Concentration: Percent, Molarity, Dilution

Preparation for General Chemistry · Dr. Karmach

Dr. Karmach

By the end of this unit, you can…

  • Calculate molarity and percent concentration, use them as conversion factors, and dilute a stock solution with M₁V₁ = M₂V₂
Dr. Karmach

1 · Concentration: Percent, Molarity, Dilution

Compute a percent or molar concentration, use either one as a conversion factor between solution and solute, and dilute a stock with C₁V₁ = C₂V₂, including the water to add.

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Three blue solutions

Copper(II) sulfate dissolves in water as a blue solution. Beaker B holds the most salt, yet it is the palest. The color follows the salt in each milliliter, not the total.

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Concentration: solute compared to solution

concentration = amount of solute ÷ amount of solution
solute: the substance dissolved · solvent: what dissolves it, usually water · solution = solute + solvent
A: 4.0 g CuSO₄ in 100 mL · B: 6.0 g in 300 mL · C: 3.0 g in 50 mL
compare each on the same footing: grams per 100 mL of solution

Rank the three beakers. Which is most concentrated, and which is most dilute?

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Concentration: solute compared to solution

concentration = amount of solute ÷ amount of solution
solute: the substance dissolved · solvent: what dissolves it, usually water · solution = solute + solvent
A: 4.0 g CuSO₄ in 100 mL · B: 6.0 g in 300 mL · C: 3.0 g in 50 mL
compare each on the same footing: grams per 100 mL of solution

Rank the three beakers. Which is most concentrated, and which is most dilute?

A 4.0 g100 mL    B 6.0 g300 mL = 2.0 g100 mL    C 3.0 g50 mL = 6.0 g100 mL

C is most concentrated, with the least salt in total. B is most dilute.

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Four ways to state a concentration

Each one is solute over the whole solution. A percent counts parts of solute per 100 parts of solution. Molarity counts moles of solute in each liter of solution.

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A concentration is an equality

0.90% (m/v) NaCl: 0.90 g NaCl = 100 mL soln · 6.0 M HCl: 6.0 mol HCl = 1 L soln
a percent's whole is 100 g or 100 mL of solution · a molarity's whole is 1 L of solution

Every equality gives a conversion factor, either way up:

0.90 g NaCl100 mL solution or 100 mL solution0.90 g NaCl     6.0 mol HCl1 L solution or 1 L solution6.0 mol HCl

Write it so the given unit cancels.

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The method

  1. Name part and whole: solute; solution = solute + solvent.
  2. Match the units: % per 100 g or mL; M in mol per L.
  3. Write the fraction: part over whole, or a factor canceling the given.
  4. Multiply and check.
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Worked example 1: grams from a percent label

Step 1 · Name part and whole

vinegar: 5.00% (m/v) acetic acid
part: acetic acid · whole: solution · given: 500. mL of solution · wanted: g acetic acid

A 500. mL bottle of vinegar is 5.00% (m/v) acetic acid. How many grams of acetic acid does it hold?

Set it up so mL of solution cancels.

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Worked example 1: solution

vinegar: 5.00% (m/v) acetic acid
part: acetic acid · whole: solution · given: 500. mL of solution · wanted: g acetic acid

One conversion factor is needed.

Step 2 · Match the units

m/v pairs grams of solute with milliliters of solution, so the label declares 5.00 g acetic acid = 100 mL solution. The given is already in mL of solution.

