Question

calculate the empirical formula of the following compounds:

1. a compound containing 8.83 g of sodium and 6.17 g of sulfur.

1. a compound that is 36.48% na, 25.41% s, and 38.11% o.

1. a compound that is 50.81% zn, 16.06% p, and 33.15% o.

calculate the molecular formula for the following compounds:

1. a compound with an empirical formula p₂o₅ and a molar mass of 284 g/mol.
2. a compound with an empirical formula ocncl and a molar mass of 232.41 g/mol.
3. a compound that is 40% c, 6.7% h, and 53.3% o. the molar mass is 60.0 g/mol.

Explanation:

Problem 7:

Step 1: Calculate moles of Na and S

Molar mass of Na ($M_{Na}$) = 22.99 g/mol, moles of Na ($n_{Na}$) = $\frac{8.83\ g}{22.99\ g/mol} \approx 0.384\ mol$
Molar mass of S ($M_{S}$) = 32.07 g/mol, moles of S ($n_{S}$) = $\frac{6.17\ g}{32.07\ g/mol} \approx 0.192\ mol$

Step 2: Find mole ratio

Divide moles by the smallest (0.192):
Ratio of Na: $\frac{0.384}{0.192} = 2$
Ratio of S: $\frac{0.192}{0.192} = 1$

Step 1: Assume 100 g sample, moles of each element

Moles of Na: $\frac{36.48\ g}{22.99\ g/mol} \approx 1.587\ mol$
Moles of S: $\frac{25.41\ g}{32.07\ g/mol} \approx 0.792\ mol$
Moles of O: $\frac{38.11\ g}{16.00\ g/mol} \approx 2.382\ mol$

Step 2: Mole ratio (divide by smallest, 0.792)

Na: $\frac{1.587}{0.792} \approx 2$
S: $\frac{0.792}{0.792} = 1$
O: $\frac{2.382}{0.792} \approx 3$

Step 1: Assume 100 g, moles of each

Moles of Zn: $\frac{50.81\ g}{65.38\ g/mol} \approx 0.777\ mol$
Moles of P: $\frac{16.06\ g}{30.97\ g/mol} \approx 0.519\ mol$
Moles of O: $\frac{33.15\ g}{16.00\ g/mol} \approx 2.072\ mol$

Step 2: Mole ratio (divide by 0.519)

Zn: $\frac{0.777}{0.519} \approx 1.5$ (×2 to eliminate decimal: 3)
P: $\frac{0.519}{0.519} = 1$ (×2: 2)
O: $\frac{2.072}{0.519} \approx 4$ (×2: 8)

Wait, correction: Recheck ratios. Let’s use 0.519 as smallest:
Zn: $0.777/0.519 ≈ 1.5$ → multiply all by 2:
Zn: 3, P: 2, O: 8 (since 2.072/0.519 ≈ 4, ×2=8). Wait, no—original division:
Wait, 0.777/0.519 ≈ 1.5, 0.519/0.519=1, 2.072/0.519≈4. So multiply by 2: 3, 2, 8.

Answer:

$Na_2S$

Problem 8: