Question

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(55.3 \% \mathrm{~K}, 14.6 \% \mathrm{P}\), and \(30.1 \% \mathrm{O}\) (b) \(24.5 \% \mathrm{Na}, 14.9 \% \mathrm{Si}\), and \(60.6 \% \mathrm{~F}\) (c) \(62.1 \% \mathrm{C}, 5.21 \% \mathrm{H}, 12.1 \% \mathrm{~N}\), and \(20.7 \% \mathrm{O}\)

Short answer

The empirical formulas for the given compounds are: (a) K₃PO₄ (b) Na₂SiF₆ (c) C₁₂H₁₂N₂O₃

Step-by-step solution

Convert percentages to grams

Assume 100g of sample, as it does not affect the final result: K: 55.3g P: 14.6g O: 30.1g

Convert grams to moles

K: \(\frac{55.3 \text{g}}{39.10 \frac{\text{g}}{\text{mol}}} = 1.414 \text{mol}\) P: \(\frac{14.6 \text{g}}{30.97 \frac{\text{g}}{\text{mol}}} = 0.471 \text{mol}\) O: \(\frac{30.1 \text{g}}{16.00 \frac{\text{g}}{\text{mol}}} = 1.881 \text{mol}\)

Divide by the smallest moles value

K: \(\frac{1.414}{0.471}\) ≈ 3 P: \(\frac{0.471}{0.471}\) ≈ 1 O: \(\frac{1.881}{0.471}\) ≈ 4

Create the empirical formula

K₃PO₄ #(b) 24.5% Na, 14.9% Si, and 60.6% F#

Convert percentages to grams

Assume 100g of sample: Na: 24.5g Si: 14.9g F: 60.6g

Convert grams to moles

Na: \(\frac{24.5 \text{g}}{22.99 \frac{\text{g}}{\text{mol}}} = 1.065 \text{mol}\) Si: \(\frac{14.9 \text{g}}{28.09 \frac{\text{g}}{\text{mol}}} = 0.530 \text{mol}\) F: \(\frac{60.6 \text{g}}{19.00 \frac{\text{g}}{\text{mol}}} = 3.189 \text{mol}\)

Divide by the smallest moles value

Na: \(\frac{1.065}{0.530}\) ≈ 2 Si: \(\frac{0.530}{0.530}\) ≈ 1 F: \(\frac{3.189}{0.530}\) ≈ 6

Create the empirical formula

Na₂SiF₆ #(c) 62.1% C, 5.21% H, 12.1% N, and 20.7% O#

Convert percentages to grams

Assume 100g of sample: C: 62.1g H: 5.21g N: 12.1g O: 20.7g

Convert grams to moles

C: \(\frac{62.1 \text{g}}{12.01 \frac{\text{g}}{\text{mol}}} = 5.171 \text{mol}\) H: \(\frac{5.21 \text{g}}{1.01 \frac{\text{g}}{\text{mol}}} = 5.158 \text{mol}\) N: \(\frac{12.1 \text{g}}{14.01 \frac{\text{g}}{\text{mol}}} = 0.863 \text{mol}\) O: \(\frac{20.7 \text{g}}{16.00 \frac{\text{g}}{\text{mol}}} = 1.294 \text{mol}\)

Divide by the smallest moles value

C: \(\frac{5.171}{0.863}\) ≈ 6 H: \(\frac{5.158}{0.863}\) ≈ 6 N: \(\frac{0.863}{0.863}\) ≈ 1 O: \(\frac{1.294}{0.863}\) ≈ 1.5 Since we obtained 1.5 instead of whole numbers, we should multiply all values by 2 to get integers: C: 6 × 2 = 12 H: 6 × 2 = 12 N: 1 × 2 = 2 O: 1.5 × 2 = 3

Create the empirical formula

C₁₂H₁₂N₂O₃