Question

1. determine the empirical formula of the indicated compound. (4 points)

a. a sulfide of phosphorus that contains 27.9 % phosphorus and no other element except sulfur
b. a compound if a 10.98 gram sample is found to contain 1.91 grams of carbon and no other element except fluorine

Explanation:

Step1: Calculate mass of each element (A)

Mass of P: $0.279 \times m_{total}$, assume $m_{total}=100\ \text{g}$, so $m_P=27.9\ \text{g}$
Mass of S: $100 - 27.9 = 72.1\ \text{g}$

Step2: Convert mass to moles (A)

Molar mass of P = $30.97\ \text{g/mol}$, molar mass of S = $32.07\ \text{g/mol}$
$n_P = \frac{27.9}{30.97} \approx 0.901\ \text{mol}$
$n_S = \frac{72.1}{32.07} \approx 2.248\ \text{mol}$

Step3: Find mole ratio (A)

Divide by smallest mole value ($0.901$):
$\text{Ratio of P} = \frac{0.901}{0.901} = 1$
$\text{Ratio of S} = \frac{2.248}{0.901} \approx 2.5 = \frac{5}{2}$
Multiply by 2 to get whole numbers: P:S = 2:5

Step4: Calculate mass of each element (B)

Mass of C = $1.91\ \text{g}$, mass of F = $10.98 - 1.91 = 9.07\ \text{g}$

Step5: Convert mass to moles (B)

Molar mass of C = $12.01\ \text{g/mol}$, molar mass of F = $19.00\ \text{g/mol}$
$n_C = \frac{1.91}{12.01} \approx 0.159\ \text{mol}$
$n_F = \frac{9.07}{19.00} \approx 0.477\ \text{mol}$

Step6: Find mole ratio (B)

Divide by smallest mole value ($0.159$):
$\text{Ratio of C} = \frac{0.159}{0.159} = 1$
$\text{Ratio of F} = \frac{0.477}{0.159} = 3$
Ratio of C:F = 1:3

Answer:

A. $\text{P}_2\text{S}_5$
B. $\text{CF}_3$