Exercises — Density, molar mass, molar volume

WHAT bridge? We have grams, we want moles → the mass↔mole bridge $n=\frac{m}{M}$. WHY divide? Dividing grams by grams-per-mole cancels the grams and leaves mol. $n=\frac{m}{M}=\frac{32}{16}=2\text{ mol}$ Answer: $2\text{ mol}$.

WHAT bridge? Volume and density known, mass wanted → rearrange $\rho =\frac{m}{V}$ into $m=\rho V$. $m=\rho V=2.70\times 10=27\text{ g}$ Notice the units: ${\text{g cm}}^{-3}\times {\text{cm}}^3=\text{g}$. The ${\text{cm}}^3$ cancels cleanly. Answer: $27\text{ g}$.

WHAT bridge? Moles → volume for a gas: $V=n{V}_m$. $V=3\times 22.7=68.1\text{ L}$ WHY is the gas identity irrelevant? Avogadro's law — equal moles of any ideal gas take equal volume at the same $T,P$. See Avogadro's Law. Answer: $68.1\text{ L}$.

WHAT bridge? Combine molar mass and molar volume: one mole is $M$ grams and fills $V_m$ litres, so grams-per-litre is $\rho =\frac{M}{V_m}$. $\rho =\frac{M}{V_m}=\frac{32}{22.7}=1.41{\text{ g L}}^{-1}$ Answer: $1.41{\text{ g L}}^{-1}$. (Room-temperature air is $\approx 1.2{\text{ g L}}^{-1}$; oxygen is a touch heavier than air — sensible.)

WHAT bridge? Invert $\rho =\frac{M}{V_m}$ to solve for $M$: $M=\rho V_m=1.25\times 22.7=28.4{\text{ g mol}}^{-1}$ $28$ points to $N_2$ (nitrogen). Answer: $M\approx 28{\text{ g mol}}^{-1}$, the gas is $N_2$.

WHAT bridge? Mass and density known, volume wanted → $V=\frac{m}{\rho }$. $V=\frac{m}{\rho }=\frac{50}{0.789}=63.4{\text{ cm}}^3=63.4\text{ mL}$ ($1{\text{ cm}}^3=1\text{ mL}$.) Answer: $63.4\text{ mL}$.

WHAT bridges? Density → mass, then molar mass → moles, then Avogadro → molecules. Three bridges in a chain. Step 1 (density bridge): $m=\rho V=1.00\times 250=250\text{ g}$. Step 2 (mole bridge): $n=\frac{m}{M}=\frac{250}{18}=13.9\text{ mol}$. Step 3 (Avogadro): $N=n{N}_A=13.9\times 6.022\times 10^{23}=8.36\times 10^{24}$ molecules. Answer: $\approx 8.36\times 10^{24}$ molecules.

KEY insight: at fixed STP both share the same $V_m$, so $\rho =\frac{M}{V_m}$ means density is directly proportional to $M$. The ratio is just the mass ratio: $\frac{{\rho }_{C{O}_2}}{{\rho }_{He}}=\frac{M_{C{O}_2}}{M_{He}}=\frac{44}{4}=11$ Answer: $C{O}_2$ is $11\times$ denser than helium. (This is why helium balloons float and $C{O}_2$ pools on the ground.)

WHAT bridge? $V_m=\frac{V}{n}$. First find the volume of one mole (18 g) of water: $V=\frac{m}{\rho }=\frac{18\text{ g}}{1.00{\text{ g cm}}^{-3}}=18{\text{ cm}}^3=18\text{ mL}$ So $V_m^{\text{liquid}}=18{\text{ mL mol}}^{-1}$. Compared to a gas's $22.7\text{ L}=22700\text{ mL}$: $\frac{22700}{18}\approx 1260$ Answer: liquid water's molar volume is $18{\text{ mL mol}}^{-1}$, about $1260\times$ smaller than a gas's — because liquid molecules touch, gas molecules are far apart.

