Why blood tests report mmol/L
Outside the US, a blood test reports fasting glucose in mmol/L, with a reference range of roughly 3.9–6.1 mmol/L: millimoles of glucose per liter of blood. Why not use mass percent, as in 1.7.6? Chemical reactions happen in moles, but in the lab you measure liquids by volume with graduated cylinders and burets. If you know how many moles of solute are in each liter, any volume you measure tells you the moles it contains.
Molarity
The molarity物质的量浓度 of a solution is the number of moles of solute per liter of solution:
Its unit, mol/L, is written M and read “molar.” A 0.10 M NaCl solution has 0.10 mol of NaCl in every liter. Square brackets mean “molarity of”: [NaCl] = 0.10 M.
The volume is the volume of the solution, not of the solvent. Dissolving 1 mol of NaCl in 1 L of water does not give exactly 1 L of solution, so it isn’t exactly 1 M.
A solution is uniform (1.7.4), so any portion you pour out has the same molarity as the rest. Only the moles of solute in the portion change.
Ion concentrations
Strong electrolytes dissociate completely in water (2.7.5), so the dissociation equation gives each ion’s concentration:
In 0.10 M CaCl₂, [Ca²⁺] = 0.10 M and [Cl⁻] = 2 × 0.10 = 0.20 M. Ion concentrations depend only on the molarity and the formula, not on how much solution you have.
This lesson uses the atomic masses on the AP periodic table (H 1.008, C 12.01, O 16.00, Na 22.99, S 32.06, Cl 35.45, Ca 40.08, Cu 63.55).
Choose a solute and a flask, enter the mass you weigh out, and read the molarity and ion concentrations. Then switch to “Molarity → mass” to find how much to weigh:
Making a solution of known molarity
A volumetric flask容量瓶 has a long narrow neck with a single etched mark. Filled to the mark at the labeled temperature (usually 20 °C), it holds exactly its stated volume. Common sizes are 100, 250, 500, and 1000 mL. Don’t dissolve solids or store solutions in it.
Here is how to make 100.0 mL of 1.00 M NaCl. Put the steps in order:
Steps to place (click to add)
Your order (click a step to send it back)
- Calculate: mass of NaCl = 1.00 mol/L × 0.1000 L × 58.44 g/mol = 5.84 gmol NaCl = MV = 1.00 M × 0.1000 L = 0.100 mol, and mass = 0.100 mol × 58.44 g/mol = 5.84 g.
- Weigh out 5.84 g of NaCl on a balanceAn electronic balance reads to 0.01 g, so you can weigh 5.84 g directly.
- Dissolve the NaCl in a beaker with some distilled water, stirring, and let it return to room temperatureDissolving can release or absorb heat. The flask is calibrated at 20 °C, so let the solution return to room temperature first.
- Pour the solution into a 100 mL volumetric flask, using the stirring rod as a guideTouch the rod to the inside of the neck below the mark so no solution spills.
- Rinse the beaker and stirring rod 2–3 times with a little distilled water and add the rinsings to the flaskSolute clings to the beaker and rod; without rinsing, some of it never reaches the flask.
- Add water until the level is 1–2 cm below the mark, then add it drop by drop until the bottom of the meniscus touches the markSwitching to a dropper near the mark keeps you from overshooting it.
- Stopper the flask and invert it several times to mixAfter mixing, the level may sit slightly below the mark. That is normal, so don’t add more water.
- Transfer the solution to a storage bottle and label itVolumetric flasks aren’t for storage. The label should give the name and molarity.
Every mistake in this procedure changes either the moles of solute n or the volume V. Use M = n/V: decide which one the mistake changes, then whether the molarity ends up too high or too low:
The molarity of the solution will be:
When you fill to the mark, keep your eye level with it. Drag the slider to see where the meniscus stops when you look down or up, and what that does to the concentration:
RuleFilling a volumetric flask to the mark: looking down adds too little water, so the solution is too concentrated; looking up adds too much, so it’s too dilute.
WhyTo fill to the mark, you sight through the mark at the bottom of the meniscus. Looking down, that line keeps sloping down past the mark, so the meniscus seems to reach the mark before it really does, and you stop adding water too soon.
How Chinese textbooks differ
- Name and units: Chinese textbooks call molarity 物质的量浓度 (“amount-of-substance concentration”), with the symbol c(B), and always write the unit as mol/L, such as c(NaCl) = 0.10 mol/L. They never use M as a unit, because M is the symbol for molar mass. The AP equation sheet uses M for both: n = m/M for molar mass and M for molarity.
- Balances: Chinese school labs use a beam balance that reads to 0.1 g, so the example above would weigh 5.9 g. US labs use electronic balances that read to 0.01 g or better.
- Error analysis: Chinese problem sets drill the effect of every possible mistake in this procedure. US textbooks say less about it, but AP lab questions often ask whether a mistake makes the calculated molarity too high or too low, and the reasoning is exactly the same.
Example 1: copper(II) sulfate from its hydrate
How many grams of copper(II) sulfate pentahydrate, CuSO₄·5H₂O, do you need to make 500.0 mL of 0.200 M CuSO₄?
