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General Chemistry I · Measurement and Math

Solution Stoichiometry

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On this page 8 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Quick check
  8. Study tools

In 30 seconds

is moles of solute per liter of solution. lowers by adding solvent while moles stay constant, giving . In solution reactions, molarity times volume gives moles for stoichiometry. finds an 's amount by precipitating, filtering, and weighing it; finds concentration by reacting an analyte with a of known molarity to the .

Why this matters

Molarity and dilution are daily clinical realities. IV medications and stock drugs are labeled in concentration units, and nurses and pharmacists dilute stock solutions to safe strengths — the same M1V1 = M2V2 math. Clinical labs measure blood glucose, electrolytes, and drug levels with standardized solutions. Getting a dilution wrong is a patient-safety issue, so the arithmetic is double-checked.

The college version

1. Molarity (Concentration)

Molarity is the most common lab unit of concentration: M = moles of soluteliters of solution Note the denominator is liters of solution, not solvent. To make a 1.00 M solution, dissolve the solute in some water and dilute to the final volume in a volumetric flask. Molarity is also a conversion factor: L × M = mol.

2. Dilution

Dilution adds solvent, leaving moles of solute unchanged: M1 V1 = M2 V2 Here M1 and V1 are the initial (stock) molarity and volume, and M2 and V2 are the final values. Any volume unit works as long as it is the same on both sides, because the equation is really "moles before = moles after" (M × V = mol).

3. Gravimetric Analysis and Titrations

Gravimetric analysis finds an unknown amount by mass: precipitate the analyte as a known insoluble compound, filter, dry, and weigh it, then convert that mass to moles and back to the analyte's amount. A titration measures concentration by reacting a known volume of analyte with a titrant of known molarity. The equivalence point is where moles of titrant exactly match moles of analyte; the is where an indicator changes color, just past the equivalence point.

How it works

  1. Calculate molarity from moles and liters, or moles from molarity and volume.
  2. For dilution, use M1V1 = M2V2 to find the stock volume, then add solvent to the final volume.
  3. For solution reactions, convert molarity and volume to moles, then use the balanced equation's mole ratio.
  4. For gravimetric analysis, precipitate, filter, dry, and weigh, then convert mass → moles → analyte amount.
  5. For titrations, add titrant until the endpoint; at the equivalence point, moles of titrant and analyte match the mole ratio.

Common confusions

Do not confuseWithDifference
MolarityMolalityMolarity is mol/L of solution; molality is mol/kg of solvent
Equivalence pointEndpointEquivalence is the stoichiometric point; endpoint is the color change
DilutionReaction stoichiometryDilution conserves moles of one solute; stoichiometry converts between substances
Solvent volumeSolution volumeMolarity uses total solution volume, including the solute
ConcentratedStrongConcentrated = amount of solute; strong = degree of ionization

Memory aids

"Moles stay, water plays" — in any dilution the moles of solute are conserved while only the water (solvent) increases. MiVi = MfVf: you don't lose what you started with.

Quick review

Topic Recap

Molarity is moles per liter of solution and is the bridge between solution volume and moles. Dilution conserves moles and follows M1V1 = M2V2. In solution stoichiometry, convert concentration and volume to moles, apply the balanced equation's mole ratio, then convert onward. Gravimetric analysis measures an analyte by the mass of a precipitate, and titrations measure concentration at the equivalence point.

Knowledge Check

  1. How many moles are in 500.0 mL of 0.250 M NaCl?
  2. What volume of 2.00 M stock is needed to make 100.0 mL of 0.500 M solution?
  3. Why must the denominator of molarity be "liters of solution" and not "liters of water"?
  4. A titration uses 22.50 mL of 0.100 M NaOH to neutralize 25.00 mL of HNO3. What is the HNO3 molarity?
  5. In gravimetric analysis, why must the precipitate be dry before weighing?

