- The compound lithium carbonate, used to treat manic depression, has the formula Li2CO3. Calculate the number of formula units of lithium carbonate present in a 0.500-g sample of lithium carbonate.
- How many grams of oxygen are present in a 0.10-g sample of adrenaline, a hormone secreted into the bloodstream in times of stress? The formula of adrenaline is C9H13NO3.
The Correct Answer and Explanation is :
Let’s address each question in detail:
1. Calculating the Number of Formula Units in a 0.500 g Sample of Lithium Carbonate (Li₂CO₃):
To determine the number of formula units in a given mass of lithium carbonate, follow these steps:
a. Calculate the Molar Mass of Li₂CO₃:
- Lithium (Li): Atomic mass ≈ 6.94 g/mol; since there are two lithium atoms: 6.94 g/mol × 2 = 13.88 g/mol.
- Carbon (C): Atomic mass ≈ 12.01 g/mol.
- Oxygen (O): Atomic mass ≈ 16.00 g/mol; with three oxygen atoms: 16.00 g/mol × 3 = 48.00 g/mol.
Adding these together:
.88 g/mol (Li) + 12.01 g/mol (C) + 48.00 g/mol (O) = 73.89 g/mol b. Convert Mass to Moles:**
Mol of Li₂CO₃ = Mass (g) / Molar Mass (g/mol)
Mo Li₂CO₃ = 0.500 g / 73.89 g/mol ≈ 0.00677 mol
c. culate the Number of Formula Units:
Using Avodro’s number (6.022 × 10²³ formula units/mol):
Number ooula units = Moles × Avogadro’s number
Number of fu units ≈ 0.00677 mol × 6.022 × 10²³ formula units/mol
Number of form its ≈ 4.08 × 10²¹ formula units
Therefore, a 0.50 mple of lithium carbonate contains approximately 4.08 × 10²¹ formula units.
2. Determining thess of Oxygen in a 0.10 g Sample of Adrenaline (C₉H₁₃NO₃):
To find the mass of oxygen in the given sample of adrenaline:
a. Calculate the Molar Mass of Adrenaline:
- Carbon (C): Atomic mass ≈ 12.01 g/mol; with nine carbon atoms: 12.01 g/mol × 9 = 108.09 g/mol.
- Hydrogen (H): Atomic mass ≈ 1.008 g/mol; with thirteen hydrogen atoms: 1.008 g/mol × 13 = 13.104 g/mol.
- Nitrogen (N): Atomic mass ≈ 14.01 g/mol; with one nitrogen atom: 14.01 g/mol.
- Oxygen (O): Atomic mass ≈ 16.00 g/mol; with three oxygen atoms: 16.00 g/mol × 3 = 48.00 g/mol.
Adding these together:
108.09 g/mol (C) + 13.10g/mol (H) + 14.01 g/mol (N) + 48.00 g/mol (O) = 183.204 g/mol
b. Determine the Massaction of Oxygen:
Mass fraction of oxygen = Tal mass of oxygen in the formula / Molar mass of adrenaline
Mass fraction of oxygen = 0g/mol / 183.204 g/mol ≈ 0.2619
c. Calculate the Mass of Oxn in the Sample:
Mass of oxygen = Mass fraction ofxygen × Sample mass
Mass of oxygen ≈ 0.2619 × 0.10 g0262 g
Therefore, a 0.10 g sample of adrenncontains approximately 0.0262 g of oxygen.
Explanation:
In both problems, thpproach involves dermining the molar mass of the compound, converting the given mass to moles, and then using stoichiometric relationships to find the desired quantity. For lithium carbonate, after finding the m present in the 0.500 g sample, Avogadro’s number is used to convert moles to the number of formula units, since one mole of any substance contains Avogadro’s number of entities (atoms, molecules, or formula units). In the case of adrenaline, the focus is on txygen content. By calculating the mass fraction of oxygen in mole of adrenaline, we can determine how much oxygen is present in any given mass of the compound by simple multiplication. These methods are fundamental in chemistry for qifying the relationships between mass, moles, and the number of particles in a substance.