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Worked example 1: solution

vinegar: 5.00% (m/v) acetic acid
part: acetic acid · whole: solution · given: 500. mL of solution · wanted: g acetic acid
Step 2 · Match the units Step 3 · Write the fraction

Two orientations exist. Only one cancels mL solution:

5.00 g acetic acid100 mL solution cancels mL solution ✓    100 mL solution5.00 g acetic acid cancels nothing ✗
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Worked example 1: solution

vinegar: 5.00% (m/v) acetic acid
part: acetic acid · whole: solution · given: 500. mL of solution · wanted: g acetic acid
Step 2 · Match the units Step 3 · Write the fraction Step 4 · Multiply and check
500. mL solution × 5.00 g acetic acid100 mL solution = 25.0 g acetic acid
5.00 g in every 100 mL, and the bottle holds five hundreds: 25.0 g ✓
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Worked example 1: the route on the map

vinegar: 5.00% (m/v) acetic acid
given: 500. mL of solution · found: 25.0 g acetic acid

The given is already the label's whole, mL of solution. One move: the label factor carries solution to solute. ✓
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Worked example 2: molarity from grams

Step 1 · Name part and whole

M = mol solute ÷ L solution
part: 34.8 g KNO₃ (101.11 g/mol) · whole: 225 mL of solution · wanted: M

What is the molarity of 225 mL of a potassium nitrate solution that contains 34.8 g of KNO₃?

Count the moves: grams to moles, mL to L, then divide.

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Worked example 2: solution

M = mol solute ÷ L solution
part: 34.8 g KNO₃ (101.11 g/mol) · whole: 225 mL of solution · wanted: M

Two conversions come first, then one division.

Step 2 · Match the units

34.8 g KNO₃ × 1 mol KNO₃101.11 g KNO₃ = 0.344 mol KNO₃    225 mL × 1 L1000 mL = 0.225 L
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Worked example 2: solution

M = mol solute ÷ L solution
part: 34.8 g KNO₃ (101.11 g/mol) · whole: 225 mL of solution · wanted: M
Step 2 · Match the units
34.8 g KNO₃ × 1 mol KNO₃101.11 g KNO₃ = 0.344 mol KNO₃    225 mL × 1 L1000 mL = 0.225 L
Step 3 · Write the fraction

Part over whole: moles on top, liters of solution on the bottom.

M = 0.344 mol KNO₃0.225 L soln = 1.53 M KNO₃
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Worked example 2: solution

M = mol solute ÷ L solution
part: 34.8 g KNO₃ (101.11 g/mol) · whole: 225 mL of solution · wanted: M
Step 2 · Match the units
34.8 g KNO₃ × 1 mol KNO₃101.11 g KNO₃ = 0.344 mol KNO₃    225 mL × 1 L1000 mL = 0.225 L
Step 3 · Write the fraction
M = 0.344 mol KNO₃0.225 L soln = 1.53 M KNO₃
Step 4 · Multiply and check. A bit under a quarter liter holds 0.344 mol, so a liter holds over four times that, over 1.38 mol: 1.53 M ✓
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Worked example 2: the route on the map

M = mol solute ÷ L solution
given: 34.8 g KNO₃ · 225 mL of solution · found: 1.53 M KNO₃

Molar mass to moles, 225 mL to 0.225 L, then divide. The label 1.53 M KNO₃ now states 1.53 mol KNO₃ = 1 L of solution. ✓
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Your turn: moles from a molarity

0.400 mol NaCl = 1 L of solution
the label "0.400 M NaCl" states this equality · given: 275 mL of solution · wanted: mol NaCl

How many moles of NaCl are in 275 mL of 0.400 M NaCl?

275 mL soln × 1 L soln mL soln × mol NaCl L soln = mol NaCl

Fill each factor so the unit before it cancels, then compute.

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Your turn: moles from a molarity

0.400 mol NaCl = 1 L of solution
the label "0.400 M NaCl" states this equality · given: 275 mL of solution · wanted: mol NaCl

How many moles of NaCl are in 275 mL of 0.400 M NaCl?

275 mL soln × 1 L soln mL soln × mol NaCl L soln = mol NaCl

Fill each factor so the unit before it cancels, then compute.

275 mL soln × 1 L soln1000 mL soln × 0.400 mol NaCl1 L soln = 0.110 mol NaCl
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Adding water keeps the solute

Adding water spreads the solute through more liquid. No solute is added or removed, so concentration × volume stays fixed: C₁V₁ = C₂V₂ (M₁V₁ = M₂V₂ for molarity; the same for percent).