Answers and Rationales

  1. 0.125 mol — 0.250 M × 0.5000 L = 0.125 mol.
  2. 25.0 mL — V1 = (0.500 × 100.0) / 2.00 = 25.0 mL.
  3. Because dissolving a solute changes the total volume; molarity must reflect the final volume of the mixture.
  4. 0.0900 M — moles NaOH = 0.100 × 0.02250 = 0.002250 mol; 1:1 with HNO3, so M = 0.002250 / 0.02500 = 0.0900 M.
  5. Because water of hydration would add mass and inflate the result; only the dry precipitate should count.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Molarity is like a recipe's strength. Imagine you always make Kool-Aid in the same pitcher: one scoop per pitcher is weak, five scoops is strong. The "scoops" are moles and the "pitcher" is the liter, so molarity says how many moles are packed into each liter. Dilution is adding more water to the same pitcher — the scoops stay the same, but now they are spread through more liters, so each glass is weaker.

It stops being exact in one way: molarity counts per liter of total solution, not per liter of water added. Dump one scoop into one liter of water and the final volume is a touch more than a liter, so the true molarity is a hair lower. Precise work fills to the mark in a volumetric flask rather than adding a full liter of water to the solute.

Simple Example

Dissolve 0.50 mol of NaCl in enough water to make 1.0 L of solution: M = 0.50 mol1.0 L = 0.50 M

Worked example

Worked example 1 — molarity: How many grams of NaOH are needed to make 250.0 mL of 0.400 M NaOH (NaOH = 40.00 g/mol)? moles = M × V = 0.400 molL × 0.2500 L = 0.100 mol mass = 0.100 mol × 40.00 gmol = 4.00 g Setup error: using 250 mL instead of 0.2500 L — convert to liters first.

Worked example 2 — dilution: How many mL of 6.00 M HCl are needed to make 500.0 mL of 0.150 M HCl? V1 = M2 V2M1 = 0.150 M × 500.0 mL6.00 M = 12.5 mL Measure 12.5 mL of the stock and dilute to 500.0 mL. Setup error: inverting the ratio (gives 20,000 mL).

Worked example 3 — titration: 25.00 mL of HCl requires 34.80 mL of 0.1000 M NaOH to reach the equivalence point. Find [HCl]. moles NaOH = 0.1000 molL × 0.03480 L = 0.003480 mol HCl and NaOH react 1:1, so moles HCl = 0.003480 mol. MHCl = 0.003480 mol0.02500 L = 0.1392 M Setup error: forgetting the mole ratio — check the balanced equation first.

Worked example 4 — gravimetric analysis: Excess AgNO3 is added to a sample containing Cl−, and 1.432 g of AgCl is collected (AgCl = 143.32 g/mol). Find the mass of Cl− (35.45 g/mol). mol AgCl = 1.432 g143.32 g/mol = 0.009991 mol Each AgCl contains one Cl−, so mol Cl− = 0.009991 mol. mass Cl- = 0.009991 mol × 35.45 g/mol = 0.3542 g

Key takeaways

  • High yield: Molarity is per liter of solution, not per liter of solvent.
  • High yield: M1V1 = M2V2 works because moles of solute are conserved during dilution.
  • High yield: Molarity × volume = moles; this is the bridge from solution to stoichiometry.
  • High yield: The equivalence point is a stoichiometric concept; the endpoint is an observed color change.
  • Convert mL to L before molarity calculations.
  • Check the balanced equation's mole ratio before converting titrant to analyte moles.
  • Gravimetric analysis requires the precipitate to be filtered, dried, and weighed.

Quick check

1 question here. Answers stay hidden until you check.

Question 1 of 1

What mass of AgCl (M = 143.32 g/mol) precipitates when 50.0 mL of 0.200 M AgNO3 is added to excess NaCl?

Choose an answer, then check it.

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Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Define molarity and calculate it from the mass of solute and volume of solution.
  • Use the dilution equation M1V1 = M2V2 to prepare solutions of a desired concentration.
  • Use molarity as a conversion factor in stoichiometric calculations for solution reactions.
  • Solve gravimetric analysis and acid-base titration problems.

Key vocabulary

Molarity (M)
Moles of solute per liter of solution
Concentration
Amount of solute in a given amount of solvent/solution
Dilution
Adding solvent to lower concentration
M1V1 = M2V2
Dilution equation (moles conserved)
Gravimetric analysis
Measuring an analyte via the mass of a precipitate
Titration
Adding a known titrant to react with an analyte
Titrant
Solution of known molarity added from a buret
Analyte
The substance whose amount is unknown
Equivalence point
Point where reactants are in exact stoichiometric ratio
Endpoint
Point where the indicator changes color

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