IDNLearn.com: Where your questions are met with thoughtful and precise answers. Find the information you need quickly and easily with our comprehensive and accurate Q&A platform.
Sagot :
### Question 7.1.1
Define the term empirical formula.
An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It provides the proportion of elements in the compound but not the actual number of atoms in the molecule.
### Question 7.1.2
Determine the molecular formula of the organic acid if the molar mass of the acid is 60 g/mol.
To determine the molecular formula of the organic acid, follow these steps:
1. Determine the moles of each element in 100 g of the compound:
- Carbon: 39.9%
- Hydrogen: 6.7%
- Oxygen: 53.4%
The atomic masses are:
- Carbon (C) = 12.01 g/mol
- Hydrogen (H) = 1.008 g/mol
- Oxygen (O) = 16.00 g/mol
Calculate the moles of each element:
[tex]\[ \text{Moles of Carbon} = \frac{39.9}{12.01} \approx 3.322 \][/tex]
[tex]\[ \text{Moles of Hydrogen} = \frac{6.7}{1.008} \approx 6.647 \][/tex]
[tex]\[ \text{Moles of Oxygen} = \frac{53.4}{16.00} \approx 3.338 \][/tex]
2. Determine the simplest whole-number ratio (empirical formula):
Find the smallest number of moles among the elements:
[tex]\[ \text{Smallest number of moles} = 3.322 \][/tex]
Divide each mole value by the smallest number of moles to find the ratio:
[tex]\[ \text{Ratio of Carbon} = \frac{3.322}{3.322} \approx 1 \][/tex]
[tex]\[ \text{Ratio of Hydrogen} = \frac{6.647}{3.322} \approx 2.001 \][/tex]
[tex]\[ \text{Ratio of Oxygen} = \frac{3.338}{3.322} \approx 1.005 \][/tex]
These ratios round off to:
[tex]\[ \text{Empirical formula} = C_1H_2O_1 \][/tex]
3. Calculate the empirical formula mass:
[tex]\[ \text{Empirical mass} = (1 \times 12.01) + (2 \times 1.008) + (1 \times 16.00) = 12.01 + 2.016 + 16.00 = 30.026 \ g/mol \][/tex]
4. Determine the number of empirical units in the molecular formula:
[tex]\[ \text{Molecular units} = \frac{60}{30.026} \approx 1.998 \][/tex]
5. Calculate the molecular formula:
Since the molecular units are approximately 2, the molecular formula is determined by multiplying each subscript in the empirical formula by this factor:
[tex]\[ \text{Molecular formula} = \left(C \cdot 2, H \cdot 2 \times 2, O \cdot 2 \right) = \left(C_2H_4O_2 \right) \][/tex]
So, the molecular formula for the organic acid is [tex]\( \mathbf{C_2H_4O_2} \)[/tex].
### Question 7.2
Determine the metal X in the unknown salt [tex]\( XNO_3 \)[/tex].
Given:
- Mass of [tex]\( XNO_3 \)[/tex] = 34 g
- Volume of [tex]\( MgCl_2 \)[/tex] solution = 250 cm³ (0.250 dm³)
- Concentration of [tex]\( MgCl_2 \)[/tex] solution = 0.4 mol/dm³
- Balanced chemical equation:
[tex]\[ MgCl_2 + 2 XNO_3 \rightarrow 2 XCl + Mg(NO_3)_2 \][/tex]
1. Calculate the moles of [tex]\( MgCl_2 \)[/tex]:
[tex]\[ \text{Moles of } MgCl_2 = \text{Concentration} \times \text{Volume} = 0.4 \ \text{mol/dm}^3 \times 0.250 \ \text{dm}^3 = 0.1 \ \text{mol} \][/tex]
2. Use the stoichiometry to find moles of [tex]\( XNO_3 \)[/tex]:
According to the balanced equation, 1 mole of [tex]\( MgCl_2 \)[/tex] reacts with 2 moles of [tex]\( XNO_3 \)[/tex]:
[tex]\[ \text{Moles of } XNO_3 = 2 \times \text{Moles of } MgCl_2 = 2 \times 0.1 = 0.2 \ \text{mol} \][/tex]
3. Calculate the molar mass of [tex]\( XNO_3 \)[/tex]:
[tex]\[ \text{Molar mass of } XNO_3 = \frac{\text{Mass}}{\text{Moles}} = \frac{34 \ \text{g}}{0.2 \ \text{mol}} = 170 \ \text{g/mol} \][/tex]
4. Determine the atomic mass of [tex]\( X \)[/tex]:
The formula of [tex]\( XNO_3 \)[/tex] consists of one atom of [tex]\( X \)[/tex], one atom of nitrogen [tex]\((N, 14.01 \ \text{g/mol})\)[/tex], and three atoms of oxygen [tex]\((O, 3 \times 16.00 \ \text{g/mol}) = 48.00 \ \text{g/mol})\)[/tex]:
[tex]\[ \text{Molar mass of } XNO_3 = \text{Atomic mass of } X + 14.01 + 48.00 \][/tex]
[tex]\[ 170 = \text{Atomic mass of } X + 62.01 \][/tex]
[tex]\[ \text{Atomic mass of } X = 170 - 62.01 = 107.99 \ \text{g/mol} \][/tex]
So, the metal [tex]\( X \)[/tex] in the unknown salt [tex]\( XNO_3 \)[/tex] is Silver (Ag) with an atomic mass of approximately 107.99 g/mol.
