Unit 1: Atomic Structure and Properties
Essential Knowledge
◼ While pure substances contain molecules or formula units of a single type, mixtures contain molecules or formula units of two or more types, whose relative proportions can vary. (SPQ-2.B.1)
◼ Elemental analysis can be used to determine the relative numbers of atoms in a substance and to determine its purity (SPQ-2.B.2)
◼ While pure substances contain molecules or formula units of a single type, mixtures contain molecules or formula units of two or more types, whose relative proportions can vary. (SPQ-2.B.1)
Aluminum metal reacts with the air and forms a thin, corrosion resistant coating of aluminum oxide, Al2O3, according to the following unbalanced equation:
Al (s) + O2 (g) → Al2O3 (s)
A sample of a mixture of aluminum and aluminum oxide weighing 120.91 grams were analyzed and found to contain 120.32 grams of aluminum.
Part A: Balance the equation provided.
Part B: What mass of oxygen was in the sample?
Part C: What mass of aluminum oxide was in the mixture?
Part D: What is the mass percent of aluminum oxide in the aluminum and aluminum oxide mixture?
*** SEE BELOW FOR WORKED OUT SOLUTION ***
Part A: Balance the equation provided.
The balanced equation is: \[ 4\text{Al} (s) + 3\text{O}_2 (g) \rightarrow 2\text{Al}_2\text{O}_3 (s) \]
Part B: What Mass of Oxygen Was in the Sample?
Given:
Total mass of the mixture = 120.91 g
Mass of aluminum (Al) = 120.32 g
First, calculate the mass of aluminum oxide (Al₂O₃) in the mixture:
$\text{Mass of Al}_2\text{O}_3 = \text{Total mass} - \text{Mass of Al}$
$\text{Mass of Al}_2\text{O}_3 = 120.91 \ \text{g} - 120.32 \ \text{g} = 0.59$
Now, let's calculate the mass of oxygen in the sample. The mass of oxygen in the sample is found using the molar masses of Al and O:
Molar masses:
Al = 26.98 g/mol
O = 16.00 g/mol
Al₂O₃ = 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol
In Al₂O₃, the molar mass ratio of oxygen to the whole compound is:
$\text{Mass ratio of O in Al}_2\text{O}_3 = \frac{3 \times 16.00 \ \text{g/mol}}{101.96 \ \text{g/mol}} = \frac{48.00 \ \text{g/mol}}{101.96 \ \text{g/mol}} \approx 0.471$
Using the mass ratio:
$\text{Mass of O in the sample} = 0.471 \times 0.59 \ \text{g} \approx 0.278 \ \text{g}$
Part C: What Mass of Aluminum Oxide Was in the Mixture?
We already calculated this in Part B. The mass of aluminum oxide (Al₂O₃) in the mixture is:
$\text{Mass of Al}_2\text{O}_3 = 0.59 \ \text{g}$
Part D: What Is the Mass Percent of Aluminum Oxide in the Aluminum and Aluminum Oxide Mixture?
The mass percent of aluminum oxide in the mixture is calculated using:
$\text{Mass percent of Al}_2\text{O}_3 = \left(\frac{\text{Mass of Al}_2\text{O}_3}{\text{Total mass of mixture}}\right) \times 100\%$
Substituting the values:
$\text{Mass percent of Al}_2\text{O}_3 = \left(\frac{0.59 \ \text{g}}{120.91 \ \text{g}}\right) \times 100\% \approx 0.49\%$
Try it yourself …
Magnesium metal reacts with nitrogen gas to form magnesium nitride, Mg₃N₂, according to the following unbalanced equation: Mg(𝑠) + N2(𝑔) → Mg3N2(𝑠)
A sample of a mixture of magnesium and magnesium nitride weighing 85.50 grams was analyzed and found to contain 70.80 grams of magnesium.
Part A: Balance the equation provided.
Part B: What mass of nitrogen was in the sample?
Part C: What mass of magnesium nitride was in the mixture?
Part D: What is the mass percent of magnesium nitride in the magnesium and magnesium nitride mixture?
Check your answers here: [Part A] [Part B] [Part C] [Part D]
◼ Elemental analysis can be used to determine the relative numbers of atoms in a substance and to determine its purity (SPQ-2.B.2)
A mixture consisting only of lithium chloride (LiCl), lithium carbonate (Li2CO3), and lithium nitrate (LiNO3), was analyzed. The elemental analysis of the mixture revealed the following:
Element % Composition
Li 14.19%
Cl 10.56%
C 6.198%
O 59.06%
N 10.01%
Calculate the mass percentage of each compound in the mixture.
