Problem 55 - Chemistry: An Atoms-Focused Approach

Question

Surgical-Grade Titanium Medical implants and high-quality jewelry items for body piercings are frequently made of a material known as G23Ti, or surgical- grade titanium. The percent composition of the material is \(64.39 \%\) titanium, \(24.19 \%\) aluminum, and \(11.42 \%\) vanadium. What is the empirical formula of surgical-grade titanium?

Step-by-Step Solution

Answer

The empirical formula for surgical-grade titanium is \(Ti_6Al_4V\).

Step 1: Convert percentages to grams

Given the percent composition, we will first convert each percentage into grams. However, the percentage does not tell us the total mass. So, let's assume that we have 100 g of surgical-grade titanium. In that case, we would have: - 64.39 g of titanium (Ti) - 24.19 g of aluminum (Al) - 11.42 g of vanadium (V)

Step 2: Convert grams to moles

Next, we will convert the mass of each element to moles using their respective molar masses. The molar masses of titanium (Ti), aluminum (Al), and vanadium (V) are 47.87 g/mol, 26.98 g/mol, and 50.94 g/mol, respectively. So, after compute, we have: - Titanium (Ti): \(\frac{64.39\text{ g}}{47.87\text{ g/mol}} = 1.345\text{ mol}\) - Aluminum (Al): \(\frac{24.19\text{ g}}{26.98\text{ g/mol}} = 0.8965\text{ mol}\) - Vanadium (V): \(\frac{11.42\text{ g}}{50.94\text{ g/mol}} = 0.2242\text{ mol}\)

Step 3: Find the smallest whole number ratio

We will now find the smallest whole number ratio between the moles of each element. In this case, divide each mole value by the smallest mole value: - Ti: \(\frac{1.345}{0.2242} = 6\) - Al: \(\frac{0.8965}{0.2242} = 4\) - V: \(\frac{0.2242}{0.2242} = 1\)

Step 4: Write the empirical formula

Based on the whole number ratios found above, the empirical formula for surgical-grade titanium is: $$ Ti_6Al_4V $$ So the empirical formula for surgical-grade titanium is \(Ti_6Al_4V\).

Key Concepts

Percent Composition

Percent composition is essentially a breakdown of the component elements that make up a compound, expressed in percentages. This is crucial because it allows us to understand the proportion of each element in the compound. For example, surgical-grade titanium has a percent composition of 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.

To calculate percent composition, you typically assume you have 100 g of the compound, making it easier to convert percentages directly into grams. In the example given, this assumption means you automatically have 64.39 g of titanium, 24.19 g of aluminum, and 11.42 g of vanadium. By using the percent composition, you can proceed to further calculations like finding the empirical formula.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Knowing the molar mass of elements involved in a compound is essential for converting mass into moles—a fundamental step in chemical calculations.

For instance, in the surgical-grade titanium with titanium (Ti), aluminum (Al), and vanadium (V) as elements, their molar masses are 47.87 g/mol, 26.98 g/mol, and 50.94 g/mol, respectively. By dividing the mass of each element by its molar mass, you convert grams into moles. This step is necessary for determining the empirical formula as it transforms mass-based measurements into the molar amounts needed for stoichiometric calculations.

Chemical Calculations

Chemical calculations involve converting known values into needed quantities using mathematical methods. In the context of determining an empirical formula, you employ chemical calculations to convert mass percentages into moles.

Consider our example of surgical-grade titanium: You start with mass values computed from the percentage composition and then convert these to moles using their respective molar masses. After finding moles for each element, the next calculation involves determining the smallest whole number ratio. This is achieved by dividing the number of moles of each element by the smallest mole value obtained. These calculations allow the formulation of a precise empirical formula.

Stoichiometry

Stoichiometry is the art of chemical bookkeeping. It's about understanding the quantitative relationships in chemical reactions and compounds. In our example with G23Ti or surgical-grade titanium, stoichiometry is used to deduce the simplest mole ratio leading to the empirical formula.

It starts by identifying the amount of each element in moles. These mole values inform how many times each element's atom appears in the simplest form of the compound. For titanium, aluminum, and vanadium, the relative stoichiometric coefficients (derived from the mole ratios) are found to be 6:4:1—leading to the empirical formula of \(Ti_6Al_4V\). This approach ensures the formula accurately depicts the simplest integer ratio of atoms in the compound.