Molarity, dilutions and C1V1 = C2V2 for preparing lab solutions

To make a solution of a known molarity, weigh out mass = M × V × molar mass, where V is the final volume in liters, dissolve the solid, and make the solution up to that volume. To make a more dilute solution from a stock of known concentration, use C1 × V1 = C2 × V2, where V2 is the final volume. For 500 mL of 0.150 M sodium chloride, the mass is 4.383 g. For the same volume from a 2.00 M stock, you take 37.5 mL of stock and add water to 500 mL. Two details change the mass you weigh: the purity of the reagent and whether the compound is a hydrate.

This article explains what molarity measures, how to convert it to a mass, how hydrates and purity change that mass, how C1V1 = C2V2 works with the fill-to-volume step, how serial dilutions compound, and how to convert molarity to the other units you will see on labels. It ends with the formulas to use in a spreadsheet.

What is molarity, and why does the solution volume matter?

Molarity (M) is the amount of solute in moles divided by the volume of the solution in liters:

M = moles of solute / liters of solution

The denominator is the total volume of the solution, not the volume of water you started with. Dissolving 58.44 g of sodium chloride in 1 L of water does not give exactly 1 M, because the final volume is not exactly 1 L. Make the solution up to its final volume in a volumetric flask, which is calibrated for that purpose, and the molarity is then correct.

Moles come from mass and molar mass: moles = mass / molar mass. Molar mass is the sum of atomic masses in the formula, in grams per mole.

How do you calculate the mass to weigh for a molar solution?

Rearranging the definition gives the mass needed for a given volume:

mass (g) = M × V (L) × molar mass (g/mol)

Worked example: 500 mL of 0.150 M sodium chloride (NaCl).

• Volume in liters: 500 mL = 0.500 L
• Moles: 0.150 × 0.500 = 0.0750 mol
• Molar mass: Na 22.990 + Cl 35.45 = 58.44 g/mol
• Mass: 0.0750 × 58.44 = 4.383 g

Weigh 4.383 g on a balance that reads to 0.001 g, dissolve it in part of the water, and make the solution up to 500 mL.

Purity

A reagent labeled 99.0% pure contains 99.0% of its stated compound by mass. To get the same number of moles from it, weigh more:

mass = (M × V × molar mass) / purity

For the same 4.383 g of pure NaCl from a 99.0% reagent, weigh 4.383 / 0.990 = 4.427 g. Use the purity on the certificate of analysis for your lot, not a typical value.

Hydrates

Some salts carry water of crystallization in their formula. Copper(II) sulfate pentahydrate is CuSO4·5H2O, and its molar mass includes five water molecules. Using the anhydrous molar mass gives a different mass and a different concentration.

Atomic masses: Cu 63.546, S 32.06, O 15.999, H 1.008.

• Anhydrous CuSO4: 63.546 + 32.06 + 4 × 15.999 = 159.60 g/mol
• Water: 2 × 1.008 + 15.999 = 18.015 g/mol, so five waters are 90.075 g/mol
• Pentahydrate CuSO4·5H2O: 159.60 + 90.075 = 249.68 g/mol

To make 250 mL of 0.100 M copper sulfate pentahydrate, the moles are 0.100 × 0.250 = 0.0250 mol, and the mass is 0.0250 × 249.68 = 6.242 g. If you weigh 3.990 g instead, which is the anhydrous mass for the same moles, you have 3.990 / 249.68 = 0.0160 mol of the hydrate, and the solution is 0.064 M. Check the formula printed on the bottle before taking the molar mass from a table.

How do you use C1V1 = C2V2 to dilute a stock solution?

When you dilute a stock solution, the number of moles stays the same. Only the volume and concentration change:

C1 × V1 = C2 × V2

• C1: concentration of the stock
• V1: volume of stock to take
• C2: target concentration
• V2: final volume

The units must match on each side. If C is in molar, V can be in milliliters or liters, as long as both volumes use the same unit.

Worked example: make 500 mL of 0.150 M NaCl from a 2.00 M stock.

• V1 = (C2 × V2) / C1 = (0.150 × 500) / 2.00 = 37.5 mL
• Diluent: V2 - V1 = 500 - 37.5 = 462.5 mL

The fill-to-volume step

V2 is the final volume of the solution, not the volume of water to add. Pipette 37.5 mL of stock into a 500 mL volumetric flask, add water until the bottom of the meniscus sits on the mark, and mix. Adding 500 mL of water to the stock would give more than 500 mL, and the concentration would be wrong. The 462.5 mL figure is a guide for how much water to expect, not the amount to measure out.

How do serial dilutions lower a concentration at each step?

A serial dilution repeats the same step, so the concentration falls by the same factor each time. A common scheme transfers 10 mL into 90 mL of diluent, a factor of 10 at each step. Starting from a 1.000 M stock:

Each row is the previous row divided by 10. The cumulative dilution is the product of the step factors, so five 1:10 steps give 1:100,000. Serial dilutions reach low concentrations with few measured volumes. The cost is that any error at an early step carries through every later step.

How do you convert molarity to g/L, % w/v and mg/L?

Labels and methods use several units. For a solute with molar mass MM, a molarity M converts as follows:

• Grams per liter: g/L = M × MM
• Percent weight per volume (% w/v), grams per 100 mL: % w/v = g/L / 10
• Milligrams per liter: mg/L = g/L × 1,000

For 0.150 M NaCl, the figures are 0.150 × 58.44 = 8.766 g/L, which is 0.8766% w/v, or 8,766 mg/L.

Parts per million (ppm) is mass per mass. For a dilute aqueous solution, mg/L is treated as ppm because the density of water is close to 1 kg/L, so a liter of the solution weighs about a kilogram. That is an approximation. The exact ppm figure uses mg per kilogram of solution, and the two differ by the solution's density. As an illustration with an assumed density of 1.10 kg/L, which is not a property of this solution, 8,766 mg/L would be 8,766 / 1.10 = 7,969 mg/kg, about 9.1% lower. For dilute solutions the gap is small. For concentrated ones, use the density.

What practical points matter when making up a lab solution?

• Use volumetric flasks and calibrated pipettes for solutions where the concentration matters. A graduated cylinder is adequate for rough work, but its tolerance is much wider.
• Dissolve the solid in part of the solvent, then make up to the final volume. This keeps the final volume exact.
• When diluting a concentrated acid, add the acid to the water slowly, never the water to the acid. Follow the safety data sheet for the specific chemical.
• Record the concentration, the date, the lot number of the reagent and its purity or hydration state on the container.

Which spreadsheet formulas calculate the mass and dilution volumes?

Put the molarity in B1, the volume in milliliters in B2, the molar mass in B3 and the purity as a fraction in B4, with 1 for pure material. Then:

• Mass to weigh: =B1*B2/1000*B3/B4. With the example values, this gives 4.383 g at a purity of 1.
• Molar mass from the formula: =22.990+35.45 for NaCl, or =63.546+32.06+4*15.999+5*(2*1.008+15.999) for CuSO4·5H2O.
• Volume of stock: put the target concentration in B5, the final volume in B6 and the stock concentration in B7, then =B5*B6/B7. Use the same unit for both volumes.
• Diluent volume: =V2-V1.
• Serial dilution step: =previous/10 in each row, or =previous*0.1.
• Conversions: =M*MM for g/L, =M*MM/10 for % w/v, and =M*MM*1000 for mg/L.

The solution and dilution calculator carries these formulas for stock and final volumes, and the molar mass calculator builds the molar mass from the element counts in a formula. For the basic definition with a shorter worked example, see how to calculate molarity.