Atomic Mass, Average Atomic Mass, Molecular Mass and Formula Mass

Why Atoms Need a Relative Mass Scale

One hydrogen atom has a mass of about $1.67 \times 10^{-24}$ g. No balance can weigh a single atom, and such numbers are awkward to calculate with.

Chemistry does not need absolute atomic masses, only how heavy each atom is compared with a reference atom. The laws of chemical combination (Section 4) showed that oxygen is roughly 16 times heavier than hydrogen and carbon roughly 12 times, so chemists built a relative scale.

Hydrogen: the first standard

Hydrogen, the lightest atom, was assigned a mass of 1 (no units): carbon came out about 12, oxygen about 16, sodium about 23.

But hydrogen combines directly with few elements, so most masses came through chains of intermediate compounds and errors piled up. Oxygen (= 16) replaced it, but physicists used pure $^{16}\mathrm{O}$ while chemists used natural oxygen (a mixture of isotopes), giving two slightly different scales.

Carbon-12: the standard since 1961

In 1961 a single standard was agreed: the isotope carbon-12, $^{12}\mathrm{C}$, assigned exactly 12 atomic mass units.

Key Point (Definition): One atomic mass unit is a mass exactly equal to one-twelfth of the mass of one carbon-12 atom. $$1\ \mathrm{u} = \frac{1}{12} \times (\text{mass of one } ^{12}\mathrm{C} \text{ atom}) = 1.66056 \times 10^{-24}\ \mathrm{g}$$

Hydrogen and oxygen on the new scale

The absolute mass of a hydrogen atom is $1.6736 \times 10^{-24}$ g:

$$\text{Mass of H atom} = \frac{1.6736 \times 10^{-24}\ \mathrm{g}}{1.66056 \times 10^{-24}\ \mathrm{g\ per\ u}} = 1.0078\ \mathrm{u} \approx 1.0080\ \mathrm{u}$$

An oxygen-16 atom is 15.995 u, close to but not exactly 16. The deviation comes from nuclear binding energy, and it is why the old oxygen = 16 scale and the carbon-12 scale do not coincide exactly.

amu becomes u

The older abbreviation amu has been replaced by u, the unified mass unit. The value is identical. Biochemistry also uses the dalton (Da): 1 Da = 1 u.

[Board] For "define atomic mass unit", give the one-twelfth-of-carbon-12 definition and the value $1.66056 \times 10^{-24}$ g. Both carry marks.

Modern atomic masses are measured by mass spectrometry, which separates ions by mass-to-charge ratio. Only the name is needed at this level.

Working Between Grams and u

Converting between grams and u is ordinary factor-label conversion (Section 3).

$$1\ \mathrm{u} = 1.66056 \times 10^{-24}\ \mathrm{g} \qquad\Longleftrightarrow\qquad 1\ \mathrm{g} = \frac{1}{1.66056 \times 10^{-24}}\ \mathrm{u} = 6.022 \times 10^{23}\ \mathrm{u}$$

The second number is the Avogadro number: since $^{12}\mathrm{C}$ is exactly 12 u, 12 g of carbon-12 contains $12\ \mathrm{g} \div (12 \times 1.66056 \times 10^{-24}\ \mathrm{g}) = 6.022 \times 10^{23}$ atoms. Section 6 builds the mole on this.

The two conversions

Key Point: The mass of a single atom in grams is always of the order $10^{-24}$ to $10^{-22}$ g. Any other exponent means a slip.

Atomic mass versus mass number

The mass number is a whole number. The isotope's atomic mass is close to it but not equal (except $^{12}\mathrm{C}$, exactly 12 by definition). The average can be far from any integer, as with chlorine's 35.45.

An unqualified "atomic mass" in a question means the periodic-table (average) value; $^{35}\mathrm{Cl}$ or $^{16}\mathrm{O}$ means that isotope.

Why periodic-table atomic masses are not integers

C 12.011, Cl 35.45, Cu 63.55, Ar 39.95: none is a whole number, for two reasons.

1. Isotopes. Most elements are mixtures of isotopes; the table shows the weighted average (next block).
2. Nuclear binding energy. Even a pure isotope such as $^{16}\mathrm{O}$ (15.995 u) or $^{1}\mathrm{H}$ (1.0078 u) is not an integer, because a nucleus has slightly less mass than its separate protons and neutrons (mass defect). This small effect belongs to nuclear physics; here you calculate with isotopes.

Average Atomic Mass

Isotopes are atoms of the same element (same protons) with different numbers of neutrons and so different masses. Natural carbon is mostly $^{12}\mathrm{C}$ with a little $^{13}\mathrm{C}$ and a trace of $^{14}\mathrm{C}$. A lab sample is this mixture, so the useful atomic mass is the abundance-weighted average.

Key Point (Definition): The average atomic mass of an element is the mean of the masses of its naturally occurring isotopes, each weighted by its relative abundance (per cent occurrence): $$\bar{m} = \sum_i (\text{fractional abundance})_i \times (\text{isotopic mass})_i = \frac{\sum_i (\%\text{ abundance})_i \times m_i}{100}$$

Carbon

$$\bar{m}_{\mathrm{C}} = (0.98892)(12\ \mathrm{u}) + (0.01108)(13.00335\ \mathrm{u}) + (2 \times 10^{-12})(14.00317\ \mathrm{u})$$ $$= 11.86704\ \mathrm{u} + 0.14408\ \mathrm{u} + (\approx 3 \times 10^{-11}\ \mathrm{u}) = 12.011\ \mathrm{u}$$

The $^{14}\mathrm{C}$ term changes nothing. No single carbon atom has mass 12.011 u; each is 12 u or 13.00335 u (rarely 14.00317 u). The number describes the population.

Chlorine

$$\bar{m}_{\mathrm{Cl}} = (0.7577)(34.9689) + (0.2423)(36.9659) = 26.4959 + 8.9568 = 35.4527 \approx 35.45\ \mathrm{u}$$

Hence 35.45 (or 35.5) for chlorine. Three atoms in four are the lighter isotope, so the average sits three-quarters of the way from 37 down to 35.

Argon

$$\bar{M}_{\mathrm{Ar}} = (0.00337)(35.96755) + (0.00063)(37.96272) + (0.99600)(39.9624) = 0.1212 + 0.0239 + 39.8025 = 39.948 \approx 39.95\ \mathrm{g\ mol^{-1}}$$

Masses in $\mathrm{g\ mol^{-1}}$ give the same arithmetic: molar mass in $\mathrm{g\ mol^{-1}}$ equals atomic mass in u numerically (Section 6).

Key Point: Periodic-table atomic masses are average atomic masses: the abundance-weighted average of the isotopic masses relative to $^{12}\mathrm{C} = 12$ u.

Average Atomic Mass: Checks, Shortcuts and Reverse Problems

The forward problem (abundances and masses given, find the average) is arithmetic. Reverse problems need one formula and two checks.

Check 1: the average lies between the extremes

A weighted average cannot exceed the heaviest isotope or fall below the lightest. Chlorine's 35.45 lies between 34.97 and 36.97. An answer outside the range means a wrong percentage.

Check 2: the average leans towards the abundant isotope

Chlorine is 75.77% $^{35}\mathrm{Cl}$, so the average must sit closer to 35: 35.45 is 0.48 above 34.97 but 1.52 below 36.97. This eliminates wrong MCQ options fast.

The two-isotope formula (reverse problems)

For two isotopes of masses $m_1$ and $m_2$, with the lighter at fractional abundance $x$ and the heavier at $(1-x)$:

$$\bar{m} = x\, m_1 + (1-x)\, m_2 \qquad\Longrightarrow\qquad x = \frac{m_2 - \bar{m}}{m_2 - m_1}$$

Read it as: the distance of the average from the heavy isotope, as a fraction of the gap between the isotopes.

Boron: $^{10}\mathrm{B}$ 10.01 u, $^{11}\mathrm{B}$ 11.01 u, average 10.81 u.

$$x_{^{10}\mathrm{B}} = \frac{11.01 - 10.81}{11.01 - 10.01} = \frac{0.20}{1.00} = 0.20$$

So boron is 20% $^{10}\mathrm{B}$ and 80% $^{11}\mathrm{B}$; 10.81 does sit much closer to 11.

[JEE Main] Reverse problems often use mass numbers instead of exact isotopic masses (35 and 37 for chlorine, giving $x = (37 - 35.5)/2 = 0.75$). Use the numbers given.

Abundance as a ratio

If $x = 0.75$ for $^{35}\mathrm{Cl}$, then $^{35}\mathrm{Cl} : ^{37}\mathrm{Cl} = 0.75 : 0.25 = 3 : 1$. Boron: $^{10}\mathrm{B} : ^{11}\mathrm{B} = 1 : 4$.

Three isotopes with one unknown

If two abundances are given, the third is $100\% - (\text{sum of the other two})$; then it is a forward problem. If two are unknown, the question supplies another relation (such as a ratio between them).

Mass spectrometry

A mass spectrometer vaporises and ionises the sample, accelerates the ions and bends their paths with a magnetic field. Heavier ions bend less, so each isotope lands at a different spot on the detector, and each peak's height is proportional to that isotope's abundance. The average atomic mass is computed from the spectrum with the formula above.

Key Point: Forward problem: average $= \sum(\text{fraction} \times \text{mass})$. Reverse problem (two isotopes): $x = \dfrac{m_2 - \bar{m}}{m_2 - m_1}$. Check that the answer lies between the isotopic masses and leans towards the abundant one.

Molecular Mass

Key Point (Definition): Molecular mass is the sum of the atomic masses of the elements present in a molecule: multiply the atomic mass of each element by the number of its atoms in the molecule and add.

Read the atom counts from the formula, look up each (average) atomic mass, multiply, add.

Methane, water and glucose

$\mathrm{CH_4}$: one C, four H.

$$M(\mathrm{CH_4}) = (12.011\ \mathrm{u}) + 4(1.008\ \mathrm{u}) = 12.011 + 4.032 = 16.043\ \mathrm{u}$$

$\mathrm{H_2O}$: two H, one O.

$$M(\mathrm{H_2O}) = 2(1.008\ \mathrm{u}) + 16.00\ \mathrm{u} = 2.016 + 16.00 = 18.02\ \mathrm{u}$$

$\mathrm{C_6H_{12}O_6}$: six C, twelve H, six O.

$$M(\mathrm{C_6H_{12}O_6}) = 6(12.011\ \mathrm{u}) + 12(1.008\ \mathrm{u}) + 6(16.00\ \mathrm{u})$$ $$= 72.066\ \mathrm{u} + 12.096\ \mathrm{u} + 96.00\ \mathrm{u} = 180.162\ \mathrm{u}$$

Reading a formula correctly

Errors come from counting atoms, especially with brackets and hydrates.

The bracket in $\mathrm{Ca(OH)_2}$ multiplies everything inside by 2. The dot in $\mathrm{CuSO_4 \cdot 5H_2O}$ adds five water molecules (ten H, five O); it does not multiply.

Which atomic masses to use

H = 1.008, C = 12.011, O = 16.00 give 16.043 u for methane and 180.162 u for glucose. Exam questions usually round: H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, giving 16, 18 and 180. Use the values given; otherwise the rounded set for JEE and NEET, the three-decimal values for board answers.

Molecular mass of glucose = 180 u is the mass of one molecule; molar mass = 180 $\mathrm{g\ mol^{-1}}$ is the mass of one mole (Section 6). Same number, different meaning.

[JEE Main] For a hydrocarbon $\mathrm{C_xH_y}$ with integer atomic masses, the molecular mass is $12x + y$; this backs out $x$ and $y$ quickly (used again in Section 7).

Formula Mass: When There Is No Molecule to Weigh

Molecular mass assumes discrete molecules: one methane molecule is a carbon bonded to four hydrogens and nothing else. Not every substance is like that.

Sodium chloride has no molecules

In solid $\mathrm{NaCl}$, $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions are packed in a three-dimensional lattice that extends to the edge of the crystal. Each $\mathrm{Na^+}$ is surrounded by six $\mathrm{Cl^-}$ and each $\mathrm{Cl^-}$ by six $\mathrm{Na^+}$ (6:6 coordination). No chlorine belongs to a particular sodium, so there is no Na–Cl pair to call a molecule.

The formula $\mathrm{NaCl}$ gives the simplest whole-number ratio of ions, one $\mathrm{Na^+}$ per $\mathrm{Cl^-}$, which keeps the crystal neutral. This ratio is a formula unit.

Key Point (Definition): For substances without discrete molecules (ionic compounds such as $\mathrm{NaCl}$, $\mathrm{CaCl_2}$, $\mathrm{Na_2SO_4}$; also network solids like $\mathrm{SiO_2}$), we calculate the formula mass instead of the molecular mass: the sum of the atomic masses of all atoms in one formula unit.

$$\text{Formula mass of NaCl} = \text{atomic mass of Na} + \text{atomic mass of Cl} = 23.0\ \mathrm{u} + 35.5\ \mathrm{u} = 58.5\ \mathrm{u}$$

Same arithmetic, different meaning

"Molecular mass of NaCl" is loose usage; in a board answer write formula mass for ionic compounds.

More formula masses

[Board] "Why formula mass rather than molecular mass for sodium chloride?" (1) Solid $\mathrm{NaCl}$ has no discrete entity; its ions form a three-dimensional lattice, each surrounded by six of opposite charge. (2) So $\mathrm{NaCl}$ gives only the simplest ion ratio (the formula unit), whose mass is the formula mass, 58.5 u.

Both quantities are in u and both become molar mass in $\mathrm{g\ mol^{-1}}$ with the same number; stoichiometry (Sections 6 to 9) treats them identically.

The section in one table

Solved Examples

Question 1: From grams to u — the hydrogen atom

The mass of one hydrogen atom is $1.6736 \times 10^{-24}$ g. Express this in unified mass units, given 1 u $= 1.66056 \times 10^{-24}$ g.

Answer: The factor taking grams to u is $\dfrac{1\ \mathrm{u}}{1.66056 \times 10^{-24}\ \mathrm{g}}$, so

$$m_{\mathrm{H}} = 1.6736 \times 10^{-24}\ \mathrm{g} \times \frac{1\ \mathrm{u}}{1.66056 \times 10^{-24}\ \mathrm{g}} = \frac{1.6736}{1.66056}\ \mathrm{u}$$

The powers of ten cancel: $1.6736 \div 1.66056 = 1.00785 \approx 1.0078$ u.

Ans: 1.0078 u (usually rounded to 1.0080 u).

Question 2: From u to grams — the oxygen-16 atom

The atomic mass of $^{16}\mathrm{O}$ is 15.995 u. What is the mass of one $^{16}\mathrm{O}$ atom in grams?

Answer: From u to grams I multiply by the mass of 1 u:

$$m(^{16}\mathrm{O}) = 15.995\ \mathrm{u} \times 1.66056 \times 10^{-24}\ \mathrm{g\ u^{-1}} = 26.561 \times 10^{-24}\ \mathrm{g}$$

In proper notation, $26.561 \times 10^{-24} = 2.6561 \times 10^{-23}$ g. The exponent $-23$ is in the expected $10^{-24}$ to $10^{-22}$ range for one atom.

Ans: $2.656 \times 10^{-23}$ g.

Watch out: Keep one non-zero digit before the decimal point; $26.56 \times 10^{-24}$ loses marks.

Question 3: Why the choice of standard matters

On the old hydrogen scale, hydrogen was assigned exactly 1. On that scale, what would be the relative mass of a $^{12}\mathrm{C}$ atom? Use $m(^{1}\mathrm{H}) = 1.0078$ u on the carbon-12 scale.

Answer: A relative scale only fixes ratios, and the ratio of a $^{12}\mathrm{C}$ atom to a $^{1}\mathrm{H}$ atom does not depend on the scale: $\dfrac{12\ \mathrm{u}}{1.0078\ \mathrm{u}} = 11.907$.

If hydrogen is exactly 1, carbon-12 is $11.907 \times 1 = 11.907$.

