Some Basic Concepts of Chemistry
The grammar of chemistry — matter, measurement, the mole and the balanced equation that every later calculation rests upon
- How matter is classified into elements, compounds and mixtures.
- The SI units, the rules of significant figures, and the factor-label method of dimensional analysis.
- The five laws of chemical combination and how Dalton's atomic theory explains them.
- The central idea of the mole and the bridge \(n=\dfrac{m}{M}=\dfrac{N}{N_A}=\dfrac{V}{V_m}\).
- How to find an empirical and molecular formula from percentage composition.
- How to do stoichiometry, identify the limiting reagent, and express concentration as molarity, molality and mole fraction.
Matter and Its Classification
Matter is anything that has mass and occupies space. Physically it appears in three common states — solid, liquid and gas — distinguished by how tightly its particles are held and how freely they move. Chemically, the more important question is whether a sample is pure or a mixture.
A pure substance has a fixed composition and a unique set of properties. It is either an element (one kind of atom, e.g. \(\ce{Na}\), \(\ce{O2}\)) or a compound (two or more elements chemically combined in a fixed mass ratio, e.g. \(\ce{H2O}\)). A mixture contains two or more substances in any proportion, separable by physical means; it is homogeneous (uniform throughout, like salt solution) or heterogeneous (non-uniform, like sand in water).
Measurement and the SI Units
Chemistry is a quantitative science, so every property is a number with a unit. The SI system rests on seven base units; everything else is derived from them. Mass in kilograms, length in metres and amount of substance in moles are the three that dominate this chapter.
| Quantity | SI base unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Temperature | kelvin | K |
| Amount of substance | mole | mol |
| Electric current | ampere | A |
| Luminous intensity | candela | cd |
Accuracy is how close a reading is to the true value; precision is how close repeated readings are to one another. A measurement can be precise yet inaccurate (a mis-calibrated balance) — both must be controlled before a result is trusted.
Significant Figures & Dimensional Analysis
The significant figures of a measurement are the digits known with certainty plus one estimated digit. They record how good the measurement is, and they must survive the arithmetic.
In \(\times,\div\) the answer keeps the fewest significant figures of any factor; in \(+,-\) it keeps the fewest decimal places. Exact (counted or defined) numbers carry infinite significant figures.
Dimensional analysis (the factor-label method) converts units by multiplying with ratios equal to one, so unwanted units cancel like algebraic symbols:
The Laws of Chemical Combination
Five empirical laws, discovered before atoms were understood, govern how elements combine. They are the experimental bedrock that Dalton's theory was built to explain.
(Lavoisier) Mass is neither created nor destroyed in a chemical change — total mass of reactants equals total mass of products.
(Proust) A given compound always contains the same elements in the same fixed proportion by mass, however it is prepared.
(Dalton) When two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a small whole-number ratio.
(Richter) The mass ratio in which two elements combine with a third is the same as, or a simple multiple of, the ratio in which they combine with each other.
(Gay-Lussac) Gases react in volumes that bear a simple whole-number ratio to one another and to the gaseous products, at the same temperature and pressure.
Equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules — the key that turned Gay-Lussac's volumes into molecular ratios.
Dalton's Atomic Theory
In 1808 John Dalton unified those laws with a single picture of matter built from indivisible particles.
Conservation of mass follows from atoms being conserved; definite and multiple proportions follow from fixed whole-number combining ratios. Later work refined the theory — atoms are divisible, and isotopes show atoms of an element can differ in mass — but the framework still underpins all of chemistry.
Atomic and Molecular Masses
Atoms are too light to weigh individually, so masses are quoted relative to a standard. The unified atomic mass unit \(1\,\text{u}\) is defined as exactly one-twelfth of the mass of a carbon-12 atom, so \(1\ \text{u}=1.66\times10^{-24}\ \text{g}\).
Because most elements are mixtures of isotopes, the average atomic mass is a weighted mean of the isotopic masses. The molecular mass of a substance is the sum of the atomic masses of its atoms; for ionic compounds, where no discrete molecule exists, we use the formula mass instead.