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The dilution method

  1. List the knowns: three of C₁, V₁, C₂, V₂; mark the unknown.
  2. Rearrange for the unknown.
  3. Match the units: C₁ with C₂, V₁ with V₂.
  4. Substitute and solve; if asked, water added = V₂ − V₁.

Dr. Karmach

Worked example 3: a percent dilution

C₁V₁ = C₂V₂
given: 9.00% (m/v) NaOH · 10.0 mL stock · diluted to 60.0 mL · wanted: % (m/v) after

What is the percent (m/v) of a solution prepared by diluting 10.0 mL of 9.00% (m/v) NaOH to 60.0 mL?

The stock is "before" (1). The diluted solution is "after" (2).

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Worked example 3: solution

C₁V₁ = C₂V₂
given: 9.00% (m/v) NaOH · 10.0 mL stock · diluted to 60.0 mL · wanted: C₂ in % (m/v)

Step 1 · List the knowns

C₁ = 9.00% (m/v). V₁ = 10.0 mL. V₂ = 60.0 mL, the whole diluted solution. The unknown is C₂.

Dr. Karmach

Worked example 3: solution

C₁V₁ = C₂V₂
given: 9.00% (m/v) NaOH · 10.0 mL stock · diluted to 60.0 mL · wanted: C₂ in % (m/v)
Step 1 · List the knowns Step 2 · Rearrange Step 3 · Match the units
C₁V₁ = C₂V₂ → C₂ = C₁ × V₁V₂

% with %, mL with mL: the volume ratio is a pure number, and C₂ comes out in % (m/v).

Dr. Karmach

Worked example 3: solution

C₁V₁ = C₂V₂
given: 9.00% (m/v) NaOH · 10.0 mL stock · diluted to 60.0 mL · wanted: C₂ in % (m/v)
Step 1 · List the knowns Step 2 · Rearrange Step 3 · Match the units
C₁V₁ = C₂V₂ → C₂ = C₁ × V₁V₂
Step 4 · Substitute and solve
C₂ = 9.00% × 10.0 mL60.0 mL = 1.50% (m/v)
Grams check: 10.0 mL × 9.00 g/100 mL = 0.900 g NaOH, and 0.900 g in 60.0 mL is 1.50 g per 100 mL ✓
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Worked example 3: the route on the map

C₁V₁ = C₂V₂
given: 9.00% (m/v) · 10.0 mL stock · 60.0 mL final · found: C₂ = 1.50% (m/v)

The unknown is the new concentration, so C₂ = C₁ × V₁ ÷ V₂. Percent pairs with percent, mL with mL. ✓
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Worked example 4: water to add

M₁V₁ = M₂V₂
given: 750.0 mL of 3.00 M HCl · diluted to 2.50 M · wanted: mL of water added

How much water, in mL, must be added to 750.0 mL of 3.00 M HCl to obtain a solution that is 2.50 M?

The relation gives the total volume. The question asks for the water.

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Worked example 4: solution

M₁V₁ = M₂V₂
given: 750.0 mL of 3.00 M HCl · diluted to 2.50 M · wanted: mL of water added

Step 1 · List the knowns

M₁ = 3.00 M. V₁ = 750.0 mL. M₂ = 2.50 M. The unknown is V₂, and then the water.

Dr. Karmach

Worked example 4: solution

M₁V₁ = M₂V₂
given: 750.0 mL of 3.00 M HCl · diluted to 2.50 M · wanted: mL of water added
Step 1 · List the knowns Step 2 · Rearrange Step 3 · Match the units
M₁V₁ = M₂V₂ → V₂ = V₁ × M₁M₂

M with M. V₁ is in mL, so V₂ comes out in mL.