To solve this problem, we need to determine the mass percentage of each compound (LiCl, Li₂CO₃, and LiNO₃) in the mixture based on the elemental composition provided. We'll use the following steps:
Determine the molar masses of the elements and compounds.
Express the percentage composition in terms of the compound contributions.
Set up and solve the system of equations based on the given percentages.
Step 1: Molar Masses of Elements and Compounds
Let's calculate the molar masses of each compound:
LiCl:
Li: 6.94 g/mol
Cl: 35.45 g/mol
Molar mass of LiCl: 6.94 +. 35.45 = 42.39 g/mol
Li₂CO₃:
Li: 2 × 6.94 g/mol = 13.88 g/mol
C: 12.01 g/mol
O: 3 × 16.00 g/mol = 48.00 g/mol
Molar mass of Li₂CO₃: 13.88 + 12.01 + 48.00 = 73.89 g/mol
LiNO₃:
Li: 6.94 g/mol
N: 14.01 g/mol
O: 3 × 16.00 g/mol = 48.00 g/mol
Molar mass of LiNO₃: 6.94 + 14.01 + 48.00 = 68.95 g/mol
Step 2: Set up the system of equations based on the given mass percentages
Assume the mass percentages of LiCl, Li₂CO₃, and LiNO₃ in the mixture are x, y, and z respectively. Given that the total mass percentage of all elements must sum to 100%, we have:
x + y + z = 100
Next, we write the mass percentage of each element as a function of the compound percentages:
The percentage of Li (Lithium) can be expressed as:
$\frac{6.94x}{42.39} + \frac{2(6.94)y}{73.89} + \frac{6.94z}{68.95} = 0.1419$
The percentage of Cl (Chlorine) can be expressed as:
$\frac{35.45x}{42.39} = 0.1056$
The percentage of C (Carbon) can be expressed as:
$\frac{12.01y}{73.89} = 0.06198$
The percentage of O (Oxygen) can be expressed as:
$\frac{3(16)y}{73.89} + \frac{3(16)z}{68.95} = 0.5906$
The percentage of N (Nitrogen) can be expressed as:
$\frac{14.01z}{68.95} = 0.1001$
Step 3: Solve the system of equations
Let's first solve the simpler equations to get the values for x, y, and z.
From the equation for Cl:
$x = \frac{0.1056 \times 42.39}{35.45} = 0.126 \text{ or } 12.6\%$
From the equation for C:
$y = \frac{0.06198 \times 73.89}{12.01} = 0.381 \text{ or } 38.1\%$
From the equation for N:
$z = \frac{0.1001 \times 68.95}{14.01} = 0.493 \text{ or } 49.3\%$
Step 4: Verify the solutions using the lithium and oxygen equations.
We now need to confirm these results with the lithium and oxygen percentages:
For Li:
$\frac{6.94 \times 0.126}{42.39} + \frac{2 \times 6.94 \times 0.381}{73.89} + \frac{6.94 \times 0.493}{68.95} = 0.1419 \text{ (which should match)}$
For O:
$\frac{3 \times 16 \times 0.381}{73.89} + \frac{3 \times 16 \times 0.493}{68.95} = 0.5906 \text{ (which should match)}$
Final Answer: The mass percentages of the compounds in the mixture are approximately:
LiCl: 12.62%
Li₂CO₃: 38.07%
LiNO₃: 49.25%
Try it yourself …
The following problem is unsolvable - there is a fundamental error in how the problem is set up. Identify the error.
A mixture consisting only of contains sodium chloride (NaCl), potassium nitrate (KNO₃), and magnesium sulfate (MgSO₄). Elemental analysis provided the following weight percentages:
Element with % Composition (by weight):
Na = 19.32%
K = 20.69%
Mg = 6.02%
S = 13.96%
O = 40.01%
Cl = ???
Calculate the mass percentage of each compound in the mixture.
Check your answer here.
Try it yourself …
The mixture consists of sodium fluoride (NaF), sodium sulfate (Na₂SO₄), and sodium nitrate (NaNO₃). The elemental analysis provided the following weight percentages:
Element = % Composition by mass
Na = 29.08%
F = 12.89%
S = 15.99%
O = 36.21%
N = 5.83%
Calculate the mass percentage of each compound in the mixture.
Check your answer here.