mole

Examples of mole in the following topics:

• Mole Fraction and Mole Percent

  • $x (\text{mole fraction}) = (\frac {\text{ moles } N_2}{\text{ moles } N_2 + \text{ moles } O_2})= (\frac {5.6 \text { moles}}{11.9 \text{ moles}})= 0.47$
  • We find that there are 0.138 moles of pentane, 0.116 moles of hexane, and 0.128 moles of benzene.
  • If we divide moles of hexane by the total moles, we calculate the mole fraction:
  • We have 0.833 moles urea and 0.388 moles cinnamic acid, so we have 1.22 moles total.
  • To find the mole fraction, we divide the moles of cinnamic acid by total number of moles:

• 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: $4.44 \:mol \:O_2 \cdot \frac{2 \:moles\: H_2O}{1 \:mole \:O_2} = 8.88 \:moles\: H_2O$.

• Converting between Moles and Atoms

  • $x \ moles \cdot \frac {6.022\times10^{23} atoms}{1 \ mole} = y \ atoms$
  • $6 \ moles \cdot \frac {6.022\times 10^{23} atoms}{1 \ mole} = 3.61\times 10^{24} atoms$
  • $\frac{x \ atoms}{6.022\times 10^{23} \frac{atoms}{1 \ mole}} = y \ moles$
  • $x \ atoms \cdot \frac{1 \ mole}{6.022\times 10^{23} \ atoms} = y \ moles$
  • $3.5 \times 10^{24} \ atoms \cdot \frac{1 \ mole}{6.022\times 10^{23} \ atoms} = 5.81 \ moles$

• Avogadro's Number and the Mole

  • The mole is represented by Avogadro's number, which is 6.02×1023 mol-1.
  • The solution is the concept of the mole, which is very important in quantitative chemistry.
  • The mole (abbreviated mol) is the SI measure of quantity of a "chemical entity," such as atoms, electrons, or protons.
  • For example, since one atom of oxygen will combine with two atoms of hydrogen to create one molecule of water (H2O), one mole of oxygen (6.022×1023 of O atoms) will combine with two moles of hydrogen (2 × 6.022×1023 of H atoms) to make one mole of H2O.
  • This video introduces counting by mass, the mole, and how it relates to atomic mass units (AMU) and Avogadro's number.

• Mass-to-Mole Conversions

  • The mole is the universal measurement of quantity in chemistry.
  • The molar mass value can be used as a conversion factor to facilitate mass-to-mole and mole-to-mass conversions.
  • The compound's molar mass is necessary when converting from grams to moles.
  • For example, convert 18 grams of water to moles of water.
  • If you have 34.5 g of NaCl, how many moles of NaCl do you have?

• Converting between Mass and Number of Moles

  • One mole (abbreviated mol) is equal to 6.022×1023 molecular entities (Avogadro's number), and each element has a different molar mass depending on the weight of 6.022×1023 of its atoms (1 mole).
  • How many moles of NaOH are present in 90 g of NaOH?
  • How many moles and how many atoms are contained in 10.0 g of nickel?
  • Given a sample's mass and number of moles in that sample, it is also possible to calculate the sample's molecular mass by dividing the mass by the number of moles to calculate g/mol.
  • This flowchart illustrates the relationships between mass, moles, and particles.

• Mass-to-Mass Conversions

  • The ratio of the coefficients of two of the compounds in a reaction (reactant or product) can be viewed as a conversion factor and can be used to facilitate mole-to-mole conversions within the reaction.
  • Therefore, for a mass-to-mass conversion, it is necessary to first convert one amount to moles, then use the conversion factor to find moles of the other substance, and then convert the molar value of interest back to mass.
  • With the number of moles of butane equal to 54 grams, it is possible to find the moles of O2 that can react with it.
  • This last equation shows that 6.05 moles of O2 can react with 0.929 moles of C4H10.
  • But by converting the butane mass to moles (0.929 moles) and using the molar ratio (13 moles oxygen : 2 moles butane), one can find the molar amount of oxygen (6.05 moles) that reacts with 54.0 grams of butane.

• Molality

  • $\text{ moles KCl} = 5.36 g \times (\frac{1 \text{ moles}}{74.5 g}) = 0.0719 \text{ moles KCl}$
  • $\text{ molality} = (\frac {\text{ moles}}{\text{kg solvent}}) = (\frac {0.0719 \text{ moles KCl}}{0.056\text{ kg water}})= 1.3\ m$
  • $\text{ moles KCN} = 25.0 g \times (\frac {1 \text{ moles}}{65.1 g}) = 0.38 \text{ moles}$
  • We multiply the moles by the reciprocal of the given molality (3.0 moles/kg) so that our units appropriately cancel.
  • $0.38 \text{ moles KCl} \times (\frac {\text{ kg acetic acid}}{3.0 \text{ moles KCl}}) = 0.12\text{ kg acetic acid}$

• Molarity

  • Molarity is defined as the moles of a solute per volume of total solution.
  • First, we must convert the mass of NaCl in grams into moles.
  • $10.0 \text{ grams NaCl} \times \frac{\text{1 mole}}{58.4 \text {g/mole}} = 0.17 \text{ moles NaCl}$
  • $n_i = (0.65 \text{ M})(4.0 \text{ L}) = 2.6 \text{ moles KCl}$
  • $\frac{4.0 \text{ g }BH_3 }{13.84 \text{g/mole }BH_3} = 0.29 \text{ moles }BH_3$

• Using Molarity in Calculations of Solutions

  • In chemistry, molar concentration, or molarity, is defined as moles of solute per total liters of solution.
  • It is important to distinguish moles from molarity; molarity is a measurement of concentration while moles are a measure of the amount of substance present at a given time.
  • How many moles of potassium bromide are contained in the sample?
  • What is the molarity of a solution containing 0.32 moles of NaCl in 3.4 liters of solution?
  • Also, molarity is a ratio that describes the moles of solute per liter of solution.