A certain alkaloid has `70.8%` carbon, `6.2%` hydrogen, `4.1%` nitrogen and the rest oxgen. What is its empirical formula :

Options

A. `C_(20)H_(21)NO_(4)`
B. `C_(20)H_(20)NO_(4)`
C. `C_(21)H_(20)NO_(3)`
D. `C_(20)H_(19)NO_(3)`

Explanation

To determine the empirical formula of the alkaloid based on the given percentages of its constituent elements, we will follow these steps:

### Step 1: Determine the percentage of oxygen
The total percentage of the elements provided is:
- Carbon: 70.8%
- Hydrogen: 6.2%
- Nitrogen: 4.1%

To find the percentage of oxygen, we subtract the sum of the percentages of carbon, hydrogen, and nitrogen from 100%:

\[ \text{Percentage of Oxygen} = 100\% - (70.8\% + 6.2\% + 4.1\%) = 100\% - 81.1\% = 18.9\% \]

### Step 2: Convert percentages to grams
Assuming we have 100 grams of the alkaloid, the mass of each element in grams will be equal to its percentage:

- Mass of Carbon = 70.8 g
- Mass of Hydrogen = 6.2 g
- Mass of Nitrogen = 4.1 g
- Mass of Oxygen = 18.9 g

### Step 3: Convert grams to moles
Now, we will convert the masses of each element to moles using their molar masses:

- Moles of Carbon (C):
\[ \text{Moles of C} = \frac{70.8 \, \text{g}}{12 \, \text{g/mol}} = 5.9 \, \text{mol} \]

- Moles of Hydrogen (H):
\[ \text{Moles of H} = \frac{6.2 \, \text{g}}{1 \, \text{g/mol}} = 6.2 \, \text{mol} \]

- Moles of Nitrogen (N):
\[ \text{Moles of N} = \frac{4.1 \, \text{g}}{14 \, \text{g/mol}} = 0.29 \, \text{mol} \]

- Moles of Oxygen (O):
\[ \text{Moles of O} = \frac{18.9 \, \text{g}}{16 \, \text{g/mol}} = 1.18 \, \text{mol} \]

### Step 4: Find the simplest ratio
Now we have the number of moles for each element:
- Carbon: 5.9
- Hydrogen: 6.2
- Nitrogen: 0.29
- Oxygen: 1.18

To find the simplest ratio, we will divide each mole value by the smallest number of moles, which is 0.29 (for nitrogen):

- Ratio of Carbon:
\[ \frac{5.9}{0.29} \approx 20.31 \approx 20 \]

- Ratio of Hydrogen:
\[ \frac{6.2}{0.29} \approx 21.38 \approx 21 \]

- Ratio of Nitrogen:
\[ \frac{0.29}{0.29} = 1 \]

- Ratio of Oxygen:
\[ \frac{1.18}{0.29} \approx 4.07 \approx 4 \]

### Step 5: Write the empirical formula
Based on the simplest whole number ratios, we can write the empirical formula as:
\[ \text{Empirical Formula} = C_{20}H_{21}N_{1}O_{4} \]

Thus, the empirical formula of the alkaloid is \( \text{C}_{20}\text{H}_{21}\text{N}\text{O}_{4} \).