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To determine the empirical formula of urea, we need to follow a step-by-step process:
### Step 1: Determine the masses of each element in the compound
Given:
- Mass of Nitrogen (N) = 1.121 g
- Mass of Hydrogen (H) = 0.161 g
- Mass of Carbon (C) = 0.480 g
- Mass of Oxygen (O) = 0.640 g
### Step 2: Calculate the number of moles of each element
We use the molar mass of each element to convert the given masses into moles.
Molar masses:
- Molar mass of Nitrogen (N) = 14.01 g/mol
- Molar mass of Hydrogen (H) = 1.008 g/mol
- Molar mass of Carbon (C) = 12.01 g/mol
- Molar mass of Oxygen (O) = 16.00 g/mol
Using the formula:
[tex]\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \][/tex]
For Nitrogen:
[tex]\[ \text{Moles of } N = \frac{1.121 \text{ g}}{14.01 \text{ g/mol}} = 0.08001427551748751 \text{ moles} \][/tex]
For Hydrogen:
[tex]\[ \text{Moles of } H = \frac{0.161 \text{ g}}{1.008 \text{ g/mol}} = 0.15972222222222224 \text{ moles} \][/tex]
For Carbon:
[tex]\[ \text{Moles of } C = \frac{0.480 \text{ g}}{12.01 \text{ g/mol}} = 0.03996669442131557 \text{ moles} \][/tex]
For Oxygen:
[tex]\[ \text{Moles of } O = \frac{0.640 \text{ g}}{16.00 \text{ g/mol}} = 0.04 \text{ moles} \][/tex]
### Step 3: Normalize the moles to the smallest number of moles
Now we find the smallest number of moles among the elements:
[tex]\[ \text{Smallest number of moles} = 0.03996669442131557 \text{ moles (for Carbon)} \][/tex]
### Step 4: Calculate the mole ratio for each element by dividing by the smallest number of moles
For Nitrogen:
[tex]\[ \text{Ratio of } N = \frac{0.08001427551748751}{0.03996669442131557} = 2 \][/tex]
For Hydrogen:
[tex]\[ \text{Ratio of } H = \frac{0.15972222222222224}{0.03996669442131557} = 4 \][/tex]
For Carbon:
[tex]\[ \text{Ratio of } C = \frac{0.03996669442131557}{0.03996669442131557} = 1 \][/tex]
For Oxygen:
[tex]\[ \text{Ratio of } O = \frac{0.04}{0.03996669442131557} = 1 \][/tex]
### Step 5: Round to the nearest whole number to get the empirical formula
The ratios are already in whole numbers:
- Nitrogen: 2
- Hydrogen: 4
- Carbon: 1
- Oxygen: 1
### Conclusion: Empirical Formula
The empirical formula of urea is [tex]\( \text{N}_2\text{H}_4\text{C}\text{O} \)[/tex].
Thus, we have determined that the empirical formula is [tex]\( \text{N}_2\text{H}_4\text{C}\text{O} \)[/tex].
### Step 1: Determine the masses of each element in the compound
Given:
- Mass of Nitrogen (N) = 1.121 g
- Mass of Hydrogen (H) = 0.161 g
- Mass of Carbon (C) = 0.480 g
- Mass of Oxygen (O) = 0.640 g
### Step 2: Calculate the number of moles of each element
We use the molar mass of each element to convert the given masses into moles.
Molar masses:
- Molar mass of Nitrogen (N) = 14.01 g/mol
- Molar mass of Hydrogen (H) = 1.008 g/mol
- Molar mass of Carbon (C) = 12.01 g/mol
- Molar mass of Oxygen (O) = 16.00 g/mol
Using the formula:
[tex]\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \][/tex]
For Nitrogen:
[tex]\[ \text{Moles of } N = \frac{1.121 \text{ g}}{14.01 \text{ g/mol}} = 0.08001427551748751 \text{ moles} \][/tex]
For Hydrogen:
[tex]\[ \text{Moles of } H = \frac{0.161 \text{ g}}{1.008 \text{ g/mol}} = 0.15972222222222224 \text{ moles} \][/tex]
For Carbon:
[tex]\[ \text{Moles of } C = \frac{0.480 \text{ g}}{12.01 \text{ g/mol}} = 0.03996669442131557 \text{ moles} \][/tex]
For Oxygen:
[tex]\[ \text{Moles of } O = \frac{0.640 \text{ g}}{16.00 \text{ g/mol}} = 0.04 \text{ moles} \][/tex]
### Step 3: Normalize the moles to the smallest number of moles
Now we find the smallest number of moles among the elements:
[tex]\[ \text{Smallest number of moles} = 0.03996669442131557 \text{ moles (for Carbon)} \][/tex]
### Step 4: Calculate the mole ratio for each element by dividing by the smallest number of moles
For Nitrogen:
[tex]\[ \text{Ratio of } N = \frac{0.08001427551748751}{0.03996669442131557} = 2 \][/tex]
For Hydrogen:
[tex]\[ \text{Ratio of } H = \frac{0.15972222222222224}{0.03996669442131557} = 4 \][/tex]
For Carbon:
[tex]\[ \text{Ratio of } C = \frac{0.03996669442131557}{0.03996669442131557} = 1 \][/tex]
For Oxygen:
[tex]\[ \text{Ratio of } O = \frac{0.04}{0.03996669442131557} = 1 \][/tex]
### Step 5: Round to the nearest whole number to get the empirical formula
The ratios are already in whole numbers:
- Nitrogen: 2
- Hydrogen: 4
- Carbon: 1
- Oxygen: 1
### Conclusion: Empirical Formula
The empirical formula of urea is [tex]\( \text{N}_2\text{H}_4\text{C}\text{O} \)[/tex].
Thus, we have determined that the empirical formula is [tex]\( \text{N}_2\text{H}_4\text{C}\text{O} \)[/tex].
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