Moles and Avogadro's Number: MDCAT Chemistry notes

Atomic mass and the carbon-12 standard

One atomic mass unit (amu) is $\frac{1}{12}$ of the mass of one atom of carbon-12. Relative atomic mass compares the average mass of an element's atoms with this standard. Most elements are mixtures of isotopes: atoms with the same number of protons and electrons (same atomic number $Z$) but different numbers of neutrons (different mass number $A$). Isotopes have the same chemical properties but different physical properties. Neutrons $=A-Z$; for example $^{66}_{30}Zn$ has 36 neutrons.

Because of isotopes, relative atomic mass is a weighted average and depends on the number of isotopes and their relative abundance. That is why no single neon atom has mass 20.18 amu, and copper's value is 63.55 amu. The type and abundance of isotopes are found by mass spectrometry, in which the sample is first vaporised and then ionised.

Worked example: boron (average 10.8) has isotopes of mass 10 and 11. If the fraction of boron-10 is $x$: $10x+11(1-x)=10.8$, so $x=0.2$, i.e. 20%.

The mole

One mole is the amount of substance containing as many particles as there are atoms in exactly 12 g of carbon-12. This number is Avogadro's number, $N_A=6.02\times10^{23}$. One mole of any substance (ethanol, ethane, NaCl) contains the same number of formula units, but not the same mass or the same number of atoms.

Key formulas

$$n=\frac{m}{M}\qquad N=n\times N_A\qquad n=\frac{V}{22.414\ \text{dm}^3}\ \text{(gas at STP)}$$

$$\text{mass of one particle}=\frac{M}{N_A}\qquad \text{Molarity}=\frac{n}{V\ (\text{dm}^3)}$$

For a given element $M$ is fixed, so moles are directly proportional to mass. Mass of one $O_2$ molecule $=\frac{32}{6.02\times10^{23}}$ g.

Quantity asked	Working	Answer
Particles in 0.25 mol $CO_2$	$0.25\times6.022\times10^{23}$	$1.505\times10^{23}$
Moles of $CO_2$ containing 16 g O	16 g O = 1 mol O atoms; 2 O per molecule	0.5 mol
H atoms in 18 g water	1 mol $H_2O$, 2 H each	$2\times6.022\times10^{23}$
Moles in 1 kg ice	1000/18	55.5 mol
Mass of 2 dm³ $O_2$ at STP	(2/22.4) × 32	2.86 g
Molarity of 85.5 g sucrose in 250 cm³	(85.5/342)/0.25	1 M

Stoichiometry

A balanced equation obeys the law of conservation of mass, and its coefficients give mole ratios. Method: convert given mass to moles, use the mole ratio, convert back.

Example: $N_2+3H_2\rightarrow2NH_3$. To make 51 g $NH_3$ (3 mol) needs $\frac{3}{2}\times3=4.5$ mol $H_2$ = 9 g. Example: 18.5 g Al in $4Al+3O_2\rightarrow2Al_2O_3$: $\frac{18.5}{27}=0.685$ mol Al gives 0.343 mol $Al_2O_3$ = 34.9 g.

Percentage composition and empirical formula

Percentage of an element $=\frac{\text{mass of element in formula}}{\text{molar mass}}\times100$. In glucose, C = $\frac{72}{180}\times100=40\%$; in $CO_2$, C = 27.3%.

The empirical formula is the simplest whole-number ratio of atoms (glucose $C_6H_{12}O_6$ → $CH_2O$). Steps: (1) find percentage composition (the first step), (2) divide by atomic masses to get gram atoms, (3) divide by the smallest to get the atomic ratio, (4) multiply to whole numbers. Molecular formula = $n\times$ empirical formula, where $n=\frac{\text{molecular mass}}{\text{empirical formula mass}}$. E.g. $CH_2Cl$ (49.5) with molar mass 99 gives $C_2H_4Cl_2$.

Common MDCAT traps

• Count atoms, not molecules, when asked "number of atoms": 10 g $O_2$ has fewer atoms than 10 g $N_2$ or $H_2$.
• Grams of oxygen in a compound refer to O atoms (16 g/mol), not $O_2$.
• Equal volumes of gases at the same T and P have equal numbers of molecules (Avogadro's law), even with different masses.
• Isotopes differ in neutrons only; isobars have the same mass number.
• Molarity uses dm³: convert 250 cm³ to 0.25 dm³.

Quick revision

• $N_A=6.02\times10^{23}$ particles per mole.
• 1 amu = 1/12 mass of a C-12 atom.
• Average atomic mass depends on isotope abundance.
• Molar volume at STP = 22.4 dm³.
• Molar mass of $CaCO_3$ = 100 g/mol.