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Moles, Molar Mass, and Molarity for AP Chemistry

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By — McGraw-Hill Professional
Updated on Apr 25, 2014

Practice problems for these concepts can be found at:

Moles and Molar Mass

The mole (mol) is the amount of a substance that contains the same number of particles as atoms in exactly 12 grams of carbon-12. This number of particles (atoms or molecules or ions) per mole is called Avogadro's number and is numerically equal to 6.022 × 1023 particles. The mole is simply a term that represents a certain number of particles, like a dozen or a pair. That relates moles to the microscopic world, but what about the macroscopic world? The mole also represents a certain mass of a chemical substance. That mass is the substance's atomic or molecular mass expressed in grams. In Chapter 5, the Basics chapter, we described the atomic mass of an element in terms of atomic mass units (amu). This was the mass associated with an individual atom. Then we described how one could calculate the mass of a compound by simply adding together the masses, in amu, of the individual elements in the compound. This is still the case, but at the macroscopic level the unit of grams is used to represent the quantity of a mole. Thus, the following relationships apply:

The mass in grams of one mole of a substance is the molar mass.

The relationship above gives a way of converting from grams to moles to particles, and vice versa. If you have any one of the three quantities, you can calculate the other two. This becomes extremely useful in working with chemical equations, as we will see later, because the coefficients in the balanced chemical equation are not only the number of individual atoms or molecules at the microscopic level, but also the number of moles at the macroscopic level.

How many moles are present in 1.20 × 1025 silver atoms?

Answer:

Molarity and Solution Calculations

We discuss solutions further in the chapter on solutions and colligative properties, but solution stoichiometry is so common on the AP exam that we will discuss it here briefly also. Solutions are homogeneous mixtures composed of a solute (substance present in smaller amount) and a solvent (substance present in larger amount). If sodium chloride is dissolved in water, the NaCl is the solute and the water the solvent.

One important aspect of solutions is their concentration, the amount of solute dissolved in the solvent. In the chapter on solutions and colligative properties we will cover several concentration units, but for the purpose of stoichiometry, the only concentration unit we will use at this time is molarity. Molarity (M) is defined as the moles of solute per liter of solution:

M = mol solute/L solution

Let's start with a simple example of calculating molarity. A solution of NaCl contains 39.12 g of this compound in 100.0 mL of solution. Calculate the molarity of NaCl.

Answer:

Knowing the volume of the solution and the molarity allows you to calculate the moles or grams of solute present.

Next, let's see how we can use molarity to calculate moles. How many moles of ammonium ions are in 0.100 L of a 0.20 M ammonium sulfate solution?

Answer:

Stoichiometry problems (including limiting-reactant problems) involving solutions can be worked in the same fashion as before, except that the volume and molarity of the solution must first be converted to moles.

If 35.00 mL of a 0.1500 M KOH solution is required to titrate 40.00 mL of a phosphoric acid solution, what is the concentration of the acid? The reaction is:

2KOH (aq) + H3PO4 (aq) → K2HPO4 (aq) + 2H2O (l)

Answer:

Practice problems for these concepts can be found at:

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