Finding Solute Amount from Molarity

Introduction

Molarity by itself is a ratio; a measured solution volume turns it into an actual amount of solute. The product n = cV is the bridge between pipette readings and balanced reaction equations. It works only when the concentration and volume describe the same solution portion.

Core explanation

If c is in mol L⁻¹ and V is in L, cV has units mol. A 25.0 mL aliquot of 0.200 M NaOH has V = 0.0250 L and n = 0.200 × 0.0250 = 0.00500 mol NaOH formula amount. The same concentration in a 100.0 mL aliquot contains four times as many moles because its volume is four times as large. The concentration does not change merely because an aliquot is removed from a well-mixed stock.

The formula can be rearranged to V = n/c when a target amount is desired. To obtain 0.0300 mol of a solute from a 0.500 M solution, one needs 0.0300/0.500 = 0.0600 L, or 60.0 mL. This calculation assumes the labeled molarity applies to the measured portion and that no reaction or loss occurs during delivery. If a concentrated stock was diluted, use the concentration that applies at the moment the aliquot was taken.

The named species matters. For 0.100 M CaCl₂, 0.200 L contains 0.0200 mol CaCl₂ formula amount. In an ideal complete-dissociation accounting, that corresponds to 0.0200 mol Ca²⁺ and 0.0400 mol Cl⁻. Multiplying by the number of ions is a separate stoichiometric step after n = cV, not a redefinition of the stock's CaCl₂ molarity.

To find solute mass, multiply moles by molar mass. A 0.0500 L portion of 0.150 M glucose contains 0.00750 mol glucose. Using about 180 g mol⁻¹ gives about 1.35 g. Be clear whether the requested mass is of an anhydrous formula, a hydrate or one element within the dissolved compound; each has a different mass per mole.

In titrations and precipitation problems, n = cV often supplies the starting reactant amount. The balanced equation then converts it to another species. For HCl + NaOH → NaCl + H₂O, 0.0100 mol NaOH is stoichiometrically matched by 0.0100 mol HCl. For H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, that same NaOH amount matches only 0.00500 mol H₂SO₄ for full neutralization. The cV calculation and reaction ratio must remain distinct.

Step-by-step reasoning

1. Identify the exact solution and solute named by the concentration. 2. Find the volume of the relevant aliquot and convert mL to L. 3. Multiply cV, showing units cancel to mol. 4. Use molar mass or a balanced coefficient only if another quantity is requested. 5. Check that the resulting moles scale sensibly with volume.

Visual explanation

Draw one 1.00 L flask marked 0.200 M and three pipettes of 10.0, 25.0 and 50.0 mL. Connect each to 0.00200, 0.00500 and 0.0100 mol. The proportional amounts show that c is common while n changes with V.

Real-world analogy

A road has a known density of trees per kilometre. Multiplying that density by a measured road length estimates the tree count for that segment. Molarity similarly multiplies amount per litre by litres of solution.

Real-world example

A technician delivering a standard solution with a pipette knows its molarity and volume. Their product gives reactant moles for a calibration reaction. A small systematic pipette-volume error directly changes the delivered amount even if the stock label is correct.

Why?

Why does a 20 mL aliquot contain twice the moles of a 10 mL aliquot from the same uniform stock? Both have the same amount per litre, so twice the volume samples twice the average number of solute entities.

Common misconception

“n = cV works with V = 25 when volume is 25 mL and c is in mol L⁻¹.” That gives a thousandfold error. Convert 25 mL to 0.025 L first.

Worked example

What mass of KNO₃, molar mass 101 g mol⁻¹, is present in 40.0 mL of 0.250 M solution? First n = 0.250 × 0.0400 = 0.0100 mol. Then m = nMᵣ = 0.0100 × 101 = 1.01 g. The 0.250 M label refers to a litre, but only a 0.0400 L portion was sampled.

Quick check

1. How many moles are in 15.0 mL of a 0.100 M solution? Answer: Convert 15.0 mL to 0.0150 L, then n = 0.100 × 0.0150 = 0.00150 mol.

Exam focus

Write volume conversion explicitly. If an equation follows, use its coefficients after finding the named solute's moles from cV.

Advanced insight

Molarity assumes the sampled solution is homogeneous and its composition stable. If a reagent decomposes, evaporates or precipitates during storage, a nominal cV product may differ from the true reactive amount.

Summary

The relation n = cV turns molarity and measured solution volume into moles of the named solute. Aliquot moles scale with volume while concentration stays the same. Molar mass and reaction coefficients are later, separate conversions.

Practice questions

1. Find moles in 80.0 mL of 0.0500 M NaCl. Answer: V = 0.0800 L, so n = 0.0500 × 0.0800 = 0.00400 mol NaCl formula amount. 2. What volume of 0.300 M solution supplies 0.0150 mol? Answer: V = n/c = 0.0150/0.300 = 0.0500 L, or 50.0 mL. 3. Why does 0.0200 mol CaCl₂ imply 0.0400 mol chloride in a simple dissociation model? Answer: Each CaCl₂ formula unit contains two chloride ions; the factor of two is a formula ratio applied after cV.