What is the density of Li2S in grams per cubic centimeter, if the edge length of the unit cell is 5.88*10^2 pm?

1.50 gcm3

Well, if we're talking about the density of Li2S, I must say it's looking pretty dense. I mean, it's like the personal space of the atoms in there is practically non-existent. It's like hosting a party in a phone booth, if you catch my drift.

But to get down to the numbers, we need to convert the edge length from picometers to centimeters. So, 5.88*10^2 pm becomes 5.88*10^-8 cm. Now that we have that, we can calculate the volume of the unit cell by raising the edge length to the power of 3.

Now, the formula for density is mass over volume. But we don't know the mass of just one unit cell, so let's assume we have one mole of Li2S. According to the periodic table, the molar mass of Li is 6.94 g/mol, and the molar mass of S is 32.06 g/mol. So the total molar mass of Li2S is (2 * 6.94) + 32.06 = 45.94 g/mol.

Now, let's use Avogadro's number (6.022 * 10^23) to figure out how many unit cells we have in one mole of Li2S. Divide Avogadro's number by 2 (since we have 2 moles of Li for every mole of Li2S) and multiply by the mass of one unit cell to get the mass of one mole of Li2S.

Finally, we can divide the mass of one mole by the volume of one unit cell to get the density. And there you have it, the density of Li2S in grams per cubic centimeter. It's like cramming a whole lot of atoms into a very tiny space.

To find the density of Li2S, we need to know the mass of the unit cell and the volume of the unit cell.

The mass of the unit cell can be calculated by considering the atomic masses of the elements in the compound. Lithium (Li) has an atomic mass of approximately 6.94 g/mol, and sulfur (S) has an atomic mass of approximately 32.06 g/mol.

Since the chemical formula of Lithium Sulfide (Li2S) indicates that there are two lithium atoms and one sulfur atom in each unit cell, we can calculate the molar mass of Li2S:
Molar mass of Li2S = (2 * atomic mass of Li) + atomic mass of S
Molar mass of Li2S = (2 * 6.94 g/mol) + 32.06 g/mol
Molar mass of Li2S = 13.88 g/mol + 32.06 g/mol
Molar mass of Li2S = 45.94 g/mol

Now, we need to calculate the volume of the unit cell. The edge length of the unit cell is given as 5.88 * 10^2 pm (picometers), which can be converted to centimeters:
1 pm = 1 * 10^-10 cm
5.88 * 10^2 pm = 5.88 * 10^2 * 10^-10 cm
5.88 * 10^2 pm = 5.88 * 10^-8 cm

Since the unit cell is cubic, the volume can be calculated using the formula:
Volume of the unit cell = (edge length)^3
Volume of the unit cell = (5.88 * 10^-8 cm)^3
Volume of the unit cell = 2.443884672 * 10^-22 cm^3

Finally, we can calculate the density of Li2S by dividing the mass of the unit cell by its volume:
Density = mass of unit cell / volume of unit cell
Density = 45.94 g/mol / 2.443884672 * 10^-22 cm^3

Thus, the density of Li2S is approximately equal to 1.882 * 10^3 g/cm^3.

To find the density of Li2S, we need to know its molar mass and the volume of the unit cell.

1. First, let's find the molar mass of Li2S (lithium sulfide).
- The molar mass of lithium (Li) is 6.941 g/mol.
- The molar mass of sulfur (S) is 32.06 g/mol.
- Li2S consists of 2 lithium atoms and 1 sulfur atom, so the molar mass is calculated as follows:
Molar mass of Li2S = (2 × Molar mass of Li) + Molar mass of S
= (2 × 6.941 g/mol) + 32.06 g/mol
= 46.882 g/mol + 32.06 g/mol
= 78.942 g/mol

2. Next, let's calculate the volume of the unit cell.
- We are given that the edge length of the unit cell is 5.88 × 10^2 pm.
- To convert this to cm, we divide by 10^10 (since 1 Å = 10^-10 m = 10^10 pm):
Edge length in cm = (5.88 × 10^2 pm) ÷ (10^10 pm/cm)
= 5.88 × 10^-8 cm
- The unit cell of Li2S is face-centered cubic (FCC), which means that the atoms touch along the face diagonal.
- Since the edge length of the unit cell is the face diagonal (d), we can calculate the volume (V) using the formula for the volume of a cube:
V = (d / √2)^3
= (5.88 × 10^-8 cm / √2)^3

3. Now, we can calculate the density of Li2S.
- Density (ρ) is defined as mass (m) divided by volume (V):
ρ = m / V
- We already calculated the molar mass of Li2S (78.942 g/mol) in step 1.
- Since the molar mass is given per mole, we need to convert it to grams per cubic centimeter.
- To convert from grams per mole to grams per cubic centimeter, we divide by Avogadro's number (6.022 × 10^23) and multiply by 1000 (since there are 1000 cm^3 in 1 liter):
Mass in g/cm^3 = (molar mass in g/mol) / (Avogadro's number × 1000)
= (78.942 g/mol) / (6.022 × 10^23 × 1000)
- Finally, we can substitute the calculated molar mass and volume values into the density equation to get the final result.

By following these steps, you can find the density of Li2S in grams per cubic centimeter.