Problem 180
Question
Determine the mass percent of anhydrous sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) and water in sodium carbonate decahydrate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\right)\)
Step-by-Step Solution
Verified Answer
In sodium carbonate decahydrate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2}
\mathrm{O}\right)\), the mass percent of anhydrous sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) is 37.03%, while the mass percent of water (H2O) is 62.97%.
1Step 1: Determine the molar mass of anhydrous sodium carbonate (Na2CO3)
To find the molar mass, sum up the molar masses of the individual elements in the compound. For Na2CO3:
Molar Mass of Na = 22.99 g/mol
Molar Mass of C = 12.01 g/mol
Molar Mass of O = 16.00 g/mol
Molar mass of Na2CO3 = (2 × 22.99) + 12.01 + (3 × 16.00) = 105.99 g/mol
2Step 2: Determine the molar mass of sodium carbonate decahydrate (Na2CO3·10H2O)
We will determine the molar mass of Na2CO3·10H2O by adding the molar mass of Na2CO3 and 10 times the molar mass of H2O.
Molar Mass of H2O = (2 × 1.01) + 16.00 = 18.02 g/mol
Molar mass of Na2CO3·10H2O = 105.99 + (10 × 18.02) = 105.99 + 180.20 = 286.19 g/mol
3Step 3: Calculate the mass percentage of anhydrous sodium carbonate (Na2CO3) in sodium carbonate decahydrate
To find the mass percentage of Na2CO3 in Na2CO3·10H2O, divide the molar mass of Na2CO3 by the molar mass of Na2CO3·10H2O and multiply by 100:
Mass percentage of Na2CO3 = (Molar mass of Na2CO3 / Molar mass of Na2CO3·10H2O) × 100 = (105.99 / 286.19) × 100 = 37.03%
4Step 4: Calculate the mass percentage of water (H2O) in sodium carbonate decahydrate
To find the mass percentage of H2O in Na2CO3·10H2O, divide the total molar mass of H2O by the molar mass of Na2CO3·10H2O and multiply by 100:
Mass percentage of H2O = (Total molar mass of H2O / Molar mass of Na2CO3·10H2O) × 100 = (180.20 / 286.19) × 100 = 62.97%
So, the mass percent of anhydrous sodium carbonate (Na2CO3) in sodium carbonate decahydrate is 37.03%, and the mass percent of water (H2O) in sodium carbonate decahydrate is 62.97%.
Key Concepts
Molar Mass CalculationSodium Carbonate DecahydrateAnhydrous Sodium CarbonateMolecular Formula
Molar Mass Calculation
Molar mass is a crucial concept in chemistry, representing the mass of one mole of a substance. It is measured in grams per mole (g/mol).
Calculating the molar mass of a compound involves summing the molar masses of all the atoms that constitute the molecular formula of the compound.
This technique helps determine other values needed to find the mass percent composition in more complex problems.
Calculating the molar mass of a compound involves summing the molar masses of all the atoms that constitute the molecular formula of the compound.
- The atomic masses of elements are usually found on the periodic table.
- Each element's molar mass is multiplied by the number of times the element appears in the formula.
This technique helps determine other values needed to find the mass percent composition in more complex problems.
Sodium Carbonate Decahydrate
Sodium carbonate decahydrate, represented as \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\), is a hydrated form of sodium carbonate.
"Deca" refers to the ten water (H₂O) molecules attached to each \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) unit.
This hydrated version changes some physical properties, like solubility and crystalline structure.
"Deca" refers to the ten water (H₂O) molecules attached to each \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) unit.
- This compound is often referred to as "washing soda" and has applications in domestic and industrial cleaning.
- The presence of water causes it to have a different molar mass than anhydrous sodium carbonate.
This hydrated version changes some physical properties, like solubility and crystalline structure.
Anhydrous Sodium Carbonate
Anhydrous sodium carbonate, commonly referred to as \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), is the base form of sodium carbonate without any water molecules attached. "Anhydrous" effectively means "without water".
Its comparative simplicity makes it invaluable for stoichiometry-based calculations and mass percent composition tasks.
- This compound appears as a white, powdery solid and is known for its applications in glass manufacturing and detergent formulations.
- Without attached water, it maintains the simple molecular weight calculated from its constituent elements.
Its comparative simplicity makes it invaluable for stoichiometry-based calculations and mass percent composition tasks.
Molecular Formula
A molecular formula represents the number and type of atoms present in a molecule of a compound. Unlike empirical formulas, which provide a simplified ratio, molecular formulas give the exact composition.
When performing calculations like determining molar mass or percentage compositions, knowing the molecular formula ensures that all components are correctly accounted for.
This guards against error and ensures precise formulation in laboratory and industrial settings.
- This formula is crucial for understanding the precise weight of the molecule because it tells us each element's contribution to the mass.
- By using the molecular formula, chemists can easily identify and quantify the possible interactions and compounds in a reaction.
When performing calculations like determining molar mass or percentage compositions, knowing the molecular formula ensures that all components are correctly accounted for.
This guards against error and ensures precise formulation in laboratory and industrial settings.
Other exercises in this chapter
Problem 178
In a laboratory setting, how would you determine if a compound was a hydrate?
View solution Problem 179
Write the formula for the following hydrates. a. nickel(II) chloride hexahydrate b. cobalt(II) chloride hexahydrate c. magnesium carbonate pentahydrate d. sodiu
View solution Problem 182
Chromium(III) nitrate forms a hydrate that is 40.50% water by mass. What is its chemical formula?
View solution Problem 183
Determine the percent composition of \(\mathrm{MgCO}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and draw a pie graph to represent the hydrate.
View solution