Question

When \(4.01 \mathrm{g}\) of mercury is strongly heated in air, the resulting oxide weighs \(4.33 \mathrm{g}\). Calculate the empirical formula of the oxide.

Short answer

The empirical formula of the mercury oxide is HgO.

Step-by-step solution

Calculate the mass of oxygen

First, we need to determine the mass of oxygen that combined with the mercury after being heated. To do this, we subtract the initial mass of the mercury from the final mass of the oxide: Mass of oxygen = Final mass of oxide - Initial mass of mercury Mass of oxygen = 4.33 g - 4.01 g Mass of oxygen = 0.32 g

Convert masses to moles

To find the mole ratio of mercury and oxygen in the oxide, we need to convert the masses of mercury and oxygen to moles. The molar mass of mercury (Hg) is 200.59 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. Moles of mercury (Hg) = (mass of mercury) / (molar mass of mercury) Moles of mercury (Hg) = 4.01 g / 200.59 g/mol Moles of mercury (Hg) = 0.02 mol Moles of oxygen (O) = (mass of oxygen) / (molar mass of oxygen) Moles of oxygen (O) = 0.32 g / 16.00 g/mol Moles of oxygen (O) = 0.02 mol

Find the mole ratio

Now, we need to find the ratio of moles of mercury to moles of oxygen. Ratio of moles (Hg/O) = (moles of Hg) / (moles of O) Ratio of moles (Hg/O) = 0.02 mol / 0.02 mol Ratio of moles (Hg/O) = 1/1

Determine the empirical formula

Since the mole ratio of mercury and oxygen is 1:1, the empirical formula of the mercury oxide must be HgO.

Key concepts

Mole Ratio

The concept of mole ratio plays a crucial role in understanding chemical formulas and reactions. In the context of this exercise, the mole ratio helps us determine the empirical formula of a compound, specifically mercury oxide. A mole ratio is the ratio between the number of moles of any two substances involved in a chemical reaction. This ratio is derived from the coefficients of a balanced chemical equation, or as in this exercise, from the moles calculated using given masses and their molar masses. Here, we calculated the moles of mercury and oxygen from their respective masses and molar masses to determine the mole ratio. The steps included:

• Finding the number of moles of mercury: 0.02 mol
• Finding the number of moles of oxygen: 0.02 mol
• Comparing these moles to establish a mole ratio.

This mole ratio of 1:1 indicates that for every mole of mercury, there is one mole of oxygen. Thus, it signals that mercury and oxygen are present in equal amounts. This 1:1 ratio ultimately leads us to the empirical formula for mercury oxide, HgO.

Molar Mass

Molar mass is a fundamental concept in chemistry, central to converting between mass and moles of a substance. It represents the mass of one mole of a given element or compound and is typically expressed in grams per mole (g/mol).In this exercise, two molar masses were key:

• Molar mass of mercury (Hg): 200.59 g/mol
• Molar mass of oxygen (O): 16.00 g/mol

To find out how many moles are present in given masses, you divide the mass by the molar mass of the element. For example, to calculate the moles of mercury and oxygen, we used the formula:\[\text{Moles} = \frac{\text{Mass of element}}{\text{Molar Mass of element}}\]This allows us to transform mass values into mole values, giving us a better understanding of the proportions of each element in a chemical reaction or compound, crucial for further steps in chemical analysis.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry involves using balanced chemical equations to quantify the relationships between reactants and products in a chemical reaction. It links the quantities of different substances involved in the reaction based on their mole ratios, making it possible to predict how much product can be formed from given reactants. Stoichiometry requires accurate mole quantities, which depend on understanding and using molar masses effectively. In this exercise, stoichiometry was applied to find the empirical formula of a compound by following these steps:

• Calculate the mass of each element in the compound.
• Convert these masses into moles using their molar masses.
• Determine the simplest ratio of moles to arrive at the empirical formula.

The process of stoichiometry helped identify that mercury and oxygen combined in a 1:1 ratio in the compound, leading to the empirical formula HgO. This reflects how stoichiometry is critical in transforming masses to moles, then to a chemical formula, effectively bridging qualitative and quantitative chemical analysis.