1.Determine the percent composition of Lithium nitrate.
2. Indicate whether each of the following compounds acts as an electrolyte when dissolved In water
a) sulfiric acid
b) ethanol
c) glucose
d) sodium nitrate
1.Determine the percent composition of Lithium nitrate.
You want percent Li in LiNO3, perdent N in LiNO3 and percent O in LiNO3. This is just like % anything; i.e., it's (grams Li/grams LiNO3)*100 = ?
%Li = (atomic mass Li/molar mass LiNO3)*100 = ?
%O = (3*atomic mass O/molar mass LiNO3)*100 = ?
%N = (atomic mass N/molar mass LiNO3)*100 = ?
2. Indicate whether each of the following compounds acts as an electrolyte when dissolved In water
a) sulfiric acid I assume you meant sulfuric acid
b) ethanol
c) glucose
d) sodium nitrate
Do you know which ones are ionic and which are covalent compounds? Ionic compounds dissolve in water and the ions conduct electricity; therefore, they are electrolytes. Covalently bonded compounds dissolve in water but they do not break apart into ions; therefore, they are non-electrolytes in solution. Solutions of a and d are electrolytes.
1. To determine the percent composition of Lithium nitrate (LiNO3), we need to calculate the mass percent of each element in the compound.
Step 1: Determine the molar mass of Lithium nitrate.
LiNO3 has 1 lithium (Li) atom, 1 nitrogen (N) atom, and 3 oxygen (O) atoms.
The atomic masses of Li, N, and O are approximately 6.94 g/mol, 14.01 g/mol, and 16.00 g/mol, respectively.
Molar mass of LiNO3 = (1 x 6.94) + (1 x 14.01) + (3 x 16.00) = 85.00 g/mol
Step 2: Calculate the mass percent of each element in Lithium nitrate.
Mass percent of Li = (Mass of Li / Molar mass of LiNO3) x 100%
Mass percent of Li = (6.94 g / 85.00 g) x 100% = 8.16%
Mass percent of N = (Mass of N / Molar mass of LiNO3) x 100%
Mass percent of N = (14.01 g / 85.00 g) x 100% = 16.48%
Mass percent of O = (Mass of O / Molar mass of LiNO3) x 100%
Mass percent of O = (3 x 16.00 g / 85.00 g) x 100% = 44.71%
Therefore, the percent composition of Lithium nitrate is approximately:
- 8.16% Lithium (Li)
- 16.48% Nitrogen (N)
- 44.71% Oxygen (O)
2. When dissolved in water, the compounds can act as electrolytes if they dissociate into ions and conduct electricity. Let's analyze each compound:
a) Sulfuric acid (H2SO4) - Sulfuric acid is a strong acid that fully ionizes in water, producing hydrogen (H+) ions and sulfate (SO4^2-) ions. Hence, it acts as an electrolyte.
b) Ethanol (C2H5OH) - Ethanol is a covalent compound and does not dissociate into ions in water. Therefore, it does not act as an electrolyte.
c) Glucose (C6H12O6) - Glucose is a covalent compound and does not dissociate into ions in water. Therefore, it does not act as an electrolyte.
d) Sodium nitrate (NaNO3) - Sodium nitrate fully dissociates into sodium (Na+) ions and nitrate (NO3-) ions in water. Hence, it acts as an electrolyte.
In summary:
a) Sulfuric acid acts as an electrolyte when dissolved in water.
b) Ethanol and glucose do not act as electrolytes when dissolved in water.
d) Sodium nitrate acts as an electrolyte when dissolved in water.
1. To determine the percent composition of Lithium nitrate, we first need to know the chemical formula for Lithium nitrate. The formula for Lithium nitrate is LiNO3.
To calculate the percent composition of each element in Lithium nitrate, we need to know the atomic masses of Lithium (Li), Nitrogen (N), and Oxygen (O).
- The atomic mass of Li is approximately 6.94 g/mol.
- The atomic mass of N is approximately 14.01 g/mol.
- The atomic mass of O is approximately 16.00 g/mol.
Next, we calculate the molar mass of Lithium nitrate:
Molar mass of LiNO3 = (Atomic mass of Li) + (Atomic mass of N) + (3 * Atomic mass of O)
= (6.94 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol)
≈ 14.01 g/mol
Now, we calculate the percent composition of each element in Lithium nitrate:
Percent composition of Li = (Number of Li atoms * Atomic mass of Li) / Molar mass of LiNO3
= (1 * 6.94 g/mol) / 14.01 g/mol
≈ 49.6%
Percent composition of N = (Number of N atoms * Atomic mass of N) / Molar mass of LiNO3
= (1 * 14.01 g/mol) / 14.01 g/mol
≈ 100%
Percent composition of O = (Number of O atoms * Atomic mass of O) / Molar mass of LiNO3
= (3 * 16.00 g/mol) / 14.01 g/mol
≈ 68.0%
Therefore, the percent composition of Lithium nitrate is approximately 49.6% Li, 100% N, and 68.0% O.
2. To determine whether each compound acts as an electrolyte when dissolved in water, we need to consider whether it dissociates into ions in water.
a) Sulfuric acid (H2SO4) is a strong acid and readily dissociates into hydrogen ions (H+) and sulfate ions (SO4^2-) when dissolved in water. Therefore, it acts as an electrolyte when dissolved in water.
b) Ethanol (C2H5OH) does not dissociate into ions when dissolved in water. It remains mostly as neutral molecules. Therefore, it does not act as an electrolyte when dissolved in water.
c) Glucose (C6H12O6) does not dissociate into ions when dissolved in water. It remains mostly as neutral molecules. Therefore, it does not act as an electrolyte when dissolved in water.
d) Sodium nitrate (NaNO3) readily dissociates into sodium ions (Na+) and nitrate ions (NO3-) when dissolved in water. Therefore, it acts as an electrolyte when dissolved in water.