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Concept 04 / 06Some Basic Concepts of Chemistry

Concentration Terms: Mass Percent, Mole Fraction, Molarity, Molality and ppm

Definitions of mass percent, volume percent, parts per million, mole fraction, molarity and molality, which terms depend on temperature, interconversion using density, dilution and mixing formulas, and the strength of solutions.

The composition of a solution can be expressed in several ways. Each suits a different purpose, and interconverting between them is a routine but error-prone skill. Mass percent (w/w): mass of solutemass of solution×100\frac{\text{mass of solute}}{\text{mass of solution}} \times 100. A 10%10\% glucose solution has 1010 g glucose in 100100 g of solution (not 100100 g of water). Volume percent (v/v) and mass-by-volume percent (w/v, grams per 100100 mL of solution) are defined similarly. Parts per million: ppm=mass of solutemass of solution×106\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6}, used for trace amounts (pollutants, hardness of water). Mole fraction: xA=nAnA+nB+⋯x_{A} = \frac{n_{A}}{n_{A} + n_{B} + \cdots}. The mole fractions of all components add to 11. It is dimensionless and is the natural variable in Raoult's law. Molarity: M=moles of solutevolume of solution in litresM = \frac{\text{moles of solute}}{\text{volume of solution in litres}} (unit mol L−1\mathrm{mol\ L^{-1}}). It is convenient for titrations because volumes are measured. Because volume changes with temperature, molarity decreases slightly when a solution is heated. Molality: m=moles of solutemass of solvent in kgm = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} (unit mol kg−1\mathrm{mol\ kg^{-1}}). Masses do not change with temperature, so molality is temperature-independent, which is why it is used for colligative properties. Temperature dependence: mass percent, ppm, mole fraction and molality are independent of temperature; molarity (and normality, w/v percent) depend on it. Interconversions. Density dd (g/mL) of the solution links volume and mass. Useful formulas (with M2M_{2} the molar mass of the solute): Molarity from mass percent ww: M=10 w dM2M = \frac{10\,w\,d}{M_{2}}. Molality from molarity: m=1000M1000d−MM2m = \frac{1000M}{1000d - MM_{2}}. Mole fraction of solute from molality in water: x2=mm+100018x_{2} = \frac{m}{m + \frac{1000}{18}}. Rather than memorising, derive them by taking a convenient basis: 11 L of solution for molarity, 11 kg of solvent for molality, or 100100 g of solution for mass percent. Dilution: moles of solute are unchanged, so M1V1=M2V2M_{1}V_{1} = M_{2}V_{2}. Mixing solutions of the same solute: Mfinal=M1V1+M2V2V1+V2M_{\text{final}} = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}} (assuming volumes are additive). Strength of a solution in grams per litre == molarity ×\times molar mass. The 'volume strength' of hydrogen peroxide is the volume of O2\mathrm{O_{2}} at STP released by 11 volume of solution: from 2H2O2→2H2O+O22\mathrm{H_{2}O_{2}} \rightarrow 2\mathrm{H_{2}O} + \mathrm{O_{2}}, volume strength =11.2×= 11.2 \times molarity. Key Takeaways: - Molarity: mol per litre of solution; molality: mol per kg of solvent. - Temperature-independent: mass %, ppm, mole fraction, molality. - M=10wdM2M = \frac{10wd}{M_{2}}; m=1000M1000d−MM2m = \frac{1000M}{1000d - MM_{2}}. - Dilution M1V1=M2V2M_{1}V_{1} = M_{2}V_{2}; mixing ∑MiVi∑Vi\frac{\sum M_{i}V_{i}}{\sum V_{i}}. - H2O2\mathrm{H_{2}O_{2}}: volume strength =11.2×= 11.2 \times molarity. - Pitfall: dividing by the volume of solvent instead of the volume of solution for molarity. Brain-Teaser 1: Commercial H2SO4\mathrm{H_{2}SO_{4}} is 98%98\% by mass with density 1.841.84 g/mL. Find its molarity and the volume needed to prepare 11 L of 0.50.5 M acid. Solution: M=10×98×1.8498=18.4M = \frac{10 \times 98 \times 1.84}{98} = 18.4 M. Dilution: 18.4×V=0.5×100018.4 \times V = 0.5 \times 1000, so V=27.2V = 27.2 mL. (Acid is added to water, never the reverse, because of the large heat of dilution.) Brain-Teaser 2: Find the molality and mole fraction of glucose in a solution of 1818 g glucose (M=180M = 180) in 500500 g water. Solution: n=0.1n = 0.1 mol, so m=0.10.5=0.2m = \frac{0.1}{0.5} = 0.2 mol/kg. Water =50018=27.78= \frac{500}{18} = 27.78 mol, so xglucose=0.127.88=0.00359x_{\text{glucose}} = \frac{0.1}{27.88} = 0.00359. Molality needs the solvent's mass only, not the solution's. Brain-Teaser 3: 200200 mL of 0.50.5 M NaCl is mixed with 300300 mL of 0.20.2 M NaCl. Find the molarity of the mixture. What is the concentration of Na+\mathrm{Na^{+}} if 500500 mL of 0.10.1 M Na2SO4\mathrm{Na_{2}SO_{4}} is then added? Solution: Moles NaCl =0.1+0.06=0.16= 0.1 + 0.06 = 0.16 in 0.50.5 L, so 0.320.32 M. Adding Na2SO4\mathrm{Na_{2}SO_{4}} contributes 0.05×2=0.10.05 \times 2 = 0.1 mol Na+\mathrm{Na^{+}}; total Na+=0.26\mathrm{Na^{+}} = 0.26 mol in 1.01.0 L, so [Na+]=0.26[\mathrm{Na^{+}}] = 0.26 M. For ion concentrations, count ions per formula unit.

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DPP: Concentration Terms: Mass Percent, Mole Fraction, Molarity, Molality and ppm

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