A hydrocarbon contains 85.7% carbon. If 42 mg of the compound contain `3.01 xx 10^20` molecules, find the molecular formula of the compound.

Answer

To find the molecular formula of the hydrocarbon that contains 85.7% carbon, we can follow these steps:

### Step 1: Determine the percentage of hydrogen
Since the hydrocarbon contains only carbon and hydrogen, we can find the percentage of hydrogen by subtracting the percentage of carbon from 100%.

\[ \text{Percentage of Hydrogen} = 100\% - 85.7\% = 14.3\% \]

### Step 2: Convert percentages to grams
Assuming we have 100 g of the compound, we can convert the percentages to grams:
- Mass of Carbon = 85.7 g
- Mass of Hydrogen = 14.3 g

### Step 3: Calculate the number of moles of each element
Using the molar masses (C = 12 g/mol and H = 1 g/mol), we can calculate the number of moles of carbon and hydrogen.

\[ \text{Moles of Carbon} = \frac{85.7 \text{ g}}{12 \text{ g/mol}} = 7.14 \text{ moles} \]

\[ \text{Moles of Hydrogen} = \frac{14.3 \text{ g}}{1 \text{ g/mol}} = 14.3 \text{ moles} \]

### Step 4: Find the simplest mole ratio
To find the simplest ratio of moles of carbon to moles of hydrogen, we divide both by the smallest number of moles:

\[ \text{Ratio of Carbon} = \frac{7.14}{7.14} = 1 \]

\[ \text{Ratio of Hydrogen} = \frac{14.3}{7.14} \approx 2 \]

Thus, the simplest ratio of carbon to hydrogen is 1:2.

### Step 5: Write the empirical formula
From the mole ratio, we can write the empirical formula as:

\[ \text{Empirical Formula} = CH_2 \]

### Step 6: Determine the molecular formula
Next, we need to find the molecular formula. We know that 42 mg of the compound contains \(3.01 \times 10^{20}\) molecules. First, convert 42 mg to grams:

\[ 42 \text{ mg} = 0.042 \text{ g} \]

Using Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol), we can find the number of moles in 0.042 g:

\[ \text{Number of moles} = \frac{0.042 \text{ g}}{M} \quad \text{(where M is the molar mass)} \]

Setting this equal to the number of molecules divided by Avogadro's number:

\[ \frac{0.042 \text{ g}}{M} = \frac{3.01 \times 10^{20}}{6.022 \times 10^{23}} \]

Calculating the right side:

\[ \frac{3.01 \times 10^{20}}{6.022 \times 10^{23}} \approx 5.00 \times 10^{-4} \text{ moles} \]

Now, equating the two expressions:

\[ \frac{0.042 \text{ g}}{M} = 5.00 \times 10^{-4} \text{ moles} \]

Solving for M:

\[ M = \frac{0.042 \text{ g}}{5.00 \times 10^{-4} \text{ moles}} \approx 84 \text{ g/mol} \]

### Step 7: Determine the molecular formula
Now, we can find the molecular formula. The molar mass of the empirical formula \(CH_2\) is:

\[ \text{Molar mass of } CH_2 = 12 \text{ g/mol (C)} + 2 \times 1 \text{ g/mol (H)} = 14 \text{ g/mol} \]

Now, we divide the molar mass of the compound by the molar mass of the empirical formula:

\[ \text{n} = \frac{84 \text{ g/mol}}{14 \text{ g/mol}} = 6 \]

Thus, the molecular formula is:

\[ \text{Molecular Formula} = (CH_2)_6 = C_6H_{12} \]

### Final Answer
The molecular formula of the compound is \(C_6H_{12}\).

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