Every atomic mass is rescaled by the same factor $1/1.0078 \approx 0.9923$; oxygen-16 would become $15.995/1.0078 = 15.871$.

Ans: About 11.907 on the hydrogen = 1 scale.

Question 4: Average atomic mass of carbon

Natural carbon contains 98.892% $^{12}\mathrm{C}$ (12 u), 1.108% $^{13}\mathrm{C}$ (13.00335 u) and $2 \times 10^{-10}$% $^{14}\mathrm{C}$ (14.00317 u). Calculate the average atomic mass of carbon.

Answer: I turn the percentages into fractions (0.98892, 0.01108, $2 \times 10^{-12}$) and multiply each by its isotopic mass:

• $0.98892 \times 12 = 11.86704$ u
• $0.01108 \times 13.00335 = 0.14408$ u
• $2 \times 10^{-12} \times 14.00317 \approx 2.8 \times 10^{-11}$ u (negligible)

Sum: $11.86704 + 0.14408 + 0.00000 = 12.01112$ u, rounded to $12.011$ u for five-figure data.

Ans: 12.011 u.

Watch out: Forgetting the $\div 100$ gives 1201 u.

Question 5: Average atomic mass of chlorine

Calculate the average atomic mass of chlorine from: $^{35}\mathrm{Cl}$, 75.77%, 34.9689 u; $^{37}\mathrm{Cl}$, 24.23%, 36.9659 u.

Answer: The abundances add to $75.77 + 24.23 = 100.00$. I weight each mass:

• $0.7577 \times 34.9689 = 26.4959$ u
• $0.2423 \times 36.9659 = 8.9568$ u

Sum: $26.4959 + 8.9568 = 35.4527$ u. This lies between 34.97 and 36.97, and closer to $^{35}\mathrm{Cl}$, which is three times as abundant.

Ans: 35.45 u (35.4527 u before rounding). This is the "35.5" for chlorine: a population average, not the mass of any atom.

Question 6: Molar mass of natural argon

Calculate the molar mass of naturally occurring argon from: $^{36}\mathrm{Ar}$, 35.96755 $\mathrm{g\ mol^{-1}}$, 0.337%; $^{38}\mathrm{Ar}$, 37.96272 $\mathrm{g\ mol^{-1}}$, 0.063%; $^{40}\mathrm{Ar}$, 39.9624 $\mathrm{g\ mol^{-1}}$, 99.600%.

Answer: The abundances add to $0.337 + 0.063 + 99.600 = 100.000\%$. Fractions times masses:

• $0.00337 \times 35.96755 = 0.12121$
• $0.00063 \times 37.96272 = 0.02392$
• $0.99600 \times 39.9624 = 39.80255$

Sum: $0.12121 + 0.02392 + 39.80255 = 39.94768\ \mathrm{g\ mol^{-1}}$.

The data were molar masses, so this is the molar mass of natural argon; numerically it equals the average atomic mass in u. With one isotope at 99.6%, the average is almost that isotope's mass.

Ans: $39.95\ \mathrm{g\ mol^{-1}}$ (average atomic mass 39.95 u).

Question 7: Reverse problem — finding isotopic abundance (boron)

Boron has two isotopes, $^{10}\mathrm{B}$ (10.01 u) and $^{11}\mathrm{B}$ (11.01 u). Its average atomic mass is 10.81 u. Find the percentage abundance of each isotope.

Answer: Let the fraction of $^{10}\mathrm{B}$ be $x$, so $^{11}\mathrm{B}$ is $(1 - x)$:

$$10.01\,x + 11.01\,(1 - x) = 10.81$$

$10.01x + 11.01 - 11.01x = 10.81 \Rightarrow -1.00x = 10.81 - 11.01 = -0.20 \Rightarrow x = 0.20$.

So $^{10}\mathrm{B}$ is 20% and $^{11}\mathrm{B}$ is $100 - 20 = 80\%$.

Check: $0.20 \times 10.01 + 0.80 \times 11.01 = 2.002 + 8.808 = 10.81$ u.

