Question

In an experiment \(1.550 \mathrm{~g}\) of mercury oxide decomposed to give oxygen gas and \(1.435 \mathrm{~g}\) of liquid mercury. What is the empirical formula of mercury oxide?

Short answer

The empirical formula of mercury oxide is HgO.

Step-by-step solution

Determine the Mass of Oxygen

The mass of mercury oxide is given as 1.550 g and the mass of mercury produced is given as 1.435 g. The mass of oxygen obtained from the decomposition can be calculated by subtracting the mass of mercury from the mass of mercury oxide: \[\text{Mass of oxygen} = 1.550 \, \text{g} - 1.435 \, \text{g} = 0.115 \, \text{g}\]

Calculate Moles of Each Element

We need to calculate the moles of mercury and oxygen using their respective molar masses. 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 Hg:\[\text{Moles of Hg} = \frac{1.435 \, \text{g}}{200.59 \, \text{g/mol}} \approx 0.00716 \, \text{mol}\]Moles of O:\[\text{Moles of O} = \frac{0.115 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.00719 \, \text{mol}\]

Determine the Simplest Whole Number Ratio

To find the empirical formula, divide the number of moles of each element by the smallest number of moles calculated:\[\text{Ratio for Hg} = \frac{0.00716}{0.00716} = 1\]\[\text{Ratio for O} = \frac{0.00719}{0.00716} \approx 1\]This gives a ratio of 1:1 for Hg to O.

Write the Empirical Formula

Since the ratio of mercury to oxygen is 1:1, the empirical formula of mercury oxide is simply HgO.

Key concepts

Understanding the Mass of Oxygen

When conducting a chemical decomposition experiment, determining the components obtained from the reaction is essential. In this specific experiment, mercury oxide was decomposed to yield mercury and oxygen. To find the mass of oxygen, we used a subtraction method. The total mass of mercury oxide was reported as 1.550 g, while the resultant mercury weighed 1.435 g. By subtracting the mass of mercury from the total mass of mercury oxide, the mass of oxygen released was calculated:

• Mass of mercury oxide = 1.550 g
• Mass of mercury = 1.435 g
• Mass of oxygen = Mass of mercury oxide - Mass of mercury
• Mass of oxygen = 1.550 g - 1.435 g = 0.115 g

Clearly, the mass of oxygen from this reaction was 0.115 g. Understanding this calculation is crucial as it sets the stage for determining the empirical formula by enabling us to calculate the moles of oxygen involved.

Comprehending the Concept of Molar Mass

Molar mass is a vital concept in chemistry, as it helps in the conversion between the mass of a substance and the amount of substance in moles. The molar mass, typically expressed in grams per mole (g/mol), is equivalent to the atomic or molecular weight of a substance.

• Molar Mass of Oxygen (O): 16.00 g/mol
• Molar Mass of Mercury (Hg): 200.59 g/mol

In this exercise, knowing the molar masses allows us to convert the measured masses into moles, which is a standardized way to count particles, atoms, or molecules. This will allow us to derive an empirical formula based on mole ratios. The calculations that follow hinge crucially on understanding these molar mass values.

Grasping Moles Calculation

Calculating moles is essential in chemistry to quantify the number of entities like atoms, ions, or molecules in a substance. This is achieved using the formula:\[\text{Moles} = \frac{\text{Mass of the substance (g)}}{\text{Molar mass (g/mol)}}\]With the given molar masses, we determined the moles of mercury and oxygen:

• Moles of Hg = \(\frac{1.435 \, \text{g}}{200.59 \, \text{g/mol}} \approx 0.00716 \, \text{mol}\)
• Moles of O = \(\frac{0.115 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.00719 \, \text{mol}\)

The next step involves finding a simple whole number ratio between these moles to identify the empirical formula, contributing to our understanding of the compound's structure. Calculating moles ensures a precise approach toward deriving this ratio.