The molecular mass of water \(\ce{H2O}\) is \(2(1.008)+16.00=18.02\ \text{u}\); the formula mass of \(\ce{NaCl}\) is \(22.99+35.45=58.44\ \text{u}\).
The Mole Concept
Chemists count atoms by weighing. The bridge between the macroscopic gram and the invisible atom is the mole — the amount of substance containing as many entities as there are atoms in \(12\ \text{g}\) of carbon-12. That number is Avogadro's number:
The mass of one mole, in grams, is the molar mass \(M\), numerically equal to the atomic or molecular mass in u. The mole sits at the centre of a three-way conversion that solves nearly every quantitative problem in the chapter.
Here \(m\) is mass, \(M\) molar mass, \(N\) the number of particles, \(V\) the gas volume and \(V_m\) the molar volume. For an ideal gas, \(V_m=22.7\ \text{L mol}^{-1}\) at STP (\(273.15\ \text{K},\ 1\ \text{bar}\)), and the older \(22.4\ \text{L mol}^{-1}\) at \(1\ \text{atm}\).
Percentage Composition, Empirical & Molecular Formula
The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual numbers. They are linked by a single integer \(k\):
Stoichiometry & the Limiting Reagent
A balanced equation is a recipe in moles. Its coefficients give the exact mole ratios in which substances react and form, so any quantity of one species can be converted to any quantity of another by passing through moles.
When reactants are not supplied in the exact stoichiometric ratio, one runs out first — the limiting reagent. It alone fixes how much product forms; the others are in excess and partly remain. To find it, convert every reactant to moles, divide each by its coefficient, and the smallest quotient marks the limiting reagent.
Ways to Express Concentration
The composition of a solution can be reported in several equivalent ways. The temperature-independent ones (molality, mole fraction, mass percent) are preferred for precise work, since molarity changes as a solution expands or contracts with temperature.
| Term | Definition | Note |
|---|---|---|
| Mass percent | \(\dfrac{\text{mass of solute}}{\text{mass of solution}}\times100\) | temperature-independent |
| Mole fraction \(x\) | \(\dfrac{n_{\text{component}}}{n_{\text{total}}}\) | \(x_A+x_B=1\) |
| Molarity \(M\) | \(\dfrac{\text{moles of solute}}{\text{volume of solution (L)}}\) | varies with temperature |
| Molality \(m\) | \(\dfrac{\text{moles of solute}}{\text{mass of solvent (kg)}}\) | temperature-independent |
| ppm | \(\dfrac{\text{mass of solute}}{\text{mass of solution}}\times10^{6}\) | very dilute solutions |
Putting It to Work
Problem. A metal sample of mass \(36.6\ \text{g}\) occupies \(2.0\ \text{cm}^3\). Report its density with correct significant figures.
Solution. Density \(=\dfrac{m}{V}\). The factor \(2.0\) has only two significant figures, so the answer is limited to two:
Problem. In \(\ce{CO}\), \(12\ \text{g}\) of carbon combines with \(16\ \text{g}\) of oxygen; in \(\ce{CO2}\) with \(32\ \text{g}\). Do these obey the law?
Solution. For a fixed \(12\ \text{g}\) of carbon, the oxygen masses are in the ratio
A simple whole-number ratio — the law of multiple proportions holds.
Problem. How many molecules, and how many atoms, are present in \(4.4\ \text{g}\) of carbon dioxide?
Solution. Molar mass of \(\ce{CO2}=44\ \text{g mol}^{-1}\), so \(n=\dfrac{4.4}{44}=0.1\ \text{mol}\):
Problem. A compound is \(40\%\ \ce{C}\), \(6.7\%\ \ce{H}\) and \(53.3\%\ \ce{O}\) by mass, with molar mass \(180\ \text{g mol}^{-1}\). Find its molecular formula.
Solution. Convert percent to moles, then to a ratio:
The molecular formula is \(\ce{C6H12O6}\) — glucose.
