mole fraction

Examples of mole fraction in the following topics:

• Mole Fraction and Mole Percent

  • Mole fractions are dimensionless, and the sum of all mole fractions in a given mixture is always equal to 1.
  • So, a mole fraction of 0.60 is equal to a mole percent of 60.0%.
  • What is the mole fraction of NaCl?
  • We can now find the mole fraction of the sugar:
  • Mole fraction increases proportionally to mass fraction in a solution of sodium chloride.

• Dalton's Law of Partial Pressure

  • The mole fraction is a way of expressing the relative proportion of one particular gas within a mixture of gases.
  • We do this by dividing the number of moles of a particular gas i by the total number of moles in the mixture:
  • What is the mole fraction of neon gas?
  • We inject into that container 0.78 moles of N2 gas at 298 K.
  • As a third measurement, we inject 0.22 moles of O2 gas at 298K into the first container, which already has 0.78 moles of N2.

• Vapor Pressure of Nonelectrolyte Solutions

  • Raoult's law states that the vapor pressure of an ideal solution is dependent on the vapor pressure of the pure solvent and the mole fraction of the component present in the solution.
  • In this equation, $p^{\star}_{\rm A}$ is the vapor pressure of the pure solvent and $x_{\rm A}$ is the mole fraction of the solvent.
  • Calculate the vapor pressure of a solution consisting of 3 moles of a nonvolatile solute and 15 moles of water at 25 oC, given that the vapor pressure of pure water at 25 oC is 23.8 torr.

• Writing Formulas for Polymeric Macromolecules

  • If Y and Z represent moles of monomer and polymer respectively, Z is approximately 10-5 Y.
  • Two experimentally determined values are common: Mn , the number average molecular weight, is calculated from the mole fraction distribution of different sized molecules in a sample, and Mw , the weight average molecular weight, is calculated from the weight fraction distribution of different sized molecules.

• Molar Mass of Gas

  • Molar mass (M) is equal to the mass of one mole of a particular element or compound; as such, molar masses are expressed in units of grams per mole (g mol–1) and are often referred to as molecular weights.
  • The average molar mass of a mixture of gases is equal to the sum of the mole fractions of each gas, multiplied by their respective molar masses:
  • How to set up and solve ideal gas law problems that involve molar mass and converting between grams and moles.

• Solution Stoichiometry

  • $0.123 \text{ L of solution} \times \frac{1.00 \text{ mole}}{1.00 \text{ L of solution}} = 0.123 \text { moles NaCl}$
  • $0.123 \text{ kg of solvent} \times \frac{1.00 \text{ mole}}{1.00 \text{ kg of solvent}} = 0.123 \text { moles NaCl}$
  • We can do this by converting both of our reactants into moles:
  • Because there are fewer moles of NaCl present in solution, NaCl is our limiting reactant.
  • Calculate concentrations of solutions in molarity, molality, mole fraction and percent by mass and volume.

• Converting between Moles and Atoms

  • $x \ moles \cdot \frac {6.022\times10^{23} atoms}{1 \ mole} = y \ atoms$
  • $6 \ moles \cdot \frac {6.022\times 10^{23} atoms}{1 \ mole} = 3.61\times 10^{24} atoms$
  • $\frac{x \ atoms}{6.022\times 10^{23} \frac{atoms}{1 \ mole}} = y \ moles$
  • This can be written without a fraction in the denominator by multiplying the number of atoms by the reciprocal of Avogadro's number:
  • $x \ atoms \cdot \frac{1 \ mole}{6.022\times 10^{23} \ atoms} = y \ moles$

• Molar Ratios

  • Molar ratios, or conversion factors, identify the number of moles of each reactant needed to form a certain number of moles of each product.
  • In a balanced chemical equation, the coefficients can be used to determine the relative amount of molecules, formula units, or moles of compounds that participate in the reaction.
  • The molar ratios identify how many moles of product are formed from a certain amount of reactant, as well as the number of moles of a reactant needed to completely react with a certain amount of another reactant.
  • In other words, 1 mol of methane will produced 1 mole of carbon dioxide (as long as the reaction goes to completion and there is plenty of oxygen present).
  • These molar ratios can also be expressed as fractions.

• Strong Acid-Strong Base Titrations

  • In the case of a strong acid-strong base titration, this pH transition would take place within a fraction of a drop of actual neutralization, since the strength of the base is high.
  • Step 1: First calculate the number of moles of NaOH added during the titration.
  • Step 2: Use stoichiometry to figure out the moles of HCl in the analyte.
  • The mole ratio between HCl and NaOH in the balanced equation is 1:1.
  • $0.018 \ moles \ NaOH \times \frac{1\ mole \ HCl}{1\ mole\ NaOH} = 0.018 \ moles \ HCl$

• Mole-to-Mole Conversions

  • Mole-to-mole conversions can be facilitated by using conversion factors found in the balanced equation for the reaction of interest.
  • There is a clear relationship between O2 and H2O: for every one mole of O2, two moles of H2O are produced.
  • Therefore, the ratio is one mole of O2 to two moles of H2O, or $\frac{1\:mol\:O_2}{2\:moles\:H_2O}$.
  • Therefore, 4 moles of H2O were produced by reacting 2 moles of O2 in excess hydrogen.
  • Therefore, to calculate the number of moles of water produced: $4.44 \:mol \:O_2 \cdot \frac{2 \:moles\: H_2O}{1 \:mole \:O_2} = 8.88 \:moles\: H_2O$.