The shortcut $x = \dfrac{m_2 - \bar{m}}{m_2 - m_1} = (11.01 - 10.81)/(11.01 - 10.01) = 0.20$ does it in one step.

Ans: 20% $^{10}\mathrm{B}$ and 80% $^{11}\mathrm{B}$ (ratio $1:4$).

Question 8: A JEE-style reverse problem with less friendly numbers (silver)

Silver consists of $^{107}\mathrm{Ag}$ (106.905 u) and $^{109}\mathrm{Ag}$ (108.905 u). If the average atomic mass of silver is 107.868 u, calculate the percentage of $^{107}\mathrm{Ag}$, and express the abundance ratio $^{107}\mathrm{Ag} : ^{109}\mathrm{Ag}$ in its simplest approximate form.

Answer: With $m_1 = 106.905$, $m_2 = 108.905$, $\bar{m} = 107.868$:

$$x_{107} = \frac{m_2 - \bar{m}}{m_2 - m_1} = \frac{108.905 - 107.868}{108.905 - 106.905} = \frac{1.037}{2.000} = 0.5185$$

So $^{107}\mathrm{Ag} \approx 51.85\%$, $^{109}\mathrm{Ag} \approx 48.15\%$, and the ratio $51.85 : 48.15 \approx 1.077 : 1$, roughly $13 : 12$.

Check: the midpoint of the masses is $107.905$; the average is 0.037 below it, so the lighter isotope should be slightly more abundant. It is.

Ans: $^{107}\mathrm{Ag} \approx 51.85\%$, $^{109}\mathrm{Ag} \approx 48.15\%$; ratio $\approx 13 : 12$ (or $1.08 : 1$).

Watch out: Denominator: gap between the isotopic masses. Numerator: gap between the heavy isotope and the average.

Question 9: Molecular masses of methane, water and glucose

Using H = 1.008 u, C = 12.011 u and O = 16.00 u, calculate the molecular masses of (i) $\mathrm{CH_4}$, (ii) $\mathrm{H_2O}$, (iii) $\mathrm{C_6H_{12}O_6}$.

Answer: One C, four H: $$M(\mathrm{CH_4}) = 12.011 + 4(1.008) = 12.011 + 4.032 = 16.043\ \mathrm{u}$$

Two H, one O: $$M(\mathrm{H_2O}) = 2(1.008) + 16.00 = 2.016 + 16.00 = 18.016 \approx 18.02\ \mathrm{u}$$

Six C, twelve H, six O: $$M(\mathrm{C_6H_{12}O_6}) = 6(12.011) + 12(1.008) + 6(16.00) = 72.066 + 12.096 + 96.00 = 180.162\ \mathrm{u}$$

Check: glucose is $6 \times \mathrm{CH_2O}$ ($12.011 + 2.016 + 16.00 = 30.027$ u), and $6 \times 30.027 = 180.162$ u.

Ans: (i) 16.043 u; (ii) 18.02 u; (iii) 180.162 u.

Question 10: Molecular masses with brackets and hydrates

Calculate the molecular mass of (i) $\mathrm{NH_3}$, (ii) $\mathrm{Ca(OH)_2}$, (iii) $\mathrm{CuSO_4 \cdot 5H_2O}$. Use H = 1.008, N = 14.01, O = 16.00, S = 32.1, Ca = 40.1, Cu = 63.5 (all in u).

Answer: $\mathrm{NH_3}$: $14.01 + 3(1.008) = 14.01 + 3.024 = 17.034 \approx 17.03$ u.

$\mathrm{Ca(OH)_2}$: the bracket doubles everything inside, so Ca 1, O 2, H 2. $$40.1 + 2(16.00) + 2(1.008) = 40.1 + 32.00 + 2.016 = 74.116 \approx 74.1\ \mathrm{u}$$

$\mathrm{CuSO_4 \cdot 5H_2O}$: the dot adds five waters, so Cu 1, S 1, O $4 + 5 = 9$, H $5 \times 2 = 10$. $$63.5 + 32.1 + 9(16.00) + 10(1.008) = 63.5 + 32.1 + 144.0 + 10.08 = 249.68 \approx 249.7\ \mathrm{u}$$

Or $M(\mathrm{CuSO_4}) + 5\,M(\mathrm{H_2O}) = 159.6 + 5(18.016) = 159.6 + 90.08 = 249.7$ u.