Define the term empirical formula.
An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It provides the proportion of elements in the compound but not the actual number of atoms in the molecule.
### Question 7.1.2
Determine the molecular formula of the organic acid if the molar mass of the acid is 60 g/mol.
To determine the molecular formula of the organic acid, follow these steps:
1. Determine the moles of each element in 100 g of the compound:
- Carbon: 39.9%
- Hydrogen: 6.7%
- Oxygen: 53.4%
The atomic masses are:
- Carbon (C) = 12.01 g/mol
- Hydrogen (H) = 1.008 g/mol
- Oxygen (O) = 16.00 g/mol
Calculate the moles of each element:
[tex]\[ \text{Moles of Carbon} = \frac{39.9}{12.01} \approx 3.322 \][/tex]
[tex]\[ \text{Moles of Hydrogen} = \frac{6.7}{1.008} \approx 6.647 \][/tex]
[tex]\[ \text{Moles of Oxygen} = \frac{53.4}{16.00} \approx 3.338 \][/tex]
2. Determine the simplest whole-number ratio (empirical formula):
Find the smallest number of moles among the elements:
[tex]\[ \text{Smallest number of moles} = 3.322 \][/tex]
Divide each mole value by the smallest number of moles to find the ratio:
[tex]\[ \text{Ratio of Carbon} = \frac{3.322}{3.322} \approx 1 \][/tex]
[tex]\[ \text{Ratio of Hydrogen} = \frac{6.647}{3.322} \approx 2.001 \][/tex]
[tex]\[ \text{Ratio of Oxygen} = \frac{3.338}{3.322} \approx 1.005 \][/tex]
These ratios round off to:
[tex]\[ \text{Empirical formula} = C_1H_2O_1 \][/tex]
3. Calculate the empirical formula mass:
[tex]\[ \text{Empirical mass} = (1 \times 12.01) + (2 \times 1.008) + (1 \times 16.00) = 12.01 + 2.016 + 16.00 = 30.026 \ g/mol \][/tex]
4. Determine the number of empirical units in the molecular formula:
[tex]\[ \text{Molecular units} = \frac{60}{30.026} \approx 1.998 \][/tex]
5. Calculate the molecular formula:
Since the molecular units are approximately 2, the molecular formula is determined by multiplying each subscript in the empirical formula by this factor:
[tex]\[ \text{Molecular formula} = \left(C \cdot 2, H \cdot 2 \times 2, O \cdot 2 \right) = \left(C_2H_4O_2 \right) \][/tex]
So, the molecular formula for the organic acid is [tex]\( \mathbf{C_2H_4O_2} \)[/tex].
### Question 7.2
Determine the metal X in the unknown salt [tex]\( XNO_3 \)[/tex].
Given:
- Mass of [tex]\( XNO_3 \)[/tex] = 34 g
- Volume of [tex]\( MgCl_2 \)[/tex] solution = 250 cm³ (0.250 dm³)
- Concentration of [tex]\( MgCl_2 \)[/tex] solution = 0.4 mol/dm³
- Balanced chemical equation:
[tex]\[ MgCl_2 + 2 XNO_3 \rightarrow 2 XCl + Mg(NO_3)_2 \][/tex]
1. Calculate the moles of [tex]\( MgCl_2 \)[/tex]:
[tex]\[ \text{Moles of } MgCl_2 = \text{Concentration} \times \text{Volume} = 0.4 \ \text{mol/dm}^3 \times 0.250 \ \text{dm}^3 = 0.1 \ \text{mol} \][/tex]
2. Use the stoichiometry to find moles of [tex]\( XNO_3 \)[/tex]:
According to the balanced equation, 1 mole of [tex]\( MgCl_2 \)[/tex] reacts with 2 moles of [tex]\( XNO_3 \)[/tex]:
[tex]\[ \text{Moles of } XNO_3 = 2 \times \text{Moles of } MgCl_2 = 2 \times 0.1 = 0.2 \ \text{mol} \][/tex]
3. Calculate the molar mass of [tex]\( XNO_3 \)[/tex]:
[tex]\[ \text{Molar mass of } XNO_3 = \frac{\text{Mass}}{\text{Moles}} = \frac{34 \ \text{g}}{0.2 \ \text{mol}} = 170 \ \text{g/mol} \][/tex]
4. Determine the atomic mass of [tex]\( X \)[/tex]:
The formula of [tex]\( XNO_3 \)[/tex] consists of one atom of [tex]\( X \)[/tex], one atom of nitrogen [tex]\((N, 14.01 \ \text{g/mol})\)[/tex], and three atoms of oxygen [tex]\((O, 3 \times 16.00 \ \text{g/mol}) = 48.00 \ \text{g/mol})\)[/tex]:
[tex]\[ \text{Molar mass of } XNO_3 = \text{Atomic mass of } X + 14.01 + 48.00 \][/tex]
[tex]\[ 170 = \text{Atomic mass of } X + 62.01 \][/tex]
[tex]\[ \text{Atomic mass of } X = 170 - 62.01 = 107.99 \ \text{g/mol} \][/tex]
So, the metal [tex]\( X \)[/tex] in the unknown salt [tex]\( XNO_3 \)[/tex] is Silver (Ag) with an atomic mass of approximately 107.99 g/mol.
Your participation is crucial to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
