A mole calculation in a solution requires using the molarity formula. The volume of the solution and the solution concentration is needed.

## Molarity Definition and Formula

Molarity is the number of moles of solute per liter of solution. A solute, which can be solid, liquid or gas, is a substance that is dissolved in a solvent. The solvent is another substance that is capable of dissolving it within its intermolecular spaces. Together, the dissolved solute and the solvent make a solution.

Molarity is also considered molar concentration because it is the measure of a concentration of a solution. The formula for molarity may be expressed as:

**M _{ }*=* mol**

**/ L**

- M is molarity
- Mol is the moles of the solute
- L is the liters of solution

## Moles: Chemistry and Counting Atoms and Molecules

To fully understand molarity, the mole concept must be understood. A mole (often abbreviated to mol) is a unit of measurement. It is a certain amount. If a dozen bagels were purchased, the amount, if counted, would be 12 bagels.

A mole, like the word dozen, denotes a particular amount, too. However, the amount, called Avogadro’s number, is very large: 6.022 × 10^{23}.

^{}If a mole of bagels were purchased, they would nearly fill the interior space of the Earth. Although a mole of anything can be counted, it is usually reserved for incredibly small items, like atoms and molecules.

## Mole Concept in Solutions

One mole of any element or chemical compound is always the same number. One mole of hydrogen would mean there are 6.022 × 10^{23} atoms of hydrogen.

A mole of sodium chloride, NaCl, is the same amount, 6.022 × 10^{23}. Here, however, it is 6.022 ×10^{23} molecules. With molarity, consider the moles of solute as finding the number of molecules in solution.

## Calculate Moles in Solution

The values for concentration and liters of solution must be given or calculated.

**Example problem**: Sugar, or sucrose, easily dissolves in water. How many moles of sucrose are in a 0.02 M solution?

## Step 1: Locate Molarity and Liters of Solution

In the problem, the molar concentration, M, is given: 0.02 M. The volume is assumed to be 1 L since the definition of molarity is moles of solute per liter of solution.

## Step 2: Use Molarity Formula

Use the formula from “Molarity Definitions and Formula” (above) to solve for moles:

M _{} = mol / L

Rearranging to solve for moles of solute:

**mol _{ }*=* M**

**×**

**L**

mol = 0.02 mol/L × 1 L = **0.02 mol of sucrose,** **C _{12}H_{22}O_{11}**

There are 0.02 moles of sucrose in a 0.02 M sucrose solution.

## Using Number of Moles to Find Grams

Commonly, a question asks for grams of solute, especially if the substance must be measured in a laboratory setting. If the question asks for how many grams of sucrose must be added to make a 0.02 M solution, these additional steps may be followed:

## Step 3: Find Molar Mass

While the counted amount of a mole of any substance is 6.022 x 10^{23}, the molar mass of that substance will be different. For example, sodium chloride, NaCl, will have a different mass than table sugar, sucrose, C_{12}H_{22}O_{11}.

Every element has a different molar mass, commonly located under the symbol on a periodic table. For example, one mole of carbon (C) has a mass of 12.01 g/mol. The molar mass of hydrogen (H) is 1.01 g/mol, and oxygen (O) is 16.00 g/mol.

For example, the molar mass of sucrose would be calculated by adding the molar masses of the individual elements:

- Sucrose contains 12 carbon atoms: 12 × 12.01 g/mol = 144.12 g/mol
- Sucrose contains 22 hydrogen atoms: 22 × 1.01 = 22. 22g/mol
- Sucrose contains 11 oxygen atoms: 11 × 16.00 = 176 g/mol

Add all the individual components of sucrose together:

144.12 g/mol + 22.22 g/mol + 176 g/mol = 342.34 g/mol

The molar mass of sucrose, **C _{12}H_{22}O_{11}, is 342.34 g/mol**

## Step 4: Find Grams of Solute

Use the number of moles calculated in Step 2 and the molar mass of sucrose from Step 3 to solve for grams:

0.02 mol of C_{12}H_{22}O_{11 }× 342.34 g C_{12}H_{22}O_{11} / 1 mol C_{12}H_{22}O_{11} = 6.85 g C_{12}H_{22}O_{11}

6.85 grams of sucrose dissolved in water will make a 0.01 M solution.