(Cl2)=70.90 g/mol.
n(Al)=10.0/26.98=0.371 mol; n(Cl2)=20.0/70.90=0.282 mol. - Compare “needed” at stoichiometric ratio: From 0.371 mol Al, Cl2 required = (3/2)×0.371=0.557 mol (>0.282 available) → Cl2 is limiting.
- Use limiting reactant to find product: n(AlCl3) = (2/3)×n(Cl2) = (2/3)×0.282 = 0.188 mol.
M(AlCl3)=133.34 g/mol → mass = 0.188×133.34 ≈ 25.1 g AlCl3 (3 SF).
What you learn: Always identify the limiting reactant using moles and the mole ratio, not by eyeballing grams.
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Gas Stoichiometry (Mini Example)
At STP, how many liters of O2 form when 12.0 g KClO3 decomposes?
Reaction: 2 KClO3 → 2 KCl + 3 O2
- n(KClO3) = 12.0 g / 122.55 g·mol−1 = 0.0980 mol
- n(O2) = (3/2) × 0.0980 = 0.147 mol
- V(O2) = 0.147 mol × 22.414 L·mol−1 = 3.29 L (3 SF)
What you learn: For gases at STP, convert to moles, apply the mole ratio, then multiply by the molar volume (22.414 L·mol−1).
5) Percent Yield (Worked Example)
Problem: Theoretical yield from the previous setup is 25.1 g AlCl3. If the actual mass collected is 22.8 g, what is the percent yield?
% yield = (actual / theoretical)×100 = (22.8 / 25.1)×100 = 90.8% (3 SF).
What you learn: Percent yield compares reality to the ideal; keep sig figs consistent with the data provided.
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6) Common Mistakes (and Fixes)
- Using grams in ratios: Convert to moles before applying coefficients.
- Skipping the balance: Unbalanced equations make wrong ratios. Balance first.
- Early rounding: Carry extra digits; round at the end per instructions/sig figs.
- Unit drift: Keep units on every line; cancel visibly.
- Wrong limiting call: Compare required vs available via mole ratios, not smaller grams.
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7) Practice Set (with Answers)
- Mass–Mass: For
2 KClO3 → 2 KCl + 3 O2, how many grams of O2 from 24.5 g KClO3? (M: KClO3=122.55 g/mol; O2=32.00 g/mol) - Limiting: For
N2 + 3 H2 → 2 NH3, start with 8.00 g N2 and 1.50 g H2. Identify the limiting reactant and mass of NH3 formed. (M: N2=28.02, H2=2.016, NH3=17.031) - % Yield: For
C + O2 → CO2, theoretical 11.0 g CO2, actual 9.2 g. Find percent yield. - Solution Stoichiometry: How many liters of 0.500 M HCl are required to neutralize 25.0 g CaCO3? Reaction:
2 HCl + CaCO3 → CaCl2 + CO2 + H2O. (M: CaCO3=100.09 g/mol)
Show Answers
- n(KClO3)=24.5/122.55=0.200 mol → n(O2)=(3/2)×0.200=0.300 mol → m=0.300×32.00=9.60 g O2 (3 SF).
What you learn: Mass–mass needs g→mol → ratio → mol→g. - n(N2)=8.00/28.02=0.2857 mol; n(H2)=1.50/2.016=0.744 mol. Needed H2 for 0.2857 mol N2 is 0.857 mol → H2 limiting. n(NH3)=(2/3)×0.744=0.496 mol → m=0.496×17.031=8.45 g NH3 (3 SF).
What you learn: Identify limiting by comparing required vs available moles. - % yield = (9.2 / 11.0)×100 = 83.6%.
What you learn: Keep sig figs consistent with given data. - n(CaCO3)=25.0/100.09=0.2498 mol → n(HCl)=2×0.2498=0.4996 mol → V=n/M=0.4996/0.500=0.999 L (≈1.00 L).
What you learn: Solution stoichiometry converts via molarity after the mole ratio.
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8) Platform Notes: ALEKS • WebAssign • MyLab
- ALEKS: Balance first; follow precision requests; show units. See ALEKS Stoichiometry Answers.
- WebAssign: Often stricter on reduced fractions and sig figs. See Cengage WebAssign Answers.
- MyLab Chemistry: Stoichiometry commonly paired with molarity/dilution steps. See Pearson Mastering Chemistry Answers.
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9) How FMMC Helps
- Mass–mass, limiting/excess, percent yield, and solution stoichiometry—done right and on time.
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10) FAQs
What is stoichiometry in chemistry?
Proportional reasoning using a balanced equation to relate moles (and thus masses/volumes) of reactants and products.
Why do coefficients matter so much?
They’re the mole ratio. Every conversion between species uses coefficients as a fraction that equals 1.
How do I identify the limiting reactant?
Convert to moles, use the mole ratio to compute how much of the other reactant is required, and compare to what you have. The one that runs out first limits.
My percent yield is 0%—what happened?
Usually wrong limiting-reactant pick, units drift, or unbalanced reaction. Recheck: balance → moles → ratio → moles → mass.
Can you help on ALEKS/WebAssign/MyLab?
Yes—formatting and precision included. See ALEKS Stoichiometry Answers, WebAssign Help, and MyLab Chemistry Help.
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11) Next Reads (Internal Links)