Questions: Chemical Reactions Finding a molecular formula from molar mass and elemental analysis of binary... Compound X has a molar mass of 104.01 g/mol and the following composition: element - mass % nitrogen - 26.93 % fluorine - 73.07 % Write the molecular formula of X.

Solution Steps

First, we need to find the empirical formula of the compound. We start by assuming we have 100 grams of the compound, which means we have 26.93 grams of nitrogen and 73.07 grams of fluorine.

Next, we convert these masses to moles:

• Moles of nitrogen: \[ \frac{26.93 \, \text{g}}{14.01 \, \text{g/mol}} = 1.922 \, \text{mol} \]

• Moles of fluorine: \[ \frac{73.07 \, \text{g}}{19.00 \, \text{g/mol}} = 3.846 \, \text{mol} \]

Now, we find the simplest whole number ratio by dividing each by the smallest number of moles:

• Ratio for nitrogen: \[ \frac{1.922}{1.922} = 1 \]

• Ratio for fluorine: \[ \frac{3.846}{1.922} = 2 \]

Thus, the empirical formula is \( \text{NF}_2 \).

To find the molecular formula, we compare the molar mass of the empirical formula with the given molar mass of the compound.

• Molar mass of \( \text{NF}_2 \): \[ 14.01 \, \text{g/mol (N)} + 2 \times 19.00 \, \text{g/mol (F)} = 52.01 \, \text{g/mol} \]

• Given molar mass of compound \( X \) is 104.01 g/mol.

Now, we find the ratio of the molar mass of the compound to the molar mass of the empirical formula: \[ \frac{104.01}{52.01} \approx 2 \]

This means the molecular formula is twice the empirical formula: \( \text{N}_2\text{F}_4 \).

Final Answer