Problem 87 - Chemistry: An Atoms-Focused Approach

Question

Artificial Bones for Medical Implants The material often used to make artificial bones is the same material that gives natural bones their strength. Its common name is hydroxyapatite, and its formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) a. Propose a systematic name for this compound. b. What is the mass percentage of calcium in it? c. When treated with hydrogen fluoride, hydroxyapatite becomes fluorapatite \(\left[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F}\right]\), an even stronger substance. Does the percent mass of Ca increase or decrease as a result of this substitution?

Step-by-Step Solution

Answer

Answer: The systematic name for hydroxyapatite is pentacalcium tris(phosphate) hydroxide. The mass percentage of calcium in hydroxyapatite is higher (39.92%) compared to fluorapatite (39.75%), so the mass percentage of calcium decreases as a result of the substitution.

Step 1: a. Systematic name of hydroxyapatite

The chemical formula for hydroxyapatite is \(\mathrm{Ca}_5\left(\mathrm{PO}_4\right)_3 \mathrm{OH}\). The systematic name for this compound consists of naming the cations (calcium) and the anions (phosphate and hydroxide) with their proper stoichiometry. In this case, the systematic name for hydroxyapatite is pentacalcium tris(phosphate) hydroxide.

Step 2: b. Mass percentage of calcium in hydroxyapatite

First, we need to find the molar mass of hydroxyapatite. \(\mathrm{Ca}_5\left(\mathrm{PO}_4\right)_3 \mathrm{OH}\) = 5 * (Mass of \(\mathrm{Ca}\)) + 3 * (Mass of \(\mathrm{PO}_4\)) + (Mass of \(\mathrm{OH}\)). Molar mass of \(\mathrm{Ca}\) = 40.08 g/mol Molar mass of \(\mathrm{P}\) = 30.97 g/mol Molar mass of \(\mathrm{O}\) = 16.00 g/mol Molar mass of \(\mathrm{H}\) = 1.01 g/mol Molar mass of \(\mathrm{PO_4}\) = 1 * (Mass of \(\mathrm{P}\)) + 4 * (Mass of \(\mathrm{O}\)) = (30.97) + 4 * (16.00) = 94.97 g/mol Molar mass of \(\mathrm{OH}\) = 1 * (Mass of \(\mathrm{O}\)) + 1 * (Mass of \(\mathrm{H}\)) = 16.00 + 1.01 = 17.01 g/mol So, the molar mass of hydroxyapatite = 5 * (40.08) + 3 * (94.97) + 17.01 = 502.31 g/mol Now, let's calculate the mass percentage of \(\mathrm{Ca}\) in hydroxyapatite. Mass percentage of \(\mathrm{Ca}\) = \(\dfrac{5 * \text{Mass of } \mathrm{Ca}}{\text{Molar mass of hydroxyapatite}} * 100\) Mass percentage of \(\mathrm{Ca}\) = \(\dfrac{5 * (40.08)}{502.31} * 100 \approx 39.92\%\)

Step 3: c. Comparison of mass percentage of calcium between hydroxyapatite and fluorapatite

Let's find the molar mass of fluorapatite. \(\mathrm{Ca}_5\left(\mathrm{PO}_4\right)_3 \mathrm{F}\) = 5 * (Mass of \(\mathrm{Ca}\)) + 3 * (Mass of \(\mathrm{PO}_4\)) + (Mass of \(\mathrm{F}\)). Molar mass of \(\mathrm{F}\) = 19.00 g/mol So, the molar mass of fluorapatite = 5 * (40.08) + 3 * (94.97) + 19.00 = 504.30 g/mol. Now, let's calculate the mass percentage of \(\mathrm{Ca}\) in fluorapatite. Mass percentage of \(\mathrm{Ca}\) = \(\dfrac{5 * \text{Mass of } \mathrm{Ca}}{\text{Molar mass of fluorapatite}} * 100\) Mass percentage of \(\mathrm{Ca}\) = \(\dfrac{5 * (40.08)}{504.30} * 100 \approx 39.75\%\) Comparing the mass percentage values of calcium between hydroxyapatite (39.92%) and fluorapatite (39.75%), the mass percentage of calcium decreases as a result of the substitution.

Key Concepts

Molar Mass Calculation

To understand the chemical composition of a compound, calculating the molar mass is essential. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It acts like a bridge, connecting the atomic world to everyday measurable quantities. For compounds, it's calculated by adding up the molar masses of its constituent atoms, which are taken directly from the periodic table.
In hydroxyapatite, with the formula \(\mathrm{Ca}_5\left(\mathrm{PO}_4\right)_3 \mathrm{OH}\), each element contributes to the total molar mass:

• Calcium (Ca): As there are five calcium atoms, we multiply its atomic mass, 40.08 g/mol, by 5.
• Phosphate group (\(\mathrm{PO}_4\)): Each includes one phosphorus and four oxygens. Calculating one \(\mathrm{PO}_4\) yields 94.97 g/mol, and with three in the compound, we multiply this total by 3.
• Hydroxide (OH): Consisting of one oxygen and one hydrogen, the total is 17.01 g/mol.

Adding these together gives the molar mass of hydroxyapatite: 502.31 g/mol. By using the molar masses of the elements like this, you can determine the mass of any compound.

Mass Percentage

Mass percentage offers insight into the composition of a compound by showing how much each element contributes to the total mass. It's expressed as a percentage, making it an excellent tool for comparing components of mixtures or compounds.
The formula for mass percentage is:\[\text{Mass percentage of Element } = \frac{\text{Total Mass of the Element in One Mole of Compound}}{\text{Molar Mass of Compound}} \times 100\]Here's how we applied this to hydroxyapatite:

• We calculated the molar mass of calcium as part of the whole formula: \(5 \times 40.08 \mathrm{g/mol} = 200.4 \mathrm{g/mol}\).
• Using the previously calculated molar mass of hydroxyapatite (502.31 g/mol), we compute the percentage of calcium. Mass percentage = \(\frac{200.4}{502.31} \times 100 \approx 39.92\%\).

Understanding this percentage helps us see how significant calcium is within hydroxyapatite compared to its total mass.

Fluorapatite

Fluorapatite is a compound similar to hydroxyapatite, distinguished by containing fluorine instead of hydroxide. This substitution is a simple example of how chemical changes can impact the properties of a material, often making it stronger, as seen with fluorapatite used in enamel and artificial bone implants.
In fluorapatite \(\left(\mathrm{Ca}_5\left(\mathrm{PO}_4\right)_3 \mathrm{F}\right)\), the substitution alters the molar mass which affects the mass percentage of calcium. Here's the step-by-step:

• Replacing OH with F slightly increases the compound's total molar mass to 504.30 g/mol.
• Despite having the same quantity of calcium atoms, with the new molar mass, the calcium's mass percentage reduces slightly: \(\approx 39.75\%\).

This small decrease exemplifies how even minor substitutions in a chemical formula can adjust the proportion of elements within a compound, affecting its overall properties.