Dr. Karmach

Worked example 4: solution

M₁V₁ = M₂V₂
given: 750.0 mL of 3.00 M HCl · diluted to 2.50 M · wanted: mL of water added
Step 1 · List the knowns Step 2 · Rearrange Step 3 · Match the units
M₁V₁ = M₂V₂ → V₂ = V₁ × M₁M₂
Step 4 · Substitute and solve
V₂ = 750.0 mL × 3.00 M2.50 M = 900. mL total → water = 900. − 750.0 = 150. mL
3.00 M down to 2.50 M is a 1.2-fold drop, so the volume grows 1.2-fold: 750.0 × 1.2 = 900. mL. Water supplies the extra 150. mL ✓
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Worked example 4: the route on the map

M₁V₁ = M₂V₂
given: 750.0 mL of 3.00 M · diluted to 2.50 M · found: 900. mL total · 150. mL of water

V₂ = V₁ × M₁ ÷ M₂ gives the total, 900. mL. The question asked for the water, so the last step subtracts the 750.0 mL of stock. ✓
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Where this goes wrong

Dividing by the solvent alone. 12.0 g of NaCl in 138.0 g of water: 12.0 ÷ 138.0 × 100 = 8.70%. The salt is part of the solution: 12.0 ÷ 150.0 × 100 = 8.00%. Molarity also divides by the solution, never by the water poured in.
Dividing by milliliters. 0.344 mol KNO₃ ÷ 225 mL = 0.00153, in mol/mL, 1000 times too small. Molarity is per liter: 0.344 ÷ 0.225 L = 1.53 M.
Flipping the dilution ratio. 750.0 mL × (2.50 ÷ 3.00) = 625 mL, less liquid than the start. Diluting grows the volume: 750.0 × (3.00 ÷ 2.50) = 900. mL.
Reporting the total as the water. 900. mL is the whole diluted solution. The water added is 900. − 750.0 = 150. mL.
Dr. Karmach

Practice 1

pickling brine: 6.00% (m/m) NaCl
given: 250. g of brine · wanted: g NaCl

A 250. g jar of pickling brine is 6.00% (m/m) NaCl. How many grams of NaCl does it hold?

  1. 14.2
  2. 1.50 × 10³
  3. 15.0
  4. 235
  5. 4.17 × 10³
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Practice 1: answer C

6.00% (m/m): 6.00 g NaCl = 100 g brine
given: 250. g of brine, the whole solution · wanted: g NaCl
250. g brine × 6.00 g NaCl100 g brine = 15.0 g NaCl, answer C

A read the label per 100 g of water, as if 106 g of brine held 6.00 g: 250. × 6.00 ÷ 106 = 14.2. B used the raw percent: 250. × 6.00 = 1.50 × 10³. D found the water: 250. − 15.0 = 235. E flipped the factor: 250. × 100 ÷ 6.00 = 4.17 × 10³.

6.00 g in every 100 g of brine, and the jar holds two and a half hundreds: 15.0 g ✓
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Practice 1: the route on the map

6.00% (m/m) NaCl brine
given: 250. g of brine · found: 15.0 g NaCl

The given is the whole, grams of brine. One move: the label factor carries solution to solute. ✓
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Practice 2

0.200 M KOH
given: 350. mL of solution · molar mass KOH 56.11 g/mol

How many grams of KOH are dissolved in 350. mL of 0.200 M KOH?

  1. 3.93
  2. 98.2
  3. 3.93 × 10³
  4. 11.2
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Practice 2: answer A

0.200 mol KOH = 1 L of solution
given: 350. mL of solution · KOH 56.11 g/mol · route: mL → L → mol → g
350. mL soln × 1 L soln1000 mL soln × 0.200 mol KOH1 L soln × 56.11 g KOH1 mol KOH = 3.93 g, answer A

B flipped the molarity: 0.350 ÷ 0.200 × 56.11 = 98.2, and the units do not cancel. C skipped mL → L: 350. × 0.200 × 56.11 = 3.93 × 10³. D used one full liter: 0.200 × 56.11 = 11.2 g, the mass in 1 L.