WHY not just $M/V_m$? Because $22.7{\text{ L mol}}^{-1}$ is only valid at STP. Here $T$ and $P$ differ, so we must go back to the source: the ideal gas law. Derive the density formula. Start from $PV=nRT$ and put $n=\frac{m}{M}$: $PV=\frac{m}{M}RT⟹\frac{m}{V}=\frac{PM}{RT}⟹\rho =\frac{PM}{RT}$ Now substitute. Keep $M$ in ${\text{kg mol}}^{-1}$ so SI units give ${\text{kg m}}^{-3}$: $M=44{\text{ g mol}}^{-1}=0.044{\text{ kg mol}}^{-1}$. $\rho =\frac{(2\times 10^5)(0.044)}{8.314\times 300}=\frac{8800}{2494.2}=3.53{\text{ kg m}}^{-3}=3.53{\text{ g L}}^{-1}$ Answer: $\rho \approx 3.53{\text{ g L}}^{-1}$. (At STP it was $1.94{\text{ g L}}^{-1}$; doubling pressure and cooling toward STP both raise density — this is larger, as expected.)

WHAT tool? Invert the formula we just built: $\rho =\frac{PM}{RT}\Rightarrow M=\frac{\rho RT}{P}$. $M=\frac{(1.96)(8.314)(300)}{1.013\times 10^5}=\frac{4888.6}{101300}=0.04826{\text{ kg mol}}^{-1}\approx 48.3{\text{ g mol}}^{-1}$ $48$ points to ozone, $O_3$. Answer: $M\approx 48{\text{ g mol}}^{-1}$, the gas is $O_3$.

This chains mass difference → density → molar mass. Step 1 — mass of just the gas: $m=153.520-150.000=3.520\text{ g}$. Step 2 — its density: $\rho =\frac{m}{V}=\frac{3.520}{2.00}=1.760{\text{ g L}}^{-1}$. Step 3 — molar mass at STP: $M=\rho V_m=1.760\times 22.7=39.95{\text{ g mol}}^{-1}$. Answer: $M\approx 40{\text{ g mol}}^{-1}$ — argon, $Ar$. (The subtraction is the whole trick: it removes the bulb's mass so only gas remains.)

WHY average by mole fraction? Because equal moles occupy equal volume (Avogadro), so a "mole of air" is $0.80$ mol $N_2$ + $0.20$ mol $O_2$. Its mass is the weighted sum: $\bar{M}=0.80\times 28+0.20\times 32=22.4+6.4=28.8{\text{ g mol}}^{-1}$ Then treat air as one gas of molar mass $28.8$: $\rho =\frac{\bar{M}}{V_m}=\frac{28.8}{22.7}=1.27{\text{ g L}}^{-1}$ Answer: $\bar{M}=28.8{\text{ g mol}}^{-1}$, $\rho \approx 1.27{\text{ g L}}^{-1}$ — matches the textbook air density.

Forecast: liquid molecules touch, gas molecules are ~10 diameters apart, so we expect a shrink of order $10^3$. Verify:

• Gas volume: $V_{\text{gas}}=22.7\text{ L}=22700\text{ mL}$.
• Liquid volume: $18\text{ g}$ (one mole) at $1.00{\text{ g cm}}^{-3}$ gives $V_{\text{liq}}=18\text{ mL}$.
• Ratio: $\frac{22700}{18}=1261$. Answer: it shrinks by a factor of $\approx 1260$. ✓ Forecast confirmed — condensing steam collapses ~1000-fold, exactly why a little steam makes a lot of water.

• Given mass + molar mass, find moles ::: $n=m/M$
• Given moles of gas at STP, find volume ::: $V=n{V}_m$, $V_m=22.7{\text{ L mol}}^{-1}$
• Given gas density at STP, find molar mass ::: $M=\rho V_m$
• Given gas density at arbitrary $T,P$, find molar mass ::: $M=\rho RT/P$
• Given a liquid's mass, find its volume ::: $V=m/\rho$
• Average molar mass of a mixture ::: $\bar{M}=\sum x_i{M}_i$ over mole fractions