Essential Knowledge

◼ One cannot count particles directly while performing laboratory work. Thus, there must be a connection between the masses of substances reacting and the actual number of particles undergoing chemical changes. (SPQ-1.A.1)

◼ Avogadro’s number (NA = 6.022 x 1023 particles/mole) provides the connection between the number of moles in a pure sample of a substance and the number of constituent particles (or formula units) of that substance. (SPQ-1.A.2)

◼ Expressing the mass of an individual atom or molecule in atomic mass units (amu) is useful because the average mass in amu of one particle (atom or molecule) or formula unit of a substance will always be numerically equal to the molar mass of that substance in grams. Thus, there is a quantitative connection between the mass of a substance and the number of particles that the substance contains. (SPQ-1.A.3)

◼ One cannot count particles directly while performing laboratory work. Thus, there must be a connection between the masses of substances reacting and the actual number of particles undergoing chemical changes. (SPQ-1.A.1)

How many moles of Lead (II) iodide, PbI2, are there in a 25.0 gram sample?

We'll need two key pieces of information:

Mass of the sample: 25.0 grams PbI2 (given in the problem)
Molar mass of PbI2: This is the mass of one mole of PbI2. We don't have this value memorized, but we can find it using the periodic table.

The molar mass is essentially the sum of the atomic masses of all the atoms in a molecule. Lead (Pb) has an atomic mass of 207.2 g/mol, Iodine (I) has an atomic mass of 126.9 g/mol, and there are two Iodine atoms in each molecule of PbI2.

Molar mass of PbI2 = (1 x atomic mass of Pb) + (2 x atomic mass of I)

= (1 x 207.2 g/mol) + (2 x 126.9 g/mol)

≈ 461.0 g/mol (round to one decimal place)

Now, we can convert grams of PbI2 to moles using the following conversion factor:

Conversion factor = $\frac{1 \text{ mole PbI2}}{461.1 \text{ g PbI2}}$

Now, multiply the mass of the sample by this conversion factor to find the moles of PbI2:

moles of PbI2 = mass of PbI2 x conversion factor

= 25.0 g PbI2 x $\frac{1 \text{ mole PbI2}}{461.1 \text{ g PbI2}}$

Perform the calculation:

moles of PbI2 = 25.0 g PbI2 x $\frac{1 \text{ mole PbI_2}}{461.1 \text{ g PbI_2}}$

= 0.0542 moles PbI2 (rounded to five decimal places)

Answer: 0.0542 mol PbI2

Try it yourself …

(1.1.1)

Calculate the number of moles of $O_3$ present in $4.3 × 10^{24}$ molecules of $O_3$.

Check you answer here.

◼ Avogadro’s number (NA = 6.022 x 1023 particles/mole) provides the connection between the number of moles in a pure sample of a substance and the number of constituent particles (or formula units) of that substance. (SPQ-1.A.2)

How many atoms of lead, Pb, are in the sample of 0.320 grams?

Step 1: Convert grams of Pb to moles of Pb

We will utilize the concept of molar mass, which represents the mass of one mole of a substance. The molar mass of lead (Pb) is 207.2 grams per mole (g/mol). This information can be found on the periodic table.

Now, let's set up a conversion factor to transform grams of Pb to moles of Pb. We can write it as:

$$ \frac{1 \text{ mole Pb}}{207.2 \text{ g Pb}} $$

We have the mass of the lead sample (0.320 grams) and want to find the number of moles. So, we can multiply the mass by the conversion factor as shown below:

moles of Pb = mass of Pb $\times$ conversion factor

= 0.320 g Pb $\times$ $\frac{1 \text{ mole Pb}}{207.2 \text{ g Pb}} $

Step 2: Convert moles of Pb to atoms of Pb

Avogadro's constant (Avogadro's number) represents the number of atoms in one mole of any substance, and its value is approximately 6.022 x 10^23 atoms/mole.

We can use Avogadro's constant as a conversion factor to convert moles of Pb to atoms of Pb.

$$ \frac{6.022 \times 10^{23} \text{ atoms Pb}}{1 \text{ mole Pb}} $$

Now, multiply the moles of Pb we obtained in step 1 by this conversion factor to find the number of atoms of Pb:

atoms of Pb = moles of Pb $\times$ conversion factor

= (moles of Pb) $\times$ $\frac{6.022 \times 10^{23} \text{ atoms Pb}}{1 \text{ mole Pb}} $

Answer: There are approximately 9.30×10^20 atoms of Pb in the sample.

