Question

Sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3},\) can be decomposed quantitatively by heating. $$2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ A 0.682 -g sample of impure NaHCO \(_{3}\) yielded a solid residue (consisting of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and other solids) with a mass of \(0.467 \mathrm{g} .\) What was the mass percent of \(\mathrm{NaHCO}_{3}\) in the sample?

Short answer

The mass percent of NaHCO₃ in the sample is 61.15%.

Step-by-step solution

Calculate Mass of CO2 and H2O Released

First, determine the mass loss from the original to the residue after the decomposition reaction. This mass is due to the release of gases, CO2 and H2O.\[\text{Mass of } \mathrm{CO}_2 \text{ and } \mathrm{H}_2 \mathrm{O} = 0.682 \text{ g (original sample)} - 0.467 \text{ g (residue)} = 0.215 \text{ g}\]

Compute Moles of Released Gases

Calculate the amount of NaHCO₃ that decomposed by using the reaction and noting that 2 moles of NaHCO₃ release 1 mole each of CO₂ and H₂O. Determine moles of NaHCO₃ using the molar mass of CO₂ (44.01 g/mol) and H₂O (18.02 g/mol).\[\text{Moles of CO}_2 + \text{moles of H}_2\text{O} = \frac{0.215 \text{ g}}{44.01 \text{ g/mol} + 18.02 \text{ g/mol}} = 0.002478 \text{ moles}\]

Calculate Moles of NaHCO3

From the reaction, each mole of CO₂ and H₂O comes from the decomposition of 2 moles of NaHCO₃. Therefore, calculate moles of NaHCO₃.\[\text{Moles of } \mathrm{NaHCO}_3 = 2 \times 0.002478 = 0.004956 \text{ moles}\]

Determine Mass of NaHCO3 in Sample

Convert the moles of NaHCO₃ back to grams using the molar mass of NaHCO₃ (84.01 g/mol).\[\text{Mass of } \mathrm{NaHCO}_3 = 0.004956 \text{ moles} \times 84.01 \text{ g/mol} = 0.4168 \text{ g}\]

Calculate Mass Percent of NaHCO3

Finally, calculate the mass percent of NaHCO₃ in the original sample.\[\text{Mass percent of } \mathrm{NaHCO}_3 = \left( \frac{0.4168 \text{ g}}{0.682 \text{ g}} \right) \times 100 = 61.15\%\]

Key concepts

Mass Percent

Mass percent is a concept used to express the concentration of an element in a compound or a component in a mixture. Imagine you're baking cookies with chocolate chips. The mass percent tells you how much of your cookie's mass is just chocolate chips. This is an essential tool in chemistry for determining the composition of mixtures or compounds.
To calculate the mass percent of a substance in a sample, you must determine the mass of the substance of interest and the total mass of the sample. The formula is:\[\text{Mass percent} = \left( \frac{\text{mass of component}}{\text{total mass of sample}} \right) \times 100\]This allows chemists to describe how much of the sample consists of a particular compound, helping to analyze everything from the purity of a sample to the efficiency of a reaction. In our exercise, finding the mass percent of sodium hydrogen carbonate involves dividing the mass of NaHCOelement{3} by the total sample mass, then multiplying by 100 to get the percentage. This gives insight into how much of the initial sample was comprised of sodium hydrogen carbonate before the decomposition occurred.

Molar Mass

Molar mass is the mass of a given substance (chemical element or chemical compound) divided by its amount of substance measured in moles. It allows you to convert between grams and moles effectively. Knowing this is crucial for any chemical reactions or calculations.
For any element, the molar mass is almost exactly the same as its atomic mass on the periodic table, but in mass units of grams per mole (g/mol). For a compound, molar mass is the sum of the molar masses of its constituent elements.

• For example, the molar mass of CO₂ (carbon dioxide) is the sum of the molar masses of 1 carbon atom (12.01 g/mol) and 2 oxygen atoms (16.00 g/mol each), totaling 44.01 g/mol.
• Water (H₂O) has a molar mass of 18.02 g/mol, with 2 hydrogen atoms contributing 2.02 g/mol and 1 oxygen atom contributing 16.00 g/mol.

In the exercise, we use molar masses to determine how many moles of gases were released during the decomposition process, which in turn helps determine the amount of sodium hydrogen carbonate that was present in the sample initially.

Quantitative Decomposition

Quantitative decomposition is a chemical process in which a compound breaks down into simpler substances and the quantities of these products are measured precisely. It is akin to breaking down a complex recipe into its basic ingredients and knowing exactly how much of each ingredient you end up with.
This type of analysis involves accurately measuring the starting and ending masses and then calculating the differences. This helps in understanding the efficiency of the reaction and how complete the decomposition was.
The decomposition reaction detailed in the exercise is:\[2 \text{NaHCO}_3 (s) \rightarrow \text{Na}_2\text{CO}_3 (s) + \text{CO}_2 (g) + \text{H}_2\text{O} (g)\]Through this process, we learn the precise and systematic nature of how a compound like sodium hydrogen carbonate can be broken down. The measurement of mass loss (from gases like CO₂ and H₂O) post-decomposition also assists in determining the amount of original compound present, which is crucial for calculating the mass percent in the sample.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They indicate the substances involved in a reaction, their quantities, and the states they are in (solid, liquid, gas, or aqueous). A chemical equation is like the blueprint for a reaction, showing how reactants are transformed into products.
These equations must be balanced to follow the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction. This means the number of atoms of each element in the reactants must equal the number of atoms in the products.
Here’s how the equation works in the context of our exercise:

• 2 moles of \(\text{NaHCO}_3\) decompose into 1 mole of \(\text{Na}_2\text{CO}_3\), 1 mole of \(\text{CO}_2\), and 1 mole of \(\text{H}_2\text{O}\).
• Each side of the equation has the same number of each type of atom, maintaining a balanced condition.

Understanding chemical equations is foundational for predicting the quantities of substances consumed and produced. In this exercise, knowing the balanced equation enables the calculation of how much sodium hydrogen carbonate decompose, which relates back to the mass percent calculation.