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What are the empirical formulas of the compounds with the following compositions? (a) 40.1 percent c, 6.6 percent h, 53.3 percent o; (b) 18.4 percent c, 21.5 percent n, 60.1 percent k ?hydrogen is 1.008 solve it accurately 4 decimal point calcu may need daw ilagay na 1mol/dun sa mga 1.008 mass nila
Accepted Answer
Here's how to determine the empirical formulas for the given compounds:
Steps:
1. Assume a 100g sample: This makes the percentages directly translate to grams.
2. Convert grams to moles: Divide the mass of each element by its molar mass.
3. Find the simplest whole-number ratio: Divide each mole value by the smallest mole value obtained.
4. Write the empirical formula: Use the whole-number ratios as subscripts for each element.
Let's apply this to your examples:
(a) 40.1% C, 6.6% H, 53.3% O
- Moles of C: 40.1 g / 12.011 g/mol = 3.34 mol
- Moles of H: 6.6 g / 1.008 g/mol = 6.55 mol
- Moles of O: 53.3 g / 15.999 g/mol = 3.33 mol
- Divide by the smallest (3.33 mol): C: 1.00, H: 1.97, O: 1.00
- Round to the nearest whole number: C1H2O1
- Empirical Formula: CH2O
(b) 18.4% C, 21.5% N, 60.1% K
- Moles of C: 18.4 g / 12.011 g/mol = 1.53 mol
- Moles of N: 21.5 g / 14.007 g/mol = 1.53 mol
- Moles of K: 60.1 g / 39.098 g/mol = 1.54 mol
- Divide by the smallest (1.53 mol): C: 1.00, N: 1.00, K: 1.01
- Round to the nearest whole number: C1N1K1
- Empirical Formula: KCN
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