Topic 1: Formulae, Equations and Amount of Substance

An empirical formula gives the simplest whole-number ratio of atoms, whereas a molecular formula gives the actual number of each type of atom in one molecule. For example, glucose has empirical formula $\ce{CH2O}$ and molecular formula $\ce{C6H12O6}$.

For $0.250\,\mathrm{mol}$ of sodium chloride, the number of formula units is $0.250\times6.02\times10^{23}=1.51\times10^{23}$. It also contains that many $\ce{Na+}$ ions and that many $\ce{Cl-}$ ions, so the total number of ions is twice the number of formula units.

1. Write correct chemical formulae; never change subscripts to balance an equation.
2. Adjust coefficients until each element is conserved.
3. Add $(s)$, $(l)$, $(g)$ and $(aq)$ from the physical states.
4. For an ionic equation, split soluble aqueous ionic substances into ions.
5. Cancel spectator ions that appear unchanged on both sides, then check atoms and charge.

For $\ce{MgCl2}$, the relative formula mass is $24.3+2(35.5)=95.3$, so its molar mass is $95.3\,\mathrm{g\,mol^{-1}}$. Calling this an $M_r$ suggests discrete molecules and is inappropriate for the giant ionic lattice.

Dissolving $5.85\,\mathrm{g}$ of $\ce{NaCl}$ and making the solution up to $500\,\mathrm{cm^3}$ gives $V=0.500\,\mathrm{dm^3}$. With $M(\ce{NaCl})=58.5\,\mathrm{g\,mol^{-1}}$, $n=0.100\,\mathrm{mol}$, so $c=0.200\,\mathrm{mol\,dm^{-3}}$. The mass concentration is $5.85/0.500=11.7\,\mathrm{g\,dm^{-3}}$.

1. Treat percentages as masses in a $100\,\mathrm{g}$ sample, or use the measured masses directly.
2. Divide each mass by the element's $A_r$ to obtain moles.
3. Divide all mole values by the smallest.
4. If necessary, multiply the whole ratio to remove values such as $1.5$ or $1.33$.
5. Write the empirical formula. For a molecular formula, calculate $k=M_r/(\text{empirical-formula mass})$ and multiply every subscript by the integer $k$.

A compound containing $40.0\%$ C, $6.7\%$ H and $53.3\%$ O gives mole values $40.0/12.0$, $6.7/1.0$ and $53.3/16.0$, a ratio close to $1:2:1$. Its empirical formula is $\ce{CH2O}$. If $M_r=180$, then $k=180/30=6$ and the molecular formula is $\ce{C6H12O6}$.

For $4.80\,\mathrm{g}$ Mg, $n(\ce{Mg})=4.80/24.3=0.198\,\mathrm{mol}$. The $1:1$ coefficient ratio gives $0.198\,\mathrm{mol}$ $\ce{MgCl2}$. With $M(\ce{MgCl2})=95.3\,\mathrm{g\,mol^{-1}}$, the theoretical mass is $18.8\,\mathrm{g}$.

At room temperature and pressure, use the stated or accepted molar volume (commonly $24.0\,\mathrm{dm^3\,mol^{-1}}$). For other conditions, use $pV=nRT$: pressure in Pa, volume in $\mathrm{m^3}$, temperature in K and $R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}}$. The same equation can find the molar mass of a volatile liquid from the mass and amount of its vapour.

$2.40\,\mathrm{dm^3}$ of gas at room conditions is $2.40/24.0=0.100\,\mathrm{mol}$. Under specified non-room conditions, $120\,\mathrm{cm^3}=1.20\times10^{-4}\,\mathrm{m^3}$ and $25^\circ\mathrm{C}=298\,\mathrm{K}$ before substitution into $n=pV/(RT)$.

For $\ce{CaCO3 -> CaO + CO2}$, if $\ce{CaO}$ is desired, the atom economy is $56.1/(56.1+44.0)\times100=56.0\%$. If the theoretical yield of $\ce{CaO}$ is $5.00\,\mathrm{g}$ but $4.20\,\mathrm{g}$ is isolated, the percentage yield is $84.0\%$.

Yield can fall because reaction is incomplete, side reactions occur, or product is lost during separation. Atom economy is fixed by the chosen balanced reaction and desired product; it does not improve merely because technique or catalyst improves yield.

1. Record masses or volumes with suitable precision and identify which measured change belongs to each substance.
2. Repeat heating or reaction to constant mass where appropriate.
3. Convert each quantity to moles using molar mass or gas conditions.
4. Divide by the smallest amount and test whether the ratio is close to small whole numbers.
5. Compare the experimental ratio with the proposed formula or equation, then state whether the data support it.

Evaluate direction as well as size of error. Incomplete reaction can leave too little mass change; loss of solid can make mass loss too large; oxidation by air can add mass; gas leaks or dissolution can reduce the collected volume. Repeats expose random variation, but repeating does not remove a systematic leak or calibration error.

1. Measure the magnesium mass and calculate $n(\ce{Mg})$.
2. Place excess dilute acid in a sealed reaction vessel connected to the gas collector.
3. Add the magnesium, seal immediately and allow the reaction to finish.
4. Record gas volume, room temperature and pressure; use $\ce{Mg + 2H+ -> Mg^{2+} + H2}$ to obtain $n(\ce{H2})$.
5. Convert the volume to $\mathrm{dm^3}$ and calculate $V_m=V/n$ in $\mathrm{dm^3\,mol^{-1}}$. Repeat and compare values.

For displacement, electron transfer changes the metal and ion identities. In the acid reaction, $\ce{H+}$ becomes hydrogen gas. In precipitation, two aqueous ions form an insoluble lattice. Spectator ions remain aqueous and cancel from the ionic equation.