Q.Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate $Ca_3(PO_4)_2$.

Concept & Intuition

When we are given the percentage by weight of each element in a compound, the “simplest whole‑number ratio” is found by converting those masses into moles. Why? Because chemical formulas count atoms, not grams. The mole is the bridge between mass and number of atoms. Once we have the mole amounts, we divide by the smallest to get the smallest integer ratio.

Step‑by‑step reasoning

1. Assume a 100 g sample – Percentages become grams directly.

   • Carbon: $10.06\%$ → $10.06\ \text{g}$
   • Hydrogen: $0.84\%$ → $0.84\ \text{g}$
   • Chlorine: $89.10\%$ → $89.10\ \text{g}$

2. Convert each mass to moles using atomic masses (C = 12.01, H = 1.008, Cl = 35.45).

   • Moles of C: $\displaystyle \frac{10.06}{12.01} \approx 0.8376$
   • Moles of H: $\displaystyle \frac{0.84}{1.008} \approx 0.8333$
   • Moles of Cl: $\displaystyle \frac{89.10}{35.45} \approx 2.513$

3. Find the smallest mole value – Here it is hydrogen: $0.8333$ mol (very close to carbon’s 0.8376, but slightly smaller).

4. Divide each mole amount by the smallest to get a ratio:

   • C: $\displaystyle \frac{0.8376}{0.8333} \approx 1.005$ → essentially 1
   • H: $\displaystyle \frac{0.8333}{0.8333} = 1$ …

The key here is to understand what mole fraction actually tells you. It’s a ratio of moles — not masses. So when you’re given a mole fraction of 0.9 for H₂SO₄ in water, it means that out of every 10 total moles in the solution, 9 are H₂SO₄ and 1 is H₂O. That’s the starting point.

Mass percent, on the other hand, is a ratio of masses. So you need to convert those moles into grams using the molar masses, then find what fraction of the total mass is acid.

Let’s walk through it.

1. Interpret the mole fraction.

   Mole fraction of H₂SO₄, $x_{\text{H}_2\text{SO}_4} = 0.9$.

   This means $x_{\text{H}_2\text{O}} = 1 - 0.9 = 0.1$.

   The simplest way to work: assume a total of 1 mole of solution. Then:

   • Moles of H₂SO₄ = $0.9$ mol
   • Moles of H₂O = $0.1$ mol

   (You could also scale it to 10 total moles — same result.)

2. Find the masses.

   Molar mass of H₂SO₄:

   $2 \times 1 + 32 + 4 \times 16 = 2 + 32 + 64 = 98$ g/mol

   Mass of H₂SO₄ = $0.9 \times 98 = 88.2$ g

   Molar mass of H₂O: $2 \times 1 + 16 = 18$ g/mol

   Mass of H₂O = $0.1 \times 18 = 1.8$ g

   Total mass of solution = $88.2 + 1.8 = 90.0$ g

3. Calculate mass percent. …

The key to this problem is understanding the geometry of a hexagonal close-packed (hcp) lattice and how tetrahedral voids relate to the number of atoms in the lattice. Many students memorise formulas without seeing why they work, so let’s build the reasoning from the ground up.

In any close-packed structure — whether hcp or ccp (fcc) — each sphere in the lattice touches its neighbours in a way that leaves gaps, or voids, between them. There are two types: octahedral voids and tetrahedral voids. For every atom in a close-packed lattice, there is exactly one octahedral void and two tetrahedral voids. This is a fixed geometric fact, not a coincidence — it comes from how the layers stack.

So if element X forms an hcp lattice, the number of X atoms is the number of lattice points. Let that number be $n$. Then the number of tetrahedral voids available is $2n$.

Now, element Y occupies two-thirds of these tetrahedral voids. That means:

We now have the ratio of Y to X:

$\frac{\text{Y}}{\text{X}} = \frac{4n/3}{n} = \frac{4}{3}$ …

When elements combine to form crystalline solids, one type of atom often forms a regular lattice structure, and the other type of atom occupies the "empty spaces" or voids within that lattice. To determine the chemical formula of such a compound, we need to find the ratio of the number of atoms of each element present in the unit cell.

