Problem 28 - Chemistry and Chemical Reactivity

Question

At higher temperatures, \(\mathrm{NaHCO}_{3}\) is converted quantitatively to \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) $$2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ Heating a 1.7184 -g sample of impure \(\mathrm{NaHCO}_{3}\) gives \(0.196 \mathrm{g}\) of \(\mathrm{CO}_{2} .\) What was the mass percent of \(\mathrm{NaHCO}_{3}\) in the original 1.7184 -g sample?

Step-by-Step Solution

Answer

The mass percent of \( \mathrm{NaHCO}_3 \) in the sample is 43.51\%.

Step 1: Write the Balanced Equation

First, understand the reaction that occurs when heating \( \mathrm{NaHCO}_3 \). The balanced chemical equation for the decomposition is: \[2 \mathrm{NaHCO}_3(\mathrm{s}) \rightarrow \mathrm{Na}_2 \mathrm{CO}_3(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g})\]

Step 2: Calculate Moles of CO2 Produced

Using the molar mass of \( \mathrm{CO}_2 \) (44.01 g/mol), calculate the moles of \( \mathrm{CO}_2 \):\[\text{moles of } \mathrm{CO}_2 = \frac{0.196 \text{ g}}{44.01 \text{ g/mol}} \approx 0.00445 \text{ moles}\]

Step 3: Use Stoichiometry to Find Moles of NaHCO3

From the balanced equation, the mole ratio of \( \mathrm{NaHCO}_3 \) to \( \mathrm{CO}_2 \) is 2:1. Therefore, for every mole of \( \mathrm{CO}_2 \) produced, 2 moles of \( \mathrm{NaHCO}_3 \) are decomposed.\[\text{moles of } \mathrm{NaHCO}_3 = 2 \times 0.00445 \approx 0.00890 \text{ moles}\]

Step 4: Calculate Mass of NaHCO3

The molar mass of \( \mathrm{NaHCO}_3 \) is approximately 84.01 g/mol. Use it to find the mass of \( \mathrm{NaHCO}_3 \) decomposed:\[\text{mass of } \mathrm{NaHCO}_3 = 0.00890 \text{ moles} \times 84.01 \text{ g/mol} = 0.74772 \text{ g}\]

Step 5: Calculate Mass Percent of NaHCO3

Finally, determine the mass percent of \( \mathrm{NaHCO}_3 \) in the original sample:\[\text{Mass percent} = \left(\frac{0.74772 \text{ g}}{1.7184 \text{ g}}\right) \times 100 \approx 43.51\%\]

Step 6: Conclusion

The mass percent of \( \mathrm{NaHCO}_3 \) in the original sample is approximately 43.51\%.

Key Concepts

Mass Percent

Mass percent is a useful way to express the concentration of a particular component within a compound or a mixture.
It tells you how much of the total mass is made up of one specific substance.To calculate mass percent, you need to know two things:

• The mass of the component of interest.
• The total mass of the mixture or compound.

The formula to find mass percent is:\[\text{Mass percent} = \left(\frac{\text{mass of component}}{\text{total mass of sample}}\right) \times 100\]In the problem, the mass percent of sodium bicarbonate \((\mathrm{NaHCO}_3)\) was found by taking the 0.74772 g of \(\mathrm{NaHCO}_3\) decomposed and dividing it by the initial total mass, 1.7184 g. The result was then multiplied by 100 to convert it to a percentage, yielding approximately 43.51\%.

Chemical Decomposition

Chemical decomposition is a type of chemical reaction where a single compound breaks down into two or more simpler substances.
It's akin to splitting a compound into its components.In the exercise, sodium bicarbonate \((\mathrm{NaHCO}_3)\) undergoes decomposition when heated, chemically splitting into sodium carbonate \((\mathrm{Na}_2\mathrm{CO}_3)\), carbon dioxide \((\mathrm{CO}_2)\), and water vapor \((\mathrm{H}_2\mathrm{O})\).Decomposition reactions are crucial in various applications:

• They help in recycling metals.
• They release gases necessary in other chemical processes.
• They are fundamental in processes like respiration and photosynthesis.

Understanding how compounds decompose aids in predicting product formation and balancing chemical equations.

Mole Calculations

Mole calculations are fundamental in stoichiometry as they allow chemists to relate masses of substances to amounts in moles.
This bridges the gap between the molecular scale and measurable quantities.To solve the exercise, several mole calculations were involved:

• First, the moles of \(\mathrm{CO}_2\) were determined by dividing the mass of \(\mathrm{CO}_2\) by its molar mass (44.01 g/mol), resulting in about 0.00445 moles.
• Using the balanced equation, the mole ratio between \(\mathrm{NaHCO}_3\) and \(\mathrm{CO}_2\) was employed. A 2:1 ratio means \(\mathrm{NaHCO}_3\) was decomposed in twice the amount of moles of \(\mathrm{CO}_2\).
• Finally, the moles of \(\mathrm{NaHCO}_3\) were converted to mass using its molar mass of 84.01 g/mol.

These calculations are essential for determining quantities and anticipating yields in chemical reactions.