0.350 L of 0.200 M holds 0.0700 mol, at about 56 g per mole: about 3.9 g ✓
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Practice 2: the route on the map

0.200 mol KOH = 1 L of solution
given: 350. mL of solution · found: 3.93 g KOH

Milliliters become liters first. Then the molarity reaches moles, and the molar mass reaches grams. ✓
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Practice 3

M₁V₁ = M₂V₂
start: 200. mL of 2.40 M NaCl · target: 0.600 M

A 200. mL portion of 2.40 M NaCl is diluted with water to 0.600 M. What volume of water, in mL, goes in?

  1. 800.
  2. 50.0
  3. 480.
  4. 600.
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Practice 3: answer D

M₁V₁ = M₂V₂
given: 200. mL of 2.40 M NaCl · diluted to 0.600 M · wanted: mL of water added
V₂ = 200. mL × 2.40 M0.600 M = 800. mL total → water = 800. − 200. = 600. mL, answer D

A stopped at the total: 800. mL still counts the 200. mL of stock. B flipped the ratio: 200. × (0.600 ÷ 2.40) = 50.0 mL, less than the start. C stopped at the solute: 200. × 2.40 = 480. mmol of NaCl, not a volume.

2.40 M down to 0.600 M is a four-fold drop, so the solution grows to four times 200. mL. Water supplies 600. mL of it ✓
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Practice 3: the route on the map

M₁V₁ = M₂V₂
given: 200. mL of 2.40 M · diluted to 0.600 M · found: 600. mL of water

V₂ = V₁ × M₁ ÷ M₂ gives the total, 800. mL. The water is the total minus the stock. ✓
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Practice 4

M = mol solute ÷ L solution
measured: 46.0 g glucose · 220.0 mL of water · 250.0 mL of solution

A technician stirs 46.0 g of glucose (C₆H₁₂O₆) into 220.0 mL of water. The solution that forms measures 250.0 mL. What is its molarity?

  1. 1.16
  2. 1.02
  3. 0.255
  4. 15.7
  5. 1.02 × 10⁻³
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Practice 4: answer B

M = mol solute ÷ L solution
C₆H₁₂O₆: 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol · whole: 250.0 mL = 0.2500 L of solution
46.0 g × 1 mol180.16 g = 0.2553 mol, then 0.2553 mol C₆H₁₂O₆0.2500 L soln = 1.02 M, answer B

A divided by the water poured in: 0.2553 ÷ 0.2200 = 1.16; the dissolved glucose takes up room too. C stopped at moles: 0.255 mol. D flipped the molar mass: 180.16 ÷ 46.0 ÷ 0.2500 = 15.7. E divided by milliliters: 0.2553 ÷ 250.0 = 1.02 × 10⁻³, in mol/mL.

A quarter liter carries about a quarter mole, so a full liter carries about one mole: 1.02 M ✓
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Practice 4: the route on the map

M = mol solute ÷ L solution
given: 46.0 g glucose · 250.0 mL of solution · found: 1.02 M

The same three moves as worked example 2. The whole is 250.0 mL of solution, not 220.0 mL of water. ✓
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Check yourself

  1. A bottle reads 12.0% (m/v) sucrose. State the equality the label declares, then write the factor that turns milliliters of the solution into grams of sucrose.
  2. 50.0 mL of a stock is diluted with water to 200. mL. By what factor does the concentration drop, and what stays the same?

Concentration returns with reactions in solution: molarity × liters gives moles of a reactant, and the mole ratio carries it from there.

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Can you…?

  • ☐ calculate molarity and percent concentration, use them as conversion factors, and dilute a stock solution with M₁V₁ = M₂V₂?

If any box stays empty, the practice site has a drill for it. 🧪

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