Chemistry · Study Guide

Moles, Mass and Concentration: Worked Chemistry Calculations

The mole connects a measurable amount of a substance to the number of particles involved in a reaction. Most introductory mole calculations become clearer when you write the units first, choose one relationship, and show the conversion explicitly. This guide covers mass, solution concentration and simple reacting amounts for school chemistry.

Use mass and molar mass together

The relationship is n=m/Mn=m/M, where n is amount in moles, m is mass in grams and M is molar mass in grams per mole. For sodium chloride, using relative atomic masses Na = 23 and Cl = 35.5 gives M = 58.5 g mol⁻¹. A mass of 5.85 g therefore contains 0.100 mol. Relative formula mass is dimensionless; molar mass has units, even though the numerical values match when expressed in grams per mole.

Count atoms in the whole formula

Before dividing, check brackets and subscripts carefully. For calcium hydroxide, Ca(OH)₂ contains one calcium, two oxygen atoms and two hydrogen atoms per formula unit. Using Ca = 40, O = 16 and H = 1 gives a molar mass of 74 g mol⁻¹. The 2 multiplies everything inside the brackets, not just hydrogen. Use the relative atomic masses provided in your question if they differ from these rounded examples.

Use concentration with compatible volume units

For amount concentration, c=n/Vc=n/V and n=cVn=cV. If c is in mol dm⁻³, V must be in dm³. Because 1 dm³ = 1000 cm³, divide a volume in cm³ by 1000. Thus 25.0 cm³ is 0.0250 dm³. A 0.200 mol dm⁻³ solution contains 0.200×0.0250=0.005000.200\times0.0250=0.00500 mol in that volume. Writing the conversion prevents an answer that is wrong by a factor of 1000.

Let the balanced equation give the ratio

For 2H2+O2→2H2O2H_2+O_2\rightarrow2H_2O, one mole of oxygen reacts with two moles of hydrogen to form two moles of water. The coefficients give mole ratios, not mass ratios. First convert the known mass to moles, then use the coefficients, then convert back to mass if required. If the question gives amounts of both reactants, identify the limiting reactant before calculating the product.

Finish with a reasonableness check

Keep extra digits during intermediate steps and round your final answer to a precision appropriate to the data. Always include units. Check whether you have calculated mass, amount or concentration: those are different quantities. Where conservation of mass applies to the whole reaction, remember that the product can include atoms supplied by another reactant. Its mass need not equal the mass of just the one reactant named in your first calculation.

Worked example: convert oxygen mass into water mass

Identify
3.20 g of oxygen reacts with excess hydrogen. Calculate the mass of water formed using 2H2+O2→2H2O2H_2+O_2\rightarrow2H_2O. Use O = 16 and H = 1.
Find moles
Oxygen is O2O_2, so its molar mass is 32.0 g mol⁻¹. n(O2)=3.20/32.0=0.100n(O_2)=3.20/32.0=0.100 mol.
Use ratio
The equation gives two moles of water for each mole of oxygen. Therefore n(H2O)=2(0.100)=0.200n(H_2O)=2(0.100)=0.200 mol.
Find mass
Water has molar mass 18.0 g mol⁻¹. m(H2O)=0.200×18.0=3.60m(H_2O)=0.200\times18.0=3.60 g, to three significant figures.
Interpret
The water is heavier than the oxygen alone because it also contains hydrogen. The hydrogen is in excess, so oxygen sets the amount of product.

Common Mistakes

Practice

How many moles are in 7.40 g of Ca(OH)₂? Use Ca = 40, O = 16 and H = 1.
The molar mass is 40+2(16+1)=7440+2(16+1)=74 g mol⁻¹. The amount is 7.40/74=0.1007.40/74=0.100 mol.
Find the amount of solute in 50.0 cm³ of a 0.150 mol dm⁻³ solution.
Convert 50.0 cm³ to 0.0500 dm³. n=cV=0.150×0.0500=0.00750n=cV=0.150\times0.0500=0.00750 mol.
0.300 mol H₂ reacts with 0.100 mol O₂. Which reactant is limiting, and how much water forms?
Oxygen is limiting: 0.100 mol O₂ requires only 0.200 mol H₂. The reaction forms 0.200 mol H₂O and leaves 0.100 mol H₂.

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