The mole and Avogadro's constant

In a nutshell 

The mole is a unit for measuring the amount of a substance. A mole is a certain number of particles in a substance. The number of particles in one mole is known as Avogadro's constant $$(6.02 \, \times \, 10^{23})$$. The number of moles is calculated using the mass and molar mass or using the concentration and volume. 

Equations  

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Constants  

Variable units 

The mole and Avogadro's constant 

The mole is a unit that is used to measure the amount of a substance. A mole is a certain number of particles in a substance. 

Avogadro's constant ($$6.02 \, \times \, 10^{23}$$) is the value for the number of particles in one mole of a substance. The equation to work out the number of moles given the number of particles and Avogadro's constant is:

$$number \space of \space moles \, = \, \frac{number \space of \space particles}{Avogadro's \space constant} \\ \ \\ n \, = \, \frac{number \space of \space particles}{L}$$

This equation can be rearranged to find the number of particles given the number of moles.

​$$number \space of \space particles \, = \, number \space of \space moles \, \times \, Avogadro's \space constant \\ \ \\ number \space of \space particles \, = \, n \, \times \, L$$​​

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Example 

How many moles is there in $$3.75 \, \times \, 10^{18}$$ lithium atoms?

First, write down the equation you will use:

$$n \, = \, \frac{number \space of \space atoms}{L}$$

Write the value of Avogadro's constant:

$$6.02 \, \times \, 10^{23}$$​​

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Substitute the values into the equation:

$$n \, = \, \frac{3.75 \, \times \, 10^{18}}{6.02 \, \times \, 10^{23}} \, = \, 6.23 \, \times \, 10^{-6} \,mol$$​​

There is $$\underline{6.23 \, \times \, 10^{-6} \,moles}$$ in $$3.75 \, \times \, 10^{18}$$ lithium atoms. 

Moles 

Molar mass is the mass per mole of a substance. The molar mass can be used alongside the mass to calculate the number of moles of a substance:

​$$number \space of \space moles \, = \, \frac{mass}{molar \space mass} \newline n \, = \, \frac{m}{M}$$

This equation can be rearranged to find either the mass or molar mass of a substance using for formula triangle.  

Rearrangement using the formula triangle:

​$$mass \, = number \space of \space moles \, \times \, molar \space mass \\ \ \\ m \, = \, n \, \times \, M \\ \ \\ molar \space mass \, = \, \frac{mass}{number \space of \space moles} \\ \ \\ M \, = \, \frac{m}{n}$$​​

Example

How many moles are there in $$4.12 \, g$$ of $$Al_2(CO_3)_3$$?

First, write down the equation you will use:

$$n \, = \, \frac{m}{M}$$​​

Next, work out the molar mass of $$Al_2(CO_3)_3$$:

$$M \, (Al_2(CO_3)_3) \, = \, (27 \, \times \, 2)\, + \,(12 \, \times \, 3) \, + \, (16 \, \times \, 9)\, = 234 \, g \, mol^{-1}$$

Substitute the value for the mass and the molar mass into the equation to find the number of moles: 

$$n \, = \, \frac{4.12}{234} \, = 0.018 \, mol$$​​

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There is $$\underline{0.018 \, moles}$$ in $$4.12 \, g$$ of $$Al_2(CO_3)_3$$.

Finding the number of atoms 

Both the equations to find the number of moles and the number of particles can be linked together to find the number of atoms of a substance. 

procedure 

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Concentration

The number of moles can also be calculated using the concentration of a substance along with the volume. 

​$$number \space of \space moles \, = \, concentration \, \times \, volume \newline n \, = \, c \, \times \, V$$

Note: Using the number of moles, the mass can be calculated using the molar mass ($$M$$).

​This equation can be rearranged to find either the concentration or volume of a substance using the formula triangle.     ​

Rearrangement using the formula triangle:

​$$concentration \, = \, \frac{number \space of \space moles}{volume} \\ \ \\ c \, = \, \frac{n}{V} \\ \ \\ volume \, = \, \frac{number \space of \space moles}{concentration} \\ \ \\ V \, = \, \frac{n}{c}$$​​

Example

What is the mass of $$150 \, cm^3$$ of $$HCl$$ with a concentration of $$0.5 \, mol \, dm^{-3}$$?

First, write down your quantities in the correct units:

$$V \, = \, 150 \, cm^3 \, = \, \frac{150 \, cm^3}{1000} \, = \, 0.15 \, dm^3$$

$$c \, = \, 0.5 \, mol \, dm^{-3}$$​​

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Next, write down the equation you will use to find the number of moles:

$$n \, = \, c \, \times \, V$$

Substitute your values into the equation to find the number of moles:

$$n \, = \, 0.15 \, \times \, 0.5 \, = \, 0.075 \, mol$$​​

Write down the equation you will use to find the mass of $$HCl$$: 

$$m \, = \, n \, \times \, M$$​​

Work out the molar mass ($$M_r$$) of $$HCl$$:

$$M \,(HCl) \, = \, 1 \, + \, 35.5 \, = \, 36.5 \, g \, mol^{-1}$$​​

Substitute your values into the equation to find the mass:

$$m \, = 0.075 \, \times \, 36.5 \,= \, 2.74 \, g$$​​

The mass of $$150 \, cm^3$$​ $$HCl$$​ ($$0.5 \, mol \, dm^{-3}$$) is $$\underline{2.74 \, g}$$.​

Concentration in grams 

The concentration can also be measured using the mass.

​$$mass \, = \, concentration \, \times \, volume \newline m \, = \, c \, \times \, V$$

As with the equation involving the number of moles, this equation can be rearranged to find either the concentration or volume of a substance.

Rearrangement: 

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​$$concentration \, = \, \frac{mass}{volume} \\ \ \\ c \, = \, \frac{m}{V} \\ \ \\ volume \, = \, \frac{mass}{concentration} \\ \ \\ V \, = \, \frac{m}{c}$$​​

Note: Instead of concentration having the unit $$mol \, dm^{-3}$$, when mass is used to calculate concentration the unit is $$g \, dm^{-3}.$$

Example 

What is the mass of $$5000 \, cm^3$$​ of sodium hydroxide with a concentration of $$1.25 \, g \, dm^{-3}$$?

First, write down your quantities in the correct units:

$$V \, = \, 5000 \, cm^3 \, = \, \frac{5000 \, cm^3}{1000} \, = 5 \, dm^3$$

$$c \, = \, 1.25 \, g \, dm^{-3}$$​​

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Next, write down the equation you will use to find the mass of sodium hydroxide:

$$m \, = \, c \, \times \, V$$​​

Substitute your values into the equation to find the mass:

$$m \, = \, 5 \, \times \, 1.25 \, = \, 6.25 \ g$$

There is a mass of $$\underline{6.25 \, g}$$ sodium hydroxide ($$1.25 \, g \, dm^{-3}$$) in a volume of $$5000 \, cm^3$$.​