Problem 34 - Chemistry and Chemical Reactivity

Question

At higher temperatures \(\mathrm{NaHCO}_{3}\) is converted quantitatively to \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Heating a 1.7184 -g sample of impure 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 original sample is approximately 43.50\%.

Step 1: Write the Balanced Equation

The balanced chemical equation is given by:\[ 2 \ \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \] This equation tells us that 2 moles of \( \mathrm{NaHCO}_{3} \) decompose to yield 1 mole of \( \mathrm{CO}_{2} \).

Step 2: Calculate Moles of CO2 Produced

First, we calculate the number of moles of \( \mathrm{CO}_{2} \) produced using its molar mass (44.01 g/mol):\[ \text{Moles of } \mathrm{CO}_{2} = \frac{0.196 \ \mathrm{g}}{44.01 \ \mathrm{g/mol}} \approx 0.00445 \ \mathrm{mol} \]

Step 3: Determine Moles of NaHCO3 in the Sample

Using the stoichiometry of the reaction, we know 2 moles of \( \mathrm{NaHCO}_{3} \) produce 1 mole of \( \mathrm{CO}_{2} \). Thus, the moles of \( \mathrm{NaHCO}_{3} \) initially present are:\[ \text{Moles of } \mathrm{NaHCO}_{3} = 2 \times 0.00445 \approx 0.00890 \ \mathrm{mol} \]

Step 4: Calculate Mass of NaHCO3 in the Sample

Find the mass of \( \mathrm{NaHCO}_{3} \) using its molar mass (84.01 g/mol):\[ \text{Mass of } \mathrm{NaHCO}_{3} = 0.00890 \ \mathrm{mol} \times 84.01 \ \mathrm{g/mol} \approx 0.7471 \ \mathrm{g} \]

Step 5: Calculate Mass Percent of NaHCO3

To find the mass percent of \( \mathrm{NaHCO}_{3} \) in the original sample, use the formula:\[ \text{Mass percent of } \mathrm{NaHCO}_{3} = \left( \frac{0.7471}{1.7184} \right) \times 100 \approx 43.50\% \]

Key Concepts

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It's like the math behind chemistry!
In our problem, stoichiometry helps us understand how substances transform during the decomposition of sodium bicarbonate \(\text{NaHCO}_3\). The balanced chemical equation given is:\[ 2 \, \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]This tells us that 2 moles of \(\text{NaHCO}_3\) will decompose to form 1 mole of \(\text{CO}_2\), among other products.
By knowing this, we can calculate how many moles of a substance we start with and what it will turn into, using the ratio of coefficients from the balanced equation.
This makes it a powerful tool in predicting the amounts of products from given reactants.

• Understanding these ratios is key to solving problems in chemical reactions, and that's precisely what makes stoichiometry fundamental in chemistry.

Mass Percent Calculation

Mass percent calculation is used to determine the concentration of a component in a compound or mixture based on its mass. It answers the question, how much of your sample is a specific substance?
For our exercise, we want to find the mass percent of \(\text{NaHCO}_3\) in an impure sample. Here's the formula:\[\text{Mass Percent} = \left( \frac{\text{Mass of Desired Component}}{\text{Total Mass of Sample}} \right) \times 100\]We calculated the mass of \(\text{NaHCO}_3\) to be 0.7471 g and the total sample mass was 1.7184 g. Plugging these into the formula gives us the mass percent:\[\left( \frac{0.7471}{1.7184} \right) \times 100 \approx 43.50\%\]
Mass percent is a practical way to express concentrations and can be useful in both laboratory settings and industry to determine purity and formulations.

Chemical Reactions

Chemical reactions are processes where substances, the reactants, transform into new substances, the products. \(\text{NaHCO}_3\) decomposition is an example of a reaction breaking down a substance into simpler forms.
Recognizing the components in chemical reactions involves understanding product formation, energy changes, and the conservation of mass.
Every chemical reaction has a specific balanced equation, showing the exact proportions in which reactants and products react and form.
This helps chemists predict amounts of substances needed or produced.

• In our scenario, we see \(\text{NaHCO}_3\) change into \(\text{Na}_2\text{CO}_3\), \(\text{CO}_2\), and \(\text{H}_2\text{O}\), showcasing the law of conservation of mass, where the mass of the reactants equals the mass of the products.

Molar Mass Calculation

Molar mass is the weight of one mole of a substance and serves as a bridge between the microscopic world of atoms and the macroscopic world we can measure.
In chemistry, it's measured in grams per mole (g/mol). Knowing the molar mass helps you convert grams to moles, a necessary step in many calculations.
For example, the molar mass of \(\text{NaHCO}_3\) is 84.01 g/mol. This means one mole of \(\text{NaHCO}_3\) weighs 84.01 grams.
In this exercise, we calculated the moles of \(\text{CO}_2\) produced using its molar mass, 44.01 g/mol, helping us find the moles and then the mass of \(\text{NaHCO}_3\).

• Accurate molar mass calculations are foundational in chemistry, aiding in stoichiometric analysis and ensuring precise conversions between mass and moles.