Butyric acid contains only C, H and O. A 4.24 mg sample of butyric acid is completely burned. It gives 8.45 mg of `CO_(2)` and 3.46 mg og `H_(2)O`. The molecular mass of butyric acid was determined by experiment to be 88 amu. What is molecular formula?

Explanation

To determine the molecular formula of butyric acid from the given data, we will follow these steps:

### Step 1: Calculate the mass of Carbon in the sample
We know that when butyric acid is burned, it produces carbon dioxide (CO₂). The mass of CO₂ produced is 8.45 mg.

Using the molar mass of CO₂ (44 g/mol), we can find the mass of carbon in the CO₂ produced:

\[ \text{Mass of Carbon} = \left(\frac{12 \text{ g}}{44 \text{ g}}\right) \times 8.45 \text{ mg} \]

Convert mg to g for easier calculation:

\[ \text{Mass of Carbon} = \left(\frac{12}{44}\right) \times 8.45 \times 10^{-3} \text{ g} = 2.30 \text{ mg} \]

### Step 2: Calculate the mass of Hydrogen in the sample
Next, we calculate the mass of hydrogen from the water produced. The mass of water (H₂O) produced is 3.46 mg.

Using the molar mass of water (18 g/mol), we find the mass of hydrogen in the water:

\[ \text{Mass of Hydrogen} = \left(\frac{2 \text{ g}}{18 \text{ g}}\right) \times 3.46 \text{ mg} \]

Convert mg to g:

\[ \text{Mass of Hydrogen} = \left(\frac{2}{18}\right) \times 3.46 \times 10^{-3} \text{ g} = 0.384 \text{ mg} \]

### Step 3: Calculate the mass of Oxygen in the sample
Now, we can find the mass of oxygen by subtracting the mass of carbon and hydrogen from the total mass of butyric acid.

\[ \text{Mass of Butyric Acid} = 4.24 \text{ mg} \]
\[ \text{Mass of Oxygen} = \text{Total Mass} - (\text{Mass of Carbon} + \text{Mass of Hydrogen}) \]
\[ \text{Mass of Oxygen} = 4.24 \text{ mg} - (2.30 \text{ mg} + 0.384 \text{ mg}) = 1.63 \text{ mg} \]

### Step 4: Convert masses to grams
We have:
- Mass of Carbon = 2.30 mg = 0.00230 g
- Mass of Hydrogen = 0.384 mg = 0.000384 g
- Mass of Oxygen = 1.63 mg = 0.00163 g

### Step 5: Calculate moles of each element
Now we will convert the masses to moles using their respective molar masses:
- Molar mass of Carbon (C) = 12 g/mol
- Molar mass of Hydrogen (H) = 1 g/mol
- Molar mass of Oxygen (O) = 16 g/mol

\[ \text{Moles of Carbon} = \frac{0.00230 \text{ g}}{12 \text{ g/mol}} = 0.0001917 \text{ mol} \]
\[ \text{Moles of Hydrogen} = \frac{0.000384 \text{ g}}{1 \text{ g/mol}} = 0.000384 \text{ mol} \]
\[ \text{Moles of Oxygen} = \frac{0.00163 \text{ g}}{16 \text{ g/mol}} = 0.000101875 \text{ mol} \]

### Step 6: Determine the simplest mole ratio
To find the simplest ratio, we divide each mole value by the smallest number of moles calculated:

\[ \text{Moles of Carbon} = \frac{0.0001917}{0.000101875} \approx 1.88 \approx 2 \]
\[ \text{Moles of Hydrogen} = \frac{0.000384}{0.000101875} \approx 3.77 \approx 4 \]
\[ \text{Moles of Oxygen} = \frac{0.000101875}{0.000101875} \approx 1 \]

### Step 7: Write the empirical formula
From the ratios, we can write the empirical formula as:
\[ \text{Empirical Formula} = C_2H_4O \]

### Step 8: Calculate the empirical formula mass
\[ \text{Empirical Formula Mass} = (2 \times 12) + (4 \times 1) + (1 \times 16) = 24 + 4 + 16 = 44 \text{ g/mol} \]

### Step 9: Determine the molecular formula
Given that the molecular mass of butyric acid is 88 g/mol, we can find the factor \( n \):

\[ n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{88}{44} = 2 \]

### Step 10: Write the molecular formula
Thus, the molecular formula of butyric acid is:
\[ \text{Molecular Formula} = n \times \text{Empirical Formula} = 2 \times C_2H_4O = C_4H_8O_2 \]

### Final Answer
The molecular formula of butyric acid is \( C_4H_8O_2 \).

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