The key concepts here are:

1. Hexagonal Close-Packed (hcp) Lattice: This is a type of close-packed structure where atoms are arranged in a hexagonal pattern. In an hcp unit cell, the effective number of atoms is 6. These atoms form the basic framework of the crystal.
2. Voids in Close-Packed Structures: In any close-packed structure (like hcp or ccp/fcc), there are two main types of interstitial voids:
   • Octahedral voids: These are surrounded by 6 atoms. The number of octahedral voids is equal to the effective number of atoms in the lattice.
   • Tetrahedral voids: These are surrounded by 4 atoms. The number of tetrahedral voids is twice the effective number of atoms in the lattice.

In this problem, element B forms the hcp lattice, and element A occupies a fraction of the tetrahedral voids. By calculating the effective number of B atoms and then the number of A atoms based on the void occupation, we can establish their ratio and thus the compound's formula.

Here's how we determine the formula:

1. Determine the effective number of B atoms:

   Element B forms the hcp lattice. For an hcp unit cell, the effective number of atoms is 6.

   So, the number of B atoms per unit cell, $N_B = 6$.

2. Calculate the total number of tetrahedral voids: …

The density of a gas is directly related to its molar mass, temperature, and pressure. This relationship is derived from the ideal gas law, which describes the behavior of most gases under typical conditions. By knowing the density of a gas at specific temperature and pressure, we can determine its molar mass and, consequently, its identity.

Here's how to approach this problem:

1. Understand the Ideal Gas Law and its relation to density.

   The ideal gas law is given by $PV = nRT$, where:

   • $P$ is pressure
   • $V$ is volume
   • $n$ is the number of moles
   • $R$ is the ideal gas constant
   • $T$ is temperature in Kelvin

   We know that the number of moles ($n$) can be expressed as the mass ($m$) of the gas divided by its molar mass ($M$): $n = \frac{m}{M}$.

   Substituting this into the ideal gas law gives:

   $PV = \frac{m}{M}RT$

   Rearranging this equation to solve for density ($\rho = \frac{m}{V}$):

   $P = \frac{m}{V} \frac{RT}{M}$

   $P = \rho \frac{RT}{M}$

   Finally, we can express density in terms of molar mass, pressure, and temperature:

   $\rho = \frac{PM}{RT}$

   Or, to find the molar mass:

   $M = \frac{\rho RT}{P}$

2. Identify the given values and standard conditions.

   The problem provides the following information:

   • Density ($\rho$) $= 1.24~\mathrm{g/L}$
   • Temperature ($T$) $= 0~^\circ\mathrm{C}$
   • Pressure ($P$) $= 1~\mathrm{atm}$

   These conditions ($0~^\circ\mathrm{C}$ and $1~\mathrm{atm}$) are known as Standard Temperature and Pressure (STP).

   Important

   For calculations involving the ideal gas law, temperature must always be in Kelvin.

   $T(\mathrm{K}) = T(^\circ\mathrm{C}) + 273.15$

   So, $T = 0~^\circ\mathrm{C} + 273.15 = 273.15~\mathrm{K}$.

   We need to choose the appropriate value for the ideal gas constant ($R$). Since pressure is in atmospheres ($\mathrm{atm}$) and volume is implied in liters ($\mathrm{L}$) from the density unit ($\mathrm{g/L}$), we use:

   $R = 0.0821~\mathrm{L \cdot atm / (mol \cdot K)}$

3. Calculate the molar mass ($M$) of the gas.

   Using the rearranged formula $M = \frac{\rho RT}{P}$:

   $M = \frac{(1.24~\mathrm{g/L}) \times (0.0821~\mathrm{L \cdot atm / (mol \cdot K)}) \times (273.15~\mathrm{K})}{(1~\mathrm{atm})}$

   Let's perform the calculation:

   $M = 1.24 \times 0.0821 \times 273.15~\mathrm{g/mol}$

   $M \approx 27.79~\mathrm{g/mol}$ …