A compound contains 40% carbon 6.6% hydrogen and the rest oxygen. If 100 ml of its decimolar solution contains 1.8 gms of it, how many emperical units are present in its molecule?

Answer

To solve the problem step by step, we will determine the empirical formula of the compound and then find out how many empirical units are present in its molecule.

### Step 1: Determine the percentage composition of the elements
We know the percentages of carbon and hydrogen:
- Carbon (C): 40%
- Hydrogen (H): 6.6%

To find the percentage of oxygen (O), we subtract the sum of the percentages of carbon and hydrogen from 100%:
\[ \text{Percentage of Oxygen} = 100\% - (40\% + 6.6\%) = 100\% - 46.6\% = 53.4\% \]

### Step 2: Create a table for the elements
We will create a table to summarize the information:

| Element | Percentage (%) | Atomic Mass (g/mol) | Molar Ratio (mol) |
|---------|----------------|----------------------|--------------------|
| Carbon | 40 | 12 | \(\frac{40}{12} = 3.33\) |
| Hydrogen| 6.6 | 1 | \(\frac{6.6}{1} = 6.6\) |
| Oxygen | 53.4 | 16 | \(\frac{53.4}{16} = 3.34\) |

### Step 3: Calculate the molar ratios
Now we calculate the molar ratios for each element:
- Molar ratio of Carbon: \(\frac{40}{12} \approx 3.33\)
- Molar ratio of Hydrogen: \(\frac{6.6}{1} = 6.6\)
- Molar ratio of Oxygen: \(\frac{53.4}{16} \approx 3.34\)

### Step 4: Determine the simplest ratio
Next, we need to find the simplest ratio among the molar ratios. The smallest value among 3.33, 6.6, and 3.34 is approximately 3.33. We will divide all molar ratios by this smallest value:

- For Carbon: \(\frac{3.33}{3.33} = 1\)
- For Hydrogen: \(\frac{6.6}{3.33} \approx 2\)
- For Oxygen: \(\frac{3.34}{3.33} \approx 1\)

### Step 5: Write the empirical formula
From the ratios, we can conclude that the empirical formula is:
\[ \text{Empirical Formula} = C_1H_2O_1 = CH_2O \]

### Step 6: Calculate the empirical formula mass
Now we calculate the empirical formula mass:
\[ \text{Empirical Formula Mass} = (1 \times 12) + (2 \times 1) + (1 \times 16) = 12 + 2 + 16 = 30 \text{ g/mol} \]

### Step 7: Determine the molecular weight from the solution
We know that 100 ml of a decimolar solution contains 1.8 grams of the compound. A decimolar solution means:
\[ \text{Concentration} = 0.1 \text{ mol/L} \]
So, for 100 ml (0.1 L):
\[ \text{Moles of compound} = 0.1 \text{ mol/L} \times 0.1 \text{ L} = 0.01 \text{ mol} \]
Given that 0.01 moles weigh 1.8 grams, we can find the molar mass:
\[ \text{Molar Mass} = \frac{1.8 \text{ g}}{0.01 \text{ mol}} = 180 \text{ g/mol} \]

### Step 8: Calculate the number of empirical units in the molecule
To find the number of empirical formula units in the molecule, we divide the molar mass by the empirical formula mass:
\[ n = \frac{\text{Molecular Weight}}{\text{Empirical Formula Mass}} = \frac{180 \text{ g/mol}}{30 \text{ g/mol}} = 6 \]

### Final Answer
The number of empirical formula units present in the molecule is **6**.

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