Try it yourself …

(1.1.2)

Calculate the number of formula units of $Na_2SO_4$ present in 0.248 g $Na_2SO_4$.

Check you answer here.

◼ Expressing the mass of an individual atom or molecule in atomic mass units (amu) is useful because the average mass in amu of one particle (atom or molecule) or formula unit of a substance will always be numerically equal to the molar mass of that substance in grams. Thus, there is a quantitative connection between the mass of a substance and the number of particles that the substance contains. (SPQ-1.A.3)

Given 5.0 gram samples of NaF, NaBr, NaI and NaCl, place them in order of least to greatest number of atoms of sodium, Na.

To arrange the samples from least to greatest in terms of the number of atoms of Na, you must first determine the number of moles of Na first. Since Na is in each compound as 1:1 ratio, calculating the number of moles of the compound will be sufficient for analysis. (An example of Na not in 1:1 ratio in Na2O .. where there are 2 Na for every 1 O.) Although the question asks for the number of atoms of Na - we can just compare the moles of Na atoms because it is proportional to the number of moles and the same size sample is taken. There are 6.022 x 1023 atoms in 1 mole. See the work below.

$ \text{Number of moles} = \frac{\text{ Mass }}{\text{ Molar mass }} $

NaF, its molar mass = 41.98 g/mol

$\text{number of moles} = \frac{5.0 \, g}{41.98 \, \frac{g}{\text{mol}}} = 0.119 \, \text{mol} \, \text{NaF} $

$0.119 \, \text{mol} \, \text{NaF} \times \frac{6.022 \cdot 10^{23} \, \text{molecules} \, \text{NaF}}{1 \, \text{mol} \, \text{NaF}} \times \frac{1 \, \text{atom} \, \text{Na}}{1 \, \text{molecule} \, \text{NaF}} = 7.66 \times 10^{22} \, \text{atoms} \, \text{Na}$

NaBr, its molar mass = 102.89 g/mol

$\text{number of moles} = \frac{5.0 \, g}{102.89 \, \frac{g}{\text{mol}}} = 0.0485 \, \text{mol} \, \text{NaBr} $

$0.0485 \, \text{mol} \, \text{NaBr} \times \frac{6.022 \cdot 10^{23} \, \text{molecules} \, \text{NaBr}}{1 \, \text{mol} \, \text{NaBr}} \times \frac{1 \, \text{atom} \, \text{Na}}{1 \, \text{molecule} \, \text{NaBr}} = 2.92 \times 10^{22} \, \text{atoms Na}$

NaI, its molar mass = 39.99 g/mol

$\text{number of moles} = \frac{5.0 \, g}{39.99 \, \frac{g}{\text{mol}}} = 0.125 \, \text{mol} \, \text{NaI} $

$0.125 \, \text{mol} \, \text{NaI} \times \frac{6.022 \cdot 10^{23} \, \text{molecules} \, \text{NaI}}{1 \, \text{mol} \, \text{NaI}} \times \frac{1 \, \text{atom} \, \text{Na}}{1 \, \text{molecule} \, \text{NaI}} = 7.53 \times 10^{22} \, \text{atoms Na}$

NaCl, its molar mass = 58.44 g/mol

$\text{number of moles} = \frac{5.0 \, g}{58.44 \, \frac{g}{\text{mol}}} = 0.0855 \, \text{mol} \, \text{NaCl} $

$0.0855 \, \text{mol} \, \text{NaI} \times \frac{6.022 \cdot 10^{23} \, \text{molecules} \, \text{NaCl}}{1 \, \text{mol} \, \text{NaCl}} \times \frac{1 \, \text{atom} \, \text{Na}}{1 \, \text{molecule} \, \text{NaCl}} = 5.15 \times 10^{22} \, \text{atoms Na}$

Therefore, just looking at moles of the compound, the number of Na atoms in each compound is proportional to this number. Therefore it can be ordered from least to greatest to determine the solution to this problem: NaI < NaBr < NaCl < NaF

In green shows the full work in converting exactly to atoms of Na. The same numbers in this calculation would be used for each of the other lines (the formula of the compound would need to change accordingly).

Try it yourself …

(1.1.3)

Calculate the number of atoms of Si present in 35.0 mol Si.

Check you answer here.

1.1 Moles and Molar Mass

Essential Knowledge