Question

Workings of An organic compound contains 40% carbon 6.7% of hydrogen and the third component is oxygen.If the relative molecular mass of the compound is 60.
Calculate the empirical formular and molecular formular

Answer 1

To find the empirical formula and molecular formula of the organic compound, we can follow these steps:

### Step 1: Determine the percentage of oxygen
Given:
- Carbon (C) = 40%
- Hydrogen (H) = 6.7%
- Oxygen (O) = 100% - (40% + 6.7%) = 100% - 46.7% = 53.3%

### Step 2: Convert the percentages to grams
Let's assume we have 100 grams of the compound, which means:
- Mass of C = 40 g
- Mass of H = 6.7 g
- Mass of O = 53.3 g

### Step 3: Convert grams to moles
To find the number of moles of each element, we use their molar masses:
- Molar mass of C = 12 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of O = 16 g/mol

Now convert grams to moles:
- Moles of C = \( \frac{40 \text{ g}}{12 \text{ g/mol}} \approx 3.33 \, \text{mol} \)
- Moles of H = \( \frac{6.7 \text{ g}}{1 \text{ g/mol}} = 6.7 \, \text{mol} \)
- Moles of O = \( \frac{53.3 \text{ g}}{16 \text{ g/mol}} \approx 3.33 \, \text{mol} \)

### Step 4: Find the simplest ratio
Next, we need to find the simplest whole number ratio by dividing all mole values by the smallest number of moles (which is 3.33 moles in this case):

- Ratio of C: \( \frac{3.33}{3.33} = 1 \)
- Ratio of H: \( \frac{6.7}{3.33} \approx 2.01 \approx 2 \)
- Ratio of O: \( \frac{3.33}{3.33} = 1 \)

This gives us a ratio of \( C: 1, H: 2, O: 1 \).

### Step 5: Empirical formula
The empirical formula is therefore \( \text{CH}_2\text{O} \).

### Step 6: Calculate the empirical formula mass
To determine if the empirical formula can be used to find the molecular formula, we calculate the molar mass of the empirical formula:

- Molar mass of \( \text{CH}_2\text{O} = 12 + (2 \times 1) + 16 = 30 \, \text{g/mol} \)

### Step 7: Determine the molecular formula
Given the relative molecular mass of the compound is 60 g/mol, we find the ratio:

- Ratio = \( \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{60 \, \text{g/mol}}{30 \, \text{g/mol}} = 2 \)

### Step 8: Calculate the molecular formula
To find the molecular formula, multiply the subscripts in the empirical formula by this ratio (2):

- Molecular formula: \( \text{C}_{(1 \times 2)}\text{H}_{(2 \times 2)}\text{O}_{(1 \times 2)} = \text{C}_2\text{H}_4\text{O}_2 \)

### Summary
- **Empirical formula**: \( \text{CH}_2\text{O} \)
- **Molecular formula**: \( \text{C}_2\text{H}_4\text{O}_2 \)