Question

What mass of sodium hydroxide has the same number of oxygen atoms as \(100.0 \mathrm{~g}\) of ammonium carbonate?

Short answer

The mass of sodium hydroxide (\(\text{NaOH}\)) with the same number of oxygen atoms as \(100.0 \mathrm{~g}\) of ammonium carbonate (\((\text{NH}_4)_2\text{CO}_3\)) is \(124.8 \mathrm{~g}\).

Step-by-step solution

Determine the molecular formulas

Chemical formulas of given compounds: Sodium hydroxide: NaOH Ammonium carbonate: (NH4)2CO3

Calculate the moles of ammonium carbonate

Now we will find the moles of ammonium carbonate from its given 100.0 g mass, using its molar mass. Molar mass of ammonium carbonate: \(2\times(14.01 + 4\times1.01) + 12.01 + 3 \times 16.00\) = 96.11 g/mol Moles of ammonium carbonate = \(\frac{100.0g}{96.11 g/mol}\) = 1.0407 moles

Determine the number of oxygen atoms in ammonium carbonate

Ammonium carbonate has 3 oxygen atoms in its chemical formula. Therefore, Number of oxygen atoms in 1.0407 moles of ammonium carbonate = \(1.0407\times 3\times 6.022\times 10^{23}\) atoms

Calculate the corresponding number of oxygen atoms in sodium hydroxide

Number of oxygen atoms in sodium hydroxide is 1 per formula unit. So, the same number of oxygen atoms from ammonium carbonate = \(1.0407\times 3\times 6.022\times 10^{23}\) atoms

Convert the number of oxygen atoms to moles of sodium hydroxide

To find the number of moles of sodium hydroxide with the given number of oxygen atoms, we divide the number of oxygen atoms by Avogadro's constant. Moles of sodium hydroxide = \(\frac{1.0407\times 3\times 6.022\times 10^{23}}{6.022\times 10^{23}}\) = 3.122 moles

Convert the moles of sodium hydroxide to grams

Lastly, to find the mass of sodium hydroxide, we multiply the number of moles by the molar mass of sodium hydroxide. Molar mass of sodium hydroxide: 22.99 + 15.99 + 1.01 = 39.99 g/mol Mass of sodium hydroxide = 3.122 moles × 39.99 g/mol = 124.8 g So, the mass of sodium hydroxide with the same number of oxygen atoms as 100.0 g of ammonium carbonate is 124.8 grams.

Key concepts

Stoichiometry

Stoichiometry is the methodology used within chemistry that relates the quantities of substances involved in chemical reactions. It derived from two Greek words that mean 'element' and 'measure', and it acts much like a recipe for a chemical reaction, telling you the proportions of reactants and the expected yield of products.

For example, in the given exercise, stoichiometry allows us to compare the number of oxygen atoms in ammonium carbonate with the number needed in sodium hydroxide. This involves understanding the relationship between the mass of substances and the moles, which then ties into Avogadro's constant to translate these relationships into actual particles or atoms.

• Determine chemical formulas and molar masses of reactants and products
• Convert mass to moles using the molar mass
• Use the mole ratio to find the relationship between substances
• Convert moles back to grams if necessary to find the mass

These are the steps we use in stoichiometry to balance the needs and products of a chemical reaction.

Avogadro's Constant

Named after the scientist Amedeo Avogadro, Avogadro's constant is a key number in chemistry that represents the number of particles found in one mole of a substance. Its value is approximately 6.022 x 10^23 particles per mole and it is crucial for translating between the microscopic scale of atoms and molecules to the macroscopic scale that we can measure in the laboratory.

In our problem, once we have calculated the moles of ammonium carbonate, we can use Avogadro's constant to find out exactly how many oxygen atoms are in those moles. It is this understanding that allows chemists to work with atoms and molecules in a quantitative manner. Avogadro's constant bridges the gap between the sub-atomic world and the human-scale by providing a way to count out atoms through their mass.

Molar Mass

Molar mass is the weight of one mole (6.022 x 10^23 particles) of any chemical substances. This mass is usually indicated in grams per mole (g/mol) and it is the sum of the atomic masses of all atoms present in the molecular formula of the compound, as found on the periodic table.

For instance, in the step-by-step solution, ammonium carbonate has a molar mass of 96.11 g/mol and sodium hydroxide has a molar mass of 39.99 g/mol. These figures are crucial because they enable the conversion of grams into moles, which, as explained earlier, is a fundamental step in stoichiometry. Knowing the molar mass gives us a direct path from the tangible quantity (weight) to the conceptual entity of moles, allowing for calculations involving the actual number of molecules involved in a reaction.

Chemical Formula

Every chemical compound has a specific chemical formula that serves as a symbolic representation of its composition. It tells us which elements and how many atoms of each are present in a molecule. The formula provides vital information for stoichiometric calculations because the proportions of elements within dictate the quantitative relationships in reactions.

In our exercise, the chemical formulas are NaOH for sodium hydroxide and (NH4)2CO3 for ammonium carbonate. By understanding these formulas, you can determine that one molecule of sodium hydroxide contains one atom of oxygen, while one formula unit of ammonium carbonate contains three oxygen atoms. Using the chemical formulas in conjunction with other stoichiometric calculations allows us to find out the equal number of oxygen atoms in different compounds, as seen when seeking the mass of sodium hydroxide corresponding to a certain mass of ammonium carbonate.