Question

Liquid ethyl mercaptan, ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{S},$ has a density of $0.84\mathrm{g}/\mathrm{mL}.$ Assuming that the combustion of this compound produces only ${\mathrm{CO}}_2,{\mathrm{H}}_2\mathrm{O},$ and ${\mathrm{SO}}_2,$ what masses of each of these three products would be produced in the combustion of $3.15\mathrm{mL}$ of ethyl mercaptan?

Short answer

The masses of each combustion product are: $C{O}_2:3.74\mathrm{g}$, $H_{2}O:2.30\mathrm{g}$, $S{O}_2:2.72\mathrm{g}$

Step-by-step solution

Balance the chemical equation

First, we write down the unbalanced chemical equation: ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{S}+{\mathrm{O}}_2\to {\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2$. We then balance this equation: ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{S}+5{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2$

Find the molar mass of ethyl mercaptan

Using the periodic table, we determine the molar masses of each atom and sum them together to find the molar mass of ethyl mercaptan. Ethyl mercaptan ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{S}$ has a molar mass of $(2\ast 12.01)+(6\ast 1.008)+32.07=62.146\mathrm{g}/\mathrm{mol}$

Convert volume to mass to moles

We know that the density is the mass divided by the volume i.e., $density=\frac{mass}{volume}$. We can solve this for mass and substitute the given volume and density into this equation. We therefore have: $mass=0.84\mathrm{g}/\mathrm{mL}\ast 3.15\mathrm{mL}=2.646\mathrm{g}$. Then, we convert this mass to moles using the molar mass from step 2 i.e., $moles=\frac{mass}{molarmass}=\frac{2.646\mathrm{g}}{62.146\mathrm{g}/\mathrm{mol}}=0.0425\mathrm{mol}$

Use stoichiometry to find amounts of products

From the balanced chemical equation, we know that each mol of ethyl mercaptan ($C_2{H}_{6}S$) produces 2 mol of carbon dioxide ($C{O}_2$), 3 mol of water ($H_{2}O$) and 1 mol of sulfur dioxide ($S{O}_2$). So, we calculate the amounts of each product respectively: $0.0425\mathrm{mol}\ast 2=0.0850\mathrm{mol}C{O}_2$, $0.0425\mathrm{mol}\ast 3=0.1275\mathrm{mol}{H}_{2}O$, and $0.0425\mathrm{mol}\ast 1=0.0425\mathrm{mol}S{O}_2$

Convert moles of products to gram

Using the molar masses of each compound ($C{O}_2:44.01\mathrm{g}/\mathrm{mol}$, $H_{2}O:18.02\mathrm{g}/\mathrm{mol}$, $S{O}_2:64.07\mathrm{g}/\mathrm{mol}$), we calculate the mass of each product: $0.0850\mathrm{mol}C{O}_2\ast 44.01\mathrm{g}/\mathrm{mol}=3.74\mathrm{g}C{O}_2$, $0.1275\mathrm{mol}{H}_{2}O\ast 18.02\mathrm{g}/\mathrm{mol}=2.30\mathrm{g}{H}_{2}O$, and $0.0425\mathrm{mol}S{O}_2\ast 64.07\mathrm{g}/\mathrm{mol}=2.72\mathrm{g}S{O}_2$. These are the masses of each combustion product.

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It tells us the amounts of each ingredient needed and the quantities of products we can expect. By using stoichiometry, we can predict how much of each substance will form in a chemical reaction. This process involves several steps:

• First, identify the mole ratio between the reactants and products based on the balanced chemical equation.
• Convert known quantities (like mass or volume) to moles using molar mass or density information. This helps us compare and predict amounts accurately since chemical reactions occur in molar quantities.
• Use the mole ratio from the balanced equation to calculate the amount of products formed from the reactants used.

In the given exercise, stoichiometry allows us to determine how 3.15 mL of ethyl mercaptan translates to precise amounts of carbon dioxide, water, and sulfur dioxide produced during combustion.

Chemical Equation Balancing

Balancing a chemical equation is essential because it reflects the law of conservation of mass. This law states that matter cannot be created or destroyed in a chemical reaction. Thus, each element present in the reactants must also be accounted for in the products, maintaining their total number.
To balance a chemical equation:

• List all reactants and products on their respective sides of the reaction equation.
• Count the number of atoms for each element on both sides.
• Adjust coefficients (numbers in front of compounds) until both sides have an equal number of each atom.

For instance, the balanced equation for the combustion of ethyl mercaptan is: $${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{S}+5{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2$$This ensures that carbon, hydrogen, sulfur, and oxygen atoms are equal on both sides, confirming mass conservation.

Molar Mass Calculation

Molar mass is crucial because it connects mass, a measure we can easily determine, to moles, which are used for stoichiometric calculations. It's like the bridge that links the measurable to the countable.
To calculate the molar mass of a compound:

• Identify the number of each type of atom within a molecule.
• Use the periodic table to find the atomic mass of each element (usually in grams per mole).
• Multiply the atomic mass by the number of corresponding atoms and add all the results together to get the total molar mass.

For ethyl mercaptan (${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{S}$), the molar mass was calculated by summing: $$(2\times 12.01)\text{ for carbon}+(6\times 1.008)\text{ for hydrogen}+32.07\text{ for sulfur}=62.146\mathrm{g}/\mathrm{mol}$$Thus, the molar mass provides the calculations with the necessary conversion factor from mass to moles.