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Sagot :
To determine the molecular formula of a compound given its empirical formula and molar mass, we need to follow a series of steps. Let’s go through the process step-by-step.
### Step 1: Calculate the Molar Mass of the Empirical Formula
First, we need to find the molar mass of the empirical formula, which is [tex]\(C_6H_6NO\)[/tex].
- The atomic mass of carbon (C) is approximately 12 g/mol.
- The atomic mass of hydrogen (H) is approximately 1 g/mol.
- The atomic mass of nitrogen (N) is approximately 14 g/mol.
- The atomic mass of oxygen (O) is approximately 16 g/mol.
Using these atomic masses, we calculate the molar mass of the empirical formula:
[tex]\[ \text{Molar mass of } C_6H_6NO = 6(12) + 6(1) + 14 + 16 \][/tex]
Calculating each component:
[tex]\[ 6 \times 12 = 72 \][/tex]
[tex]\[ 6 \times 1 = 6 \][/tex]
[tex]\[ 1 \times 14 = 14 \][/tex]
[tex]\[ 1 \times 16 = 16 \][/tex]
Adding these together gives:
[tex]\[ 72 + 6 + 14 + 16 = 108 \][/tex]
So, the molar mass of the empirical formula [tex]\(C_6H_6NO\)[/tex] is [tex]\(108 \text{ g/mol}\)[/tex].
### Step 2: Determine the Ratio of Molar Masses
Next, we compare the given molar mass of the compound to the molar mass of the empirical formula to find the ratio.
Given molar mass of the compound:
[tex]\[ 216 \text{ g/mol} \][/tex]
Molar mass of the empirical formula:
[tex]\[ 108 \text{ g/mol} \][/tex]
The ratio of the molar mass of the compound to the molar mass of the empirical formula is:
[tex]\[ \text{Ratio} = \frac{216}{108} = 2 \][/tex]
### Step 3: Calculate the Molecular Formula
The molecular formula is a multiple of the empirical formula. The ratio calculated above indicates how many times the empirical formula's mass fits into the compound's molar mass.
Thus, we multiply each subscript in the empirical formula [tex]\(C_6H_6NO\)[/tex] by this ratio (2) to get the molecular formula:
[tex]\[ C_{6 \times 2}H_{6 \times 2}N_{1 \times 2}O_{1 \times 2} \][/tex]
Simplifying:
[tex]\[ C_{12}H_{12}N_2O_2 \][/tex]
Therefore, the molecular formula of the compound is:
[tex]\[ \boxed{C_{12}H_{12}N_2O_2} \][/tex]
### Conclusion
By step-by-step calculation, we have determined that the molecular formula of the compound with a molar mass of 216 g/mol and an empirical formula [tex]\(C_6H_6NO\)[/tex] is [tex]\(C_{12}H_{12}N_2O_2\)[/tex]. This matches the second option provided.
### Step 1: Calculate the Molar Mass of the Empirical Formula
First, we need to find the molar mass of the empirical formula, which is [tex]\(C_6H_6NO\)[/tex].
- The atomic mass of carbon (C) is approximately 12 g/mol.
- The atomic mass of hydrogen (H) is approximately 1 g/mol.
- The atomic mass of nitrogen (N) is approximately 14 g/mol.
- The atomic mass of oxygen (O) is approximately 16 g/mol.
Using these atomic masses, we calculate the molar mass of the empirical formula:
[tex]\[ \text{Molar mass of } C_6H_6NO = 6(12) + 6(1) + 14 + 16 \][/tex]
Calculating each component:
[tex]\[ 6 \times 12 = 72 \][/tex]
[tex]\[ 6 \times 1 = 6 \][/tex]
[tex]\[ 1 \times 14 = 14 \][/tex]
[tex]\[ 1 \times 16 = 16 \][/tex]
Adding these together gives:
[tex]\[ 72 + 6 + 14 + 16 = 108 \][/tex]
So, the molar mass of the empirical formula [tex]\(C_6H_6NO\)[/tex] is [tex]\(108 \text{ g/mol}\)[/tex].
### Step 2: Determine the Ratio of Molar Masses
Next, we compare the given molar mass of the compound to the molar mass of the empirical formula to find the ratio.
Given molar mass of the compound:
[tex]\[ 216 \text{ g/mol} \][/tex]
Molar mass of the empirical formula:
[tex]\[ 108 \text{ g/mol} \][/tex]
The ratio of the molar mass of the compound to the molar mass of the empirical formula is:
[tex]\[ \text{Ratio} = \frac{216}{108} = 2 \][/tex]
### Step 3: Calculate the Molecular Formula
The molecular formula is a multiple of the empirical formula. The ratio calculated above indicates how many times the empirical formula's mass fits into the compound's molar mass.
Thus, we multiply each subscript in the empirical formula [tex]\(C_6H_6NO\)[/tex] by this ratio (2) to get the molecular formula:
[tex]\[ C_{6 \times 2}H_{6 \times 2}N_{1 \times 2}O_{1 \times 2} \][/tex]
Simplifying:
[tex]\[ C_{12}H_{12}N_2O_2 \][/tex]
Therefore, the molecular formula of the compound is:
[tex]\[ \boxed{C_{12}H_{12}N_2O_2} \][/tex]
### Conclusion
By step-by-step calculation, we have determined that the molecular formula of the compound with a molar mass of 216 g/mol and an empirical formula [tex]\(C_6H_6NO\)[/tex] is [tex]\(C_{12}H_{12}N_2O_2\)[/tex]. This matches the second option provided.
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