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The plan. Molarity and volume give moles of solute; then convert to the mass of what you actually weigh, the hydrate: M, V → mol → g.
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Moles of solute needed. Volume must be in liters: 500.0 mL = 0.5000 L.
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Switch to the hydrate. Each formula unit of CuSO₄·5H₂O contains one CuSO₄, so you need 0.100 mol of the hydrate.
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Molar mass of the hydrate. The water of hydration counts:
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Mass.
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A common mistake. Using the molar mass of anhydrous CuSO₄ (159.61 g/mol), you would weigh only 16.0 g. That much hydrate contains 16.0 ÷ 249.69 = 0.0641 mol CuSO₄, so the solution would be only 0.128 M, too dilute.
Example 2: ion concentrations
You dissolve 11.1 g of CaCl₂ in water to make 500.0 mL of solution. Find [CaCl₂] and [Cl⁻]. Then you pour out 50.0 mL. What are [Cl⁻] and the moles of Cl⁻ in that portion?
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Moles of CaCl₂. Molar mass = 40.08 + 2(35.45) = 110.98 g/mol:
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Molarity of the solute.
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[Cl⁻]. The dissociation equation shows that each CaCl₂ gives 2 Cl⁻:
so [Cl⁻] = 2 × 0.200 M = 0.400 M.
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The 50.0 mL portion. A solution is uniform, so pouring some out doesn’t change its molarity: [Cl⁻] is still 0.400 M. Moles are proportional to volume:
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A common mistake. Thinking that taking 1/10 of the solution makes the concentration 1/10 as large. The moles drop to 1/10; the concentration stays the same.
- 0.500 mol of NaCl in 500.0 mL of solution: [NaCl]
0.500 mol ÷ 0.5000 L = 1.00 M.
- 4.00 g of NaOH in 1.000 L of solution: [NaOH]
4.00 g ÷ 40.00 g/mol = 0.100 mol, in 1.000 L: 0.100 M.
- [Cl⁻] in 2.00 M CaCl₂
Each CaCl₂ gives 2 Cl⁻: 2 × 2.00 M = 4.00 M.
- [Na⁺] in 0.100 M Na₂SO₄
Each Na₂SO₄ gives 2 Na⁺: 0.200 M.
- 10.0 mL poured from a bottle of 1.00 M NaCl: the molarity of the 10.0 mL
A solution is uniform, so a portion has the same molarity: 1.00 M.
- 58.44 g of NaCl in 2.000 L of solution: [NaCl]
58.44 g ÷ 58.44 g/mol = 1.000 mol, and 1.000 mol ÷ 2.000 L = 0.500 M.
- [H⁺] in 1.00 M H₂SO₄, treating both H⁺ as fully ionized
Each H₂SO₄ gives 2 H⁺: 2.00 M. (In AP chemistry the second ionization is only partial, so the real value is a little lower.)
- 9.81 g of H₂SO₄ in 200.0 mL of solution: [H₂SO₄]
9.81 g ÷ 98.08 g/mol = 0.100 mol, and 0.100 mol ÷ 0.2000 L = 0.500 M.
- 0.400 mol of NaOH in 2.00 L of solution: [NaOH]
0.400 mol ÷ 2.00 L = 0.200 M.
Common mistakes
Mistake 1: “1 mol of NaCl in 1 L of water is 1 M.” Molarity uses liters of solution. Dissolve the NaCl, then add water to the mark of a 1000 mL volumetric flask.
Mistake 2: “Pouring out half the solution halves its molarity.” A solution is uniform, so the molarity stays the same. What halves is the moles of solute.
Mistake 3: “Molarity and molar mass are both M, so they’re the same kind of quantity.” Molar mass M is in g/mol; molarity M is in mol/L. Always write the units to keep them apart.
Quick check
Which of these solutions has the highest [Cl⁻]?
Ion concentrations don’t depend on volume. AlCl₃: 3 × 0.100 = 0.300 M; NaCl: 0.200 M; CaCl₂: 0.200 M; KCl: 0.250 M.
While making 100.0 mL of 1.00 M NaCl, a student looks down at the mark from above when filling the flask. The molarity of the solution will be:
Looking down, the meniscus seems to touch the mark before it gets there, so the volume is too small and M = n/V is too high.
Summary
- Molarity M = mol solute / L solution, in mol/L (M). The volume is that of the solution, and any portion has the same molarity.
- In a solution of a strong electrolyte, the formula sets each ion’s concentration: [Cl⁻] = 2[CaCl₂], whatever the volume.
- To make a solution: calculate, weigh, dissolve and cool, transfer, rinse, fill to the mark, and mix. For error analysis, use M = n/V and ask whether the mistake changed n or V.
On to AP (for later)
The AP equation sheet gives M = nsolute/Lsolution. AP topic 3.7 asks you to find molarity from a particle diagram, or to draw one: a diagram of CaCl₂ solution must show twice as many Cl⁻ ions as Ca²⁺ ions. Lab questions often ask whether an error makes a calculated molarity too high or too low, and why, which is exactly the analysis in this lesson.