Question

When \(\mathrm{M}_{2} \mathrm{~S}_{3}(s)\) is heated in air, it is converted to \(\mathrm{MO}_{2}(s)\). A \(4.000-\mathrm{g}\) sample of \(\mathrm{M}_{2} \mathrm{~S}_{3}(s)\) shows a decrease in mass of \(0.277 \mathrm{~g}\) when it is heated in air. What is the average atomic mass of M?

Short answer

The average atomic mass of M is approximately 2.87 g/mol.

Step-by-step solution

Write the balanced chemical equation.

First, we need to write the balanced chemical equation for the given reaction: \(M_2S_3(s) + \dfrac{9}{2}O_2(g) → 2MO_2(s) + 3SO_2(g)\)

Determine the mass of O2 and S in the sample.

The given mass loss of 0.277 g corresponds to the difference in mass between M2S3 and the products of the reaction. The mass loss is a result of the conversion of sulfur to SO2 and part of the mass of the metal oxide, MO2. Thus, we can calculate the mass of O2 and S in the sample: - Mass of sulfur in M2S3: 4.000 g - 0.277 g = 3.723 g - Mass of oxygen in MO2: 0.277 g - Mass of sulfur (S) = 3.723 g * (3/2) = 5.585 g

Determine the molar mass of sulfur and oxygen.

Consult the periodic table to find the molar mass of sulfur (S) and oxygen (O): - Molar mass of sulfur (S): 32.06 g/mol - Molar mass of oxygen (O): 16.00 g/mol

Calculate the moles of sulfur and oxygen in the sample.

Now, we can calculate the moles of sulfur and oxygen in the sample using their molar masses: - Moles of S in M2S3: 5.585 g / 32.06 g/mol = 0.174 moles - Moles of O2 in MO2: 0.277 g / 16.00 g/mol = 0.017 moles

Use the stoichiometry to find moles of M.

Using the stoichiometry of the balanced chemical equation, we can find the moles of M in the sample: \(Moles\ of\ M = 2 \times Moles\ of\ S = 2 \times 0.174 = 0.348\ moles\)

Calculate the average atomic mass of M.

Finally, we can calculate the average atomic mass of M using the mass of M2S3 and the moles of M: \(Average\ atomic\ mass\ of\ M = \dfrac{Mass\ of\ M_{2}S_{3}}{2 \times Moles\ of\ M}\) \(Average\ atomic\ mass\ of\ M = \dfrac{4.000\ g}{2 \times 0.348\ moles }= 2.874\ g/mol\) The average atomic mass of M is approximately 2.87 g/mol.

Key concepts

Balanced Chemical Equation

In chemical reactions, writing a balanced chemical equation is crucial for understanding how substances interact. A balanced equation ensures that the number of atoms for each element is the same on both sides of the reaction. This satisfies the law of conservation of mass, which states that matter cannot be created or destroyed.
For the reaction in our exercise, the balanced equation is:\[ M_2S_3(s) + \dfrac{9}{2}O_2(g) \rightarrow 2MO_2(s) + 3SO_2(g) \]
Here, two molecules of \(M_2S_3\) react with oxygen \(O_2\) to form \(MO_2\) and \(SO_2\). By ensuring everything is balanced, we can accurately predict and measure reactants and products.

Moles of Sulfur

Understanding how to determine the moles of a substance is vital in chemistry. A mole is a unit that measures the amount of a chemical substance, based on the concept of Avogadro's number \((6.022 \times 10^{23})\) particles per mole. To find the moles of sulfur \(S\) in this exercise, we used the given mass and the molar mass of sulfur from the periodic table.
First, calculate the mass of sulfur in \(M_2S_3\) by subtracting the mass of oxygen:

• Mass of sulfur: \(4.000\ g - 0.277\ g = 3.723\ g\)

Then, using sulfur’s molar mass (32.06 g/mol), determine the moles:\[ \text{Moles of S} = \frac{3.723\ g}{32.06\ g/mol} = 0.116\text{ moles} \]
This calculation is crucial for further stoichiometric steps.

Mass Loss Calculation

During chemical reactions, products and reactants undergo transformation, often leading to changes in mass. In this exercise, the sample mass decreases due to the conversion from \(M_2S_3\) to \(MO_2\) and \(SO_2\).
The observed mass loss is the difference between the initial mass and the remaining mass, largely attributed to the sulfur converting to gas form \(SO_2\) during heating.
- Initial mass of \(M_2S_3\): 4.000 g- Mass loss: 0.277 g
By calculating this loss, we track how much of the substance transformed, leading to crucial insights into reaction stoichiometry and atomic weight determination.

Stoichiometry

Stoichiometry is the branch of chemistry dealing with the relationships between reactants and products in a chemical reaction. It uses the balanced chemical equation to predict how much of each substance is required or produced.
In this exercise, stoichiometry lets us calculate the moles of metal \(M\) based on the stoichiometric coefficients from the balanced equation. With our previous calculation of \(0.116\) moles of \(S\), we deduced that:

• \(2\) moles of \(M\) react with \(3\) moles of \(S\).

Thus:
\[ \text{Moles of M} = \frac{2}{3} \times 0.116 = 0.077\text{ moles} \]
Applying stoichiometry helps find the average atomic mass and understand the material conversion within a reaction.