Question

You combine 1.25 g of germanium, Ge, with excess chlorine, \(\mathrm{Cl}_{2} .\) The mass of product, \(\mathrm{Ge}_{x} \mathrm{Cl}_{y},\) is \(3.69 \mathrm{g}\) What is the formula of the product, Ge, \(\mathrm{Cl}_{y}\) ?

Short answer

The formula of the product is \( \text{GeCl}_4 \).

Step-by-step solution

Determine Moles of Germanium

First, we find the molar mass of germanium (Ge), which is approximately 72.63 g/mol. Then, calculate the moles of germanium by using the formula: \( \text{moles of Ge} = \frac{\text{mass of Ge}}{\text{molar mass of Ge}} \). For 1.25 g of Ge, the moles of Ge are: \( \frac{1.25 \, \text{g}}{72.63 \, \text{g/mol}} \approx 0.0172 \, \text{mol} \).

Determine Mass of Chlorine in the Product

Subtract the mass of germanium from the total mass of the product to find the mass of chlorine used: \( 3.69 \, \text{g} \, - \, 1.25 \, \text{g} = 2.44 \, \text{g} \) of chlorine.

Calculate Moles of Chlorine

The molar mass of chlorine (Cl) is approximately 35.45 g/mol for a single atom, but since Chlorine is diatomic (\( \text{Cl}_2 \)), we use 70.90 g/mol. Therefore, calculate the moles of Cl by: \( \text{moles of Cl} = \frac{\text{mass of Cl}}{\text{molar mass of Cl}} = \frac{2.44 \, \text{g}}{70.90 \, \text{g/mol}} \approx 0.0344 \, \text{mol} \). However, note each mole of chlorine atoms used is twice the moles of \( \text{Cl}_2 \) used, so use 35.45 g/mol for calculations: \( \frac{2.44 \, \text{g}}{35.45 \, \text{g/mol}} \approx 0.0688 \, \text{mol} \).

Determine Ratio of Germanium to Chlorine

To find the formula, we need the simplest mole ratio of Ge to Cl. Divide the moles of each by the smallest number of moles calculated: \( \frac{0.0172}{0.0172} : \frac{0.0688}{0.0172} \approx 1:4 \). This indicates there's one germanium atom for every four chlorine atoms.

Write the Empirical Formula

Based on the mole ratio determined in the previous step, the empirical formula of the product is \( \text{GeCl}_4 \).

Key concepts

Mole Calculation

Mole calculation is a crucial step in many chemical reactions. Understanding it helps you determine how much of each substance is involved in a reaction. A mole is essentially a way to quantify very small particles like atoms or molecules in chemical science. To calculate moles, you need two essential pieces of information:

• The mass of the substance you have.
• The molar mass of the substance, which is the mass of one mole of the substance.

You can find the molar mass on the periodic table, measured in grams per mole (g/mol). In our exercise, we are calculating the moles of germanium, Ge, by using its molar mass (72.63 g/mol). By dividing the mass of the germanium sample (1.25 g) by its molar mass, we obtain the number of moles: \[ \text{Moles of Ge} = \frac{\text{mass of Ge}}{\text{molar mass of Ge}} = \frac{1.25 \, \text{g}}{72.63 \, \text{g/mol}} \approx 0.0172 \, \text{mol} \].
This calculation helps us understand how many Ge atoms are participating in the reaction.

Empirical Formula

The empirical formula is all about finding the simplest ratio of elements in a compound. It's super handy when you're dealing with a new chemical product formed in a reaction. Unlike the molecular formula, which tells you the actual number of atoms of each element in a molecule, the empirical formula gives you the simplest whole number ratio. To calculate it:

• Determine the number of moles of each element in the product.
• Find the simplest integer ratio of these moles.
• Express this ratio as the empirical formula.

In the exercise, we calculated moles for both germanium and chlorine. We found a 1:4 mole ratio for Ge to Cl, leading us to deduce the empirical formula as \( \text{GeCl}_4 \). That means, for every atom of germanium, there are four atoms of chlorine.

Chemical Composition

Chemical composition tells us about the constituent elements in a compound and their proportions. Understanding chemical composition is like getting inside information on how atoms are bonded in a compound. In our specific example, the chemical composition refers to how germanium and chlorine come together to form a new compound. First, determine the mass of each component in the product.
Subtracting the germanium's mass from the total mass of the compound gives us the mass of chlorine. Next, these mass values are converted to moles, which are crucial for identifying the ratio of components. Here, the chemical composition is illustrated via the mole ratio of germanium to chlorine, 1:4, in \( \text{GeCl}_4 \), reflecting the product’s foundation.

Stoichiometry

Stoichiometry, a fundamental part of chemistry, assists in calculating the relationships between reactants and products in a chemical reaction. Think of it as chemistry's way of doing a balance check. It uses the law of conservation of mass, meaning that mass is neither created nor destroyed in a chemical reaction.
The key components of stoichiometry involve:

• Balancing chemical equations to have equal number of each type of atom on both sides.
• Using mole ratios derived from these balanced equations to solve for unknown quantities.

In our exercise, after calculating moles, stoichiometry lets us deduce the empirical formula using these mole ratios. By maintaining the simplest integer ratio after dividing by the smallest number of moles, stoichiometry ensures that every reaction respects the conservation principle, resulting in the final compound \( \text{GeCl}_4 \).