Question

Analyzing Data In the laboratory, a sample of pure nickel was placed in a clean, dry, weighed crucible. The crucible was heated so that the nickel would react with the oxygen in the air. After the reaction appeared complete, the crucible was allowed to cool and the mass was determined. The crucible was reheated and allowed to cool. Its mass was then determined again to be certain that the reaction was complete. The following data were collected: Mass of crucible = 30.02 g Mass of nickel and crucible = 31.07 g Mass of nickel oxide and crucible = 31.36 g Determine the following information based on the data given above: Mass of nickel = Mass of nickel oxide = Mass of oxygen = Based on your calculations, what is the empirical formula for the nickel oxide?

Short answer

Mass of Nickel = 1.05 gMass of Nickel Oxide = 1.34 gMass of Oxygen = 0.29 gEmpirical Formula = NiO

Step-by-step solution

Determine the Mass of Nickel

Subtract the mass of the empty crucible from the mass of the crucible with the nickel. Mass of Nickel = Mass of Nickel and Crucible - Mass of Crucible Mass of Nickel = 31.07 g - 30.02 g Mass of Nickel = 1.05 g

Determine the Mass of Nickel Oxide

Subtract the mass of the empty crucible from the mass of the crucible with nickel oxide. Mass of Nickel Oxide = Mass of Nickel Oxide and Crucible - Mass of Crucible Mass of Nickel Oxide = 31.36 g - 30.02 g Mass of Nickel Oxide = 1.34 g

Determine the Mass of Oxygen

Subtract the mass of the nickel from the mass of the nickel oxide. Mass of Oxygen = Mass of Nickel Oxide - Mass of Nickel Mass of Oxygen = 1.34 g - 1.05 g Mass of Oxygen = 0.29 g

Calculate Moles of Nickel and Oxygen

Use the molar mass (atomic mass) to find the moles of Ni and O. Moles of Ni = Mass of Ni / Molar Mass of Ni Moles of Ni = 1.05 g / 58.69 g/mol Moles of Ni = 0.0179 mol Moles of O = Mass of O / Molar Mass of O Moles of O = 0.29 g / 16.00 g/mol Moles of O = 0.0181 mol

Determine the Empirical Formula

Calculate the ratio of moles of nickel to moles of oxygen to find the empirical formula. The ratio is approximately 1:1 as 0.0179 mol and 0.0181 mol are very close to each other. Therefore, the empirical formula for nickel oxide is NiO.

Key concepts

Mass Calculation

Understanding mass calculation is vital in chemistry. It's the starting point of many analyses and helps determine the quantities of substances involved in a chemical reaction. Let's break it down using the given exercise as an example.

In the problem, we start by finding the mass of nickel alone. To do this, we subtract the mass of the empty crucible from the mass of the crucible with the nickel:

Mass of Nickel = Mass of Nickel and Crucible - Mass of Crucible
Mass of Nickel = 31.07 g - 30.02 g = 1.05 g

Next, we find the mass of nickel oxide by subtracting the mass of the crucible from the mass of the crucible with nickel oxide:

Mass of Nickel Oxide = Mass of Nickel Oxide and Crucible - Mass of Crucible
Mass of Nickel Oxide = 31.36 g - 30.02 g = 1.34 g

Finally, to find the mass of oxygen that reacted with the nickel, we subtract the mass of nickel from the mass of nickel oxide:

Mass of Oxygen = Mass of Nickel Oxide - Mass of Nickel
Mass of Oxygen = 1.34 g - 1.05 g = 0.29 g

Mole Concept

The mole concept is a central idea in chemistry, enabling us to count atoms and molecules by weighing them. A mole is defined as exactly 6.02214076 × 10²³ particles (Avogadro's number), and it connects the mass of a substance to the number of particles it contains.

To apply the mole concept in the exercise, we start by converting the masses of nickel and oxygen to moles using their molar masses (atomic masses). The formula to convert mass to moles is:

Moles = Mass / Molar Mass

For nickel (Ni), with a molar mass of 58.69 g/mol:

Moles of Ni = 1.05 g / 58.69 g/mol ≈ 0.0179 mol

For oxygen (O), with a molar mass of 16.00 g/mol:

Moles of O = 0.29 g / 16.00 g/mol ≈ 0.0181 mol

These calculations tell us the number of moles, which helps us in the next step to determine the empirical formula by comparing their mole ratios.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, following the law of conservation of mass. In the example given, nickel reacts with oxygen from the air to form nickel oxide. By understanding the mass changes through the reaction, we can deduce the empirical formula of the product.

After calculating the moles of nickel and oxygen (0.0179 mol and 0.0181 mol, respectively), we compare these mole values to find the simplest whole number ratio between them:

Most ratios need to be simplified to the smallest whole numbers. However, in this case:

Ratio of Ni to O ≈ 0.0179 mol Ni : 0.0181 mol O

Since 0.0179 is very close to 0.0181, we can approximate the ratio to 1:1. Therefore, the simplest ratio or the empirical formula for nickel oxide is:

NiO

This empirical formula tells us that for each nickel atom, there is one oxygen atom bonded to it in the compound.