Ans: (i) 17.03 u; (ii) 74.1 u; (iii) 249.7 u.

Watch out: Brackets multiply, dots add. Strictly (ii) and (iii) are ionic, so call these formula masses in a board answer.

Question 11: Formula masses of ionic compounds

Calculate the formula mass of (i) $\mathrm{NaCl}$, (ii) $\mathrm{CaCl_2}$, (iii) $\mathrm{Na_2SO_4}$, (iv) $\mathrm{Al_2(SO_4)_3}$. Use Na = 23.0, Cl = 35.5, Ca = 40.1, S = 32.1, O = 16.0, Al = 27.0 (u). Why is the term "formula mass" used here rather than "molecular mass"?

Answer: These are ionic solids: the ions form a three-dimensional lattice with no discrete molecules (in $\mathrm{NaCl}$ each ion has six neighbours of opposite charge). The formula is only the simplest ion ratio, the formula unit, so I find the mass of a formula unit.

NaCl: $23.0 + 35.5 = 58.5$ u.

$\mathrm{CaCl_2}$ ($\mathrm{Ca^{2+}}$, two $\mathrm{Cl^-}$): $40.1 + 2(35.5) = 40.1 + 71.0 = 111.1$ u.

$\mathrm{Na_2SO_4}$ (two $\mathrm{Na^+}$, one $\mathrm{SO_4^{2-}}$): $2(23.0) + 32.1 + 4(16.0) = 46.0 + 32.1 + 64.0 = 142.1$ u.

$\mathrm{Al_2(SO_4)_3}$ (two $\mathrm{Al^{3+}}$, three $\mathrm{SO_4^{2-}}$; Al 2, S 3, O 12): $$2(27.0) + 3(32.1) + 12(16.0) = 54.0 + 96.3 + 192.0 = 342.3\ \mathrm{u}$$

Ans: (i) 58.5 u; (ii) 111.1 u; (iii) 142.1 u; (iv) 342.3 u.

Watch out: In $\mathrm{Al_2(SO_4)_3}$ the 3 outside the bracket multiplies the S and all four O: 12 oxygens, not 4.

Question 12: Mass of one $^{12}\mathrm{C}$ atom, and counting atoms by mass

(i) What is the mass of one $^{12}\mathrm{C}$ atom in grams? (ii) How many atoms are there in 52 u of helium (He = 4 u)? (iii) How many water molecules are there in 180.2 u of water? (iv) How many atoms are there in 52 moles of argon and in 52 g of helium? (Take $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$.)

Answer: (i) One $^{12}\mathrm{C}$ atom is exactly 12 u: $$m(^{12}\mathrm{C}) = 12 \times 1.66056 \times 10^{-24}\ \mathrm{g} = 19.927 \times 10^{-24}\ \mathrm{g} = 1.9927 \times 10^{-23}\ \mathrm{g}$$ Check: $12\ \mathrm{g} \div 6.022 \times 10^{23} = 1.9927 \times 10^{-23}$ g.

(ii) One He atom is 4 u: $\dfrac{52\ \mathrm{u}}{4\ \mathrm{u\ per\ atom}} = 13$ atoms. No Avogadro number here.

(iii) One $\mathrm{H_2O}$ is 18.02 u: $\dfrac{180.2}{18.02} = 10$ molecules.

(iv) 52 mol Ar: $52 \times 6.022 \times 10^{23} = 3.131 \times 10^{25}$ atoms. 52 g He: $52/4 = 13$ mol, so $13 \times 6.022 \times 10^{23} = 7.829 \times 10^{24}$ atoms.

Ans: (i) $1.9927 \times 10^{-23}$ g; (ii) 13 atoms; (iii) 10 molecules; (iv) $3.131 \times 10^{25}$ atoms of Ar and $7.829 \times 10^{24}$ atoms of He.

Watch out: "52 u of He" and "52 g of He" differ by a factor of $6.022 \times 10^{23}$. Read the unit first.