Question

How many atoms are present in a \(75.0 \mathrm{cm}^{3}\) sample of plumber's solder, a lead-tin alloy containing \(67 \% \mathrm{Pb}\) by mass and having a density of \(9.4 \mathrm{g} / \mathrm{cm}^{3} ?\)

Short answer

The total number of atoms in the sample is approximately \(1.37 \times 10^{24}\) atoms .

Step-by-step solution

Calculate the total mass of the sample

Multiply the density by the volume, \(9.4 \mathrm{g} / \mathrm{cm}^{3} \times 75.0 \mathrm{cm}^{3} = 705 g\). The total mass of the sample is thus 705 g.

Determine the mass of element Pb(Lead)

Multiply the percentage of lead by the total mass, \(67 \% \times 705 g = 472.35\, g\). Thus the mass of Pb in the solder is 472.35 g.

Calculate the number of moles of Pb(Lead)

Divide the mass of lead by the molar mass of lead, \(472.35 g/207.2 g/mol = 2.2806 \, moles\). The number of moles of lead therefore is 2.2806 moles.

Determine the number of atoms of Pb(Lead)

Multiply the number of moles by Avogadro's number \(2.2806 moles \times 6.022 \times 10^{23} atoms/mole = 1.37256872 \times 10^{24}\, atoms\). Hence, the total number of atoms in the sample is \(1.37256872 \times 10^{24}\, atoms\).

Key concepts

Molar Mass

Molar mass is a key concept in stoichiometry that represents the mass of a substance per mole of its entities (atoms or molecules).
It is usually expressed in grams per mole (g/mol). To calculate the molar mass of a substance, such as lead (Pb) in our exercise, it's essential to use the atomic weights found on the periodic table.
The molar mass of lead is 207.2 g/mol, which means that one mole of lead atoms has a mass of 207.2 grams.In stoichiometry, molar mass allows us to convert between the mass of a substance and the number of moles.
This conversion is imperative for calculating quantities in chemical reactions and exercises like the one we have, where we identify how many atoms are in a given mass of a substance.
Using the molar mass, the mass of lead in the sample (472.35 g) is converted into moles by dividing the given mass by the molar mass of lead: \[ \text{Moles of Pb} = \frac{\text{Mass of Pb}}{\text{Molar Mass of Pb}} = \frac{472.35 \, \text{g}}{207.2 \, \text{g/mol}} \approx 2.2806 \, \text{moles} \]

Avogadro's Number

Avogadro's number is a fundamental constant used in chemistry to relate the amount of substance in moles to the number of entities, such as atoms or molecules.
This number, which is approximately \( 6.022 \times 10^{23} \), represents the quantity of atoms, molecules, or other elementary entities in exactly one mole of a substance.In a stoichiometric calculation, once we determine the moles of a substance, Avogadro's number helps us find out how many atoms (or molecules) there are in those moles.
In our problem, Avogadro's number is used to convert the amount of moles of lead into atoms. For example, multiplying the moles of lead (2.2806 moles) by Avogadro's number gives:\[ \text{Number of Pb atoms} = 2.2806 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole} = 1.37256872 \times 10^{24} \, \text{atoms} \] This steps us into the micro world of atoms, bridging macroscopic measurements to atomic scale quantities, a vital aspect of chemistry and stoichiometry.

Density Calculation

Density is a physical property that indicates how much mass is contained in a given volume of a substance. It is expressed in grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³).
To find the mass of a substance from its volume, one can multiply the density by the volume.In this exercise, the density calculation was crucial to finding the total mass of the lead-tin alloy sample.
The density of the alloy was given as 9.4 g/cm³, and the volume was 75.0 cm³. Using these, the total mass of the alloy can be calculated as:\[ \text{Mass} = \text{Density} \times \text{Volume} = 9.4 \, \text{g/cm}^3 \times 75.0 \, \text{cm}^3 = 705 \, \text{g} \] Consequently, knowing the density and volume allowed us to convert from volume to mass, both of which are integral parts of many stoichiometry problems.
The various calculations driven by density also enable chemists to move seamlessly between macroscopic and microscopic understanding of substances.