Mass — Question 6

Question 6

Mass fractions are not volume fractions

A two-component specimen contains masses $m_A$ and $m_B$, with total mass $M=300.0\ \mathrm{g}$. The pure-component densities at the measurement conditions are $$\rho_A=2.000\ \mathrm{g\,cm^{-3}},\qquad \rho_B=3.000\ \mathrm{g\,cm^{-3}}.$$ Its measured total volume is $V=120.0\ \mathrm{cm^3}$. Treat all values as exact model data. Initially assume mass and volume are additive, so $$m_A+m_B=M,\qquad \frac{m_A}{\rho_A}+\frac{m_B}{\rho_B}=V.$$ No chemical model is needed; take these as the governing relations.

1. Find both component masses, their mass fractions, and their volume fractions.

2. A student estimates the bulk density as $w_A\rho_A+w_B\rho_B$, using mass fractions $w_i=m_i/M$. Compare this with $M/V$ and derive the correct relation between bulk density and mass fractions under the additive-volume model.

3. Now allow a specified volume contraction $\Delta V=3.0\ \mathrm{cm^3}$, so the observed volume is $$V=\frac{m_A}{\rho_A}+\frac{m_B}{\rho_B}-\Delta V.$$ Find new component masses that reproduce the same measured $M$ and $V$.

4. If the contraction is unknown, do total mass and volume alone uniquely determine composition? Explain which extra information would restore the inference.

Question 6 - Solution

Mass fractions are not volume fractions

(a) Use the two independent additive relations.

Using grams and cubic centimetres, substitute $m_B=300-m_A$: $$\frac{m_A}{2}+\frac{300-m_A}{3}=120 \quad\Longrightarrow\quad \boxed{m_A=120\ \mathrm{g}},\quad\boxed{m_B=180\ \mathrm{g}}.$$ Thus $w_A=\boxed{0.40}$ and $w_B=\boxed{0.60}$. Their volumes are $120/2=60\ \mathrm{cm^3}$ and $180/3=60\ \mathrm{cm^3}$, so both volume fractions are $\boxed{0.50}$. Equal volumes do not imply equal masses when densities differ.

(b) The correct average depends on what supplies the weights.

The proposed mass-weighted arithmetic average gives $0.40(2)+0.60(3)=2.60\ \mathrm{g\,cm^{-3}}$. But the observed bulk density is $$\rho_{\mathrm{bulk}}=M/V=300/120=\boxed{2.50\ \mathrm{g\,cm^{-3}}}.$$ Divide the additive-volume equation by $M$: $$\boxed{\frac{1}{\rho_{\mathrm{bulk}}}= \frac{w_A}{\rho_A}+\frac{w_B}{\rho_B}}.$$ Here $0.40/2+0.60/3=0.40\ \mathrm{cm^3\,g^{-1}}$, whose reciprocal is 2.50 g/cm$^3$. An arithmetic density average instead works with the component volume fractions in this model.

(c) A contraction changes the inferred composition.

The pure-component volumes must now sum to $120+3=123\ \mathrm{cm^3}$: $$\frac{m_A}{2}+\frac{300-m_A}{3}=123 \quad\Longrightarrow\quad \boxed{m_A=138\ \mathrm{g}},\quad\boxed{m_B=162\ \mathrm{g}}.$$ Their separate volumes are 69 and 54 cm$^3$; subtracting 3 cm$^3$ gives the observed 120 cm$^3$. Total mass remains 300 g.

(d) The model supplies information the measurements lack.

$$\frac{m_A}{2\ \mathrm{g\,cm^{-3}}}+\frac{300\ \mathrm g-m_A}{3\ \mathrm{g\,cm^{-3}}}=120\ \mathrm{cm^3}+\Delta V\ \Longrightarrow\ m_A=120\ \mathrm g+6\ \mathrm{g\,cm^{-3}}\Delta V.$$

Unknown contraction introduces another unknown. The two observed totals then admit different compositions, as the examples show. A measured contraction, a validated composition-dependent volume relation, or an independent measurement of one component’s mass could close the problem. Mass additivity alone does not imply volume additivity.