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Sagot :
To determine the molecular formula of a compound given its molar mass and percent composition, we follow these steps:
1. Calculate the Mass of Each Element in the Compound:
- The molar mass of the compound is 92.02 grams/mole.
- The percent composition is 30.4% nitrogen (N) and 69.6% oxygen (O).
Therefore, the mass of each element in 1 mole of the compound:
[tex]\[ \text{Mass of N} = \left( \frac{30.4}{100} \right) \times 92.02 \text{ g} = 27.97 \text{ g} \][/tex]
[tex]\[ \text{Mass of O} = \left( \frac{69.6}{100} \right) \times 92.02 \text{ g} = 64.05 \text{ g} \][/tex]
2. Convert Masses to Moles:
- The molar mass of nitrogen (N) is 14.01 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
Therefore, the number of moles of each element:
[tex]\[ \text{Moles of N} = \frac{27.97 \text{ g}}{14.01 \text{ g/mol}} = 1.997 \approx 2 \text{ moles} \][/tex]
[tex]\[ \text{Moles of O} = \frac{64.05 \text{ g}}{16.00 \text{ g/mol}} = 4.003 \approx 4 \text{ moles} \][/tex]
3. Determine the Simplest Whole Number Ratio:
- The ratio of moles of nitrogen to moles of oxygen is approximately:
[tex]\[ N : O = 2 : 4 \][/tex]
Simplifying the ratio by dividing both numbers by the smallest value (2):
[tex]\[ N : O = 1 : 2 \][/tex]
4. Establish the Empirical and Molecular Formula:
- The simplest whole number ratio gives us the empirical formula [tex]\( NO_2 \)[/tex].
- To see if this is the molecular formula, we must check if the molar mass of [tex]\( NO_2 \)[/tex] matches the given molar mass of the compound.
The molar mass of [tex]\( NO_2 \)[/tex]:
[tex]\[ \text{Molar mass of } NO_2 = 14.01 \text{ (N)} + 2 \times 16.00 \text{ (O)} = 14.01 + 32.00 = 46.01 \text{ g/mol} \][/tex]
Since the molar mass of the compound (92.02 g/mol) is twice the molar mass of [tex]\( NO_2 \)[/tex], the molecular formula must be:
[tex]\[ \text{Molecular formula} = 2 \times NO_2 = N_2O_4 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{N_2O_4} \][/tex]
1. Calculate the Mass of Each Element in the Compound:
- The molar mass of the compound is 92.02 grams/mole.
- The percent composition is 30.4% nitrogen (N) and 69.6% oxygen (O).
Therefore, the mass of each element in 1 mole of the compound:
[tex]\[ \text{Mass of N} = \left( \frac{30.4}{100} \right) \times 92.02 \text{ g} = 27.97 \text{ g} \][/tex]
[tex]\[ \text{Mass of O} = \left( \frac{69.6}{100} \right) \times 92.02 \text{ g} = 64.05 \text{ g} \][/tex]
2. Convert Masses to Moles:
- The molar mass of nitrogen (N) is 14.01 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
Therefore, the number of moles of each element:
[tex]\[ \text{Moles of N} = \frac{27.97 \text{ g}}{14.01 \text{ g/mol}} = 1.997 \approx 2 \text{ moles} \][/tex]
[tex]\[ \text{Moles of O} = \frac{64.05 \text{ g}}{16.00 \text{ g/mol}} = 4.003 \approx 4 \text{ moles} \][/tex]
3. Determine the Simplest Whole Number Ratio:
- The ratio of moles of nitrogen to moles of oxygen is approximately:
[tex]\[ N : O = 2 : 4 \][/tex]
Simplifying the ratio by dividing both numbers by the smallest value (2):
[tex]\[ N : O = 1 : 2 \][/tex]
4. Establish the Empirical and Molecular Formula:
- The simplest whole number ratio gives us the empirical formula [tex]\( NO_2 \)[/tex].
- To see if this is the molecular formula, we must check if the molar mass of [tex]\( NO_2 \)[/tex] matches the given molar mass of the compound.
The molar mass of [tex]\( NO_2 \)[/tex]:
[tex]\[ \text{Molar mass of } NO_2 = 14.01 \text{ (N)} + 2 \times 16.00 \text{ (O)} = 14.01 + 32.00 = 46.01 \text{ g/mol} \][/tex]
Since the molar mass of the compound (92.02 g/mol) is twice the molar mass of [tex]\( NO_2 \)[/tex], the molecular formula must be:
[tex]\[ \text{Molecular formula} = 2 \times NO_2 = N_2O_4 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{N_2O_4} \][/tex]
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