Problem. \(28\ \text{g}\) of \(\ce{N2}\) reacts with \(3\ \text{g}\) of \(\ce{H2}\) to form ammonia. Which is limiting, and what mass of \(\ce{NH3}\) forms?
Solution. Moles: \(\ce{N2}=\tfrac{28}{28}=1\), \(\ce{H2}=\tfrac{3}{2}=1.5\). The recipe \(\ce{N2 + 3H2 -> 2NH3}\) needs \(3\) mol \(\ce{H2}\) per mol \(\ce{N2}\); only \(1.5\) mol is present, so \(\ce{H2}\) is limiting:
Half the nitrogen \((0.5\ \text{mol},\ 14\ \text{g})\) is left over in excess.
Problem. \(5.85\ \text{g}\) of \(\ce{NaCl}\) is dissolved to make \(500\ \text{mL}\) of solution. Find its molarity, then the volume of water needed to dilute it to \(0.05\ \text{M}\).
Solution. \(n=\tfrac{5.85}{58.5}=0.1\ \text{mol}\), so
So add \(2000-500=1500\ \text{mL}\) of water.
Chapter Summary
Pure substances (elements, compounds) versus mixtures (homogeneous, heterogeneous); mixtures separate physically, compounds chemically.
Seven SI base units; significant figures track precision; dimensional analysis converts units by cancellation.
Conservation of mass, definite, multiple and reciprocal proportions, gaseous volumes — all explained by Dalton's atoms.
\(n=\dfrac{m}{M}=\dfrac{N}{N_A}=\dfrac{V}{V_m}\) with \(N_A=6.022\times10^{23}\).
Empirical is the simplest ratio; molecular \(=k\times\) empirical, with \(k=\tfrac{M}{\text{emp. mass}}\).
Coefficients are mole ratios; the limiting reagent fixes the yield; concentration is molarity, molality or mole fraction.
Problems
Identify what is asked, route everything through the mole, and keep the significant figures honest. Difficulty rises down the list. Take atomic masses as \(\ce{H}=1,\ \ce{C}=12,\ \ce{N}=14,\ \ce{O}=16,\ \ce{Na}=23,\ \ce{S}=32,\ \ce{Cl}=35.5,\ \ce{Ca}=40\).
- Express \(0.00204\ \text{kg}\) in grams and state the number of significant figures.
- How many moles, and how many atoms, are there in \(11.5\ \text{g}\) of sodium?
- Calculate the mass of \(0.25\ \text{mol}\) of \(\ce{CaCO3}\).
- A gas occupies \(5.6\ \text{L}\) at STP (\(1\ \text{atm}\)). How many moles and how many molecules does it contain?
- Copper forms two oxides containing \(79.9\%\) and \(88.8\%\) copper. Show these obey the law of multiple proportions.
- Find the empirical formula of a compound that is \(70\%\ \ce{Fe}\) and \(30\%\ \ce{O}\) by mass.
- A hydrocarbon contains \(85.7\%\) carbon and has molar mass \(56\ \text{g mol}^{-1}\). Find its molecular formula.
- What mass of \(\ce{CO2}\) is produced when \(10\ \text{g}\) of \(\ce{CaCO3}\) is completely decomposed, \(\ce{CaCO3 -> CaO + CO2}\)?
- Calculate the molarity of a solution made by dissolving \(4.9\ \text{g}\) of \(\ce{H2SO4}\) in water to make \(250\ \text{mL}\) of solution.
- How many millilitres of \(0.5\ \text{M}\) \(\ce{HCl}\) are needed to prepare \(100\ \text{mL}\) of \(0.1\ \text{M}\) \(\ce{HCl}\)?
- In the reaction \(\ce{2H2 + O2 -> 2H2O}\), \(4\ \text{g}\) of \(\ce{H2}\) is mixed with \(32\ \text{g}\) of \(\ce{O2}\). Identify the limiting reagent and find the mass of water formed.
- A \(6.8\%\) by mass aqueous solution of \(\ce{H2O2}\) has density \(1.0\ \text{g mL}^{-1}\). Find its molarity and the mole fraction of \(\ce{H2O2}\).