Question

An oxide of metal (M) has \(40 \%\) by mass of oxygen. Metal \(\mathrm{M}\) has a relative atomic mass of 24 . The empirical formula of oxide is (1) \(\mathrm{M}_{2} \mathrm{O}\) (2) \(\mathrm{M}_{2} \mathrm{O}_{3}\) (3) MO (4) \(\mathrm{M}_{3} \mathrm{O}_{4}\)

Short answer

The empirical formula of the oxide is MO.

Step-by-step solution

- Identify the given information

The problem states that an oxide of metal (M) has 40% by mass of oxygen and that metal M has a relative atomic mass of 24.

- Determine the mass ratio

Given the oxide contains 40% oxygen by mass, then the mass percentage of metal M is 60%. This means in 100g of the oxide, there are 40g of oxygen and 60g of metal M.

- Calculate the moles of metal M and oxygen

First, find the moles of each component. The atomic mass of metal M is 24. Thus, the moles of metal M in 60g is \[\text{Moles of M} = \frac{60g}{24g/mol} = 2.5 \text{ moles}\].\ Similarly, the atomic mass of oxygen is 16. The moles of oxygen in 40g is \[\text{Moles of O} = \frac{40g}{16g/mol} = 2.5 \text{ moles}\].

- Determine the mole ratio

The mole ratio of metal M to oxygen in the compound is 2.5 : 2.5 which simplifies to 1 : 1.

- Establish the empirical formula

Given the 1:1 mole ratio of metal M to oxygen, the empirical formula of the oxide is MO.

Key concepts

mass percentage calculation

Mass percentage calculation helps us know how much of each element is present in a compound by mass. In our example with the metal oxide, the question tells us there is 40% oxygen. This means out of 100 grams of the compound, 40 grams are oxygen.
To find the mass percentage of another element, subtract the given percentage of one element from 100%. For instance, if the compound is 40% oxygen, it must be 60% metal M, because 100% - 40% = 60%.

• This gives us a clear picture of how much of each element there is in any given mass of the compound.

mole ratio

A mole ratio tells us how many moles of each element are in a compound. To find the mole ratio, we first need to calculate the moles of each element. Use the mass and atomic mass of each element to do this.
For metal M, which has an atomic mass of 24, we have 60 grams. So, the moles are \[\text{{Moles of M}} = \frac{60g}{24g/mol} = 2.5 \text{{ moles}}.\]

• For oxygen, with an atomic mass of 16, we have 40 grams. The moles are \[\text{{Moles of O}} = \frac{40g}{16g/mol} = 2.5 \text{{ moles}}.\]

Next, divide the moles of each element by the smallest number of moles present in the calculation to find the simplest whole number ratio. Here, both oxygen and metal M have the same number of moles, so the mole ratio is \[\text{{2.5:2.5 = 1:1}}.\]
This tells us hence the empirical formula is MO.

atomic mass

Atomic mass is a key concept in chemistry as it helps us calculate the moles of an element. Atomic mass is the mass of an atom, usually expressed in atomic mass units (amu). Each element has a unique atomic mass.

• For example, the atomic mass of oxygen is 16 amu.
• The atomic mass of metal M given in the problem is 24 amu. This is why 24 is used in our calculations.

Knowing the atomic mass, you can convert mass to moles using the formula:
\[ \text{{Moles}} = \frac{\text{{Mass(g)}}}{\text{{Atomic mass(g/mol)}}} \]
This helps in determining how much of each element is needed to form compounds, calculate reactions, and balance chemical equations.