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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
Accepted Answer
Here are the empirical formulas for the given compounds:
(a) CH2O
(b) KNO3
Here's how to calculate them:
1. Assume a 100g sample: This makes the percentages directly translate to grams.
2. Convert grams to moles: Divide each element's mass by its molar mass:
Carbon (C): 40.1 g / 12.011 g/mol = 3.338 mol
Hydrogen (H): 6.6 g / 1.008 g/mol = 6.548 mol
Oxygen (O): 53.3 g / 15.999 g/mol = 3.331 mol
3. Find the simplest whole-number ratio: Divide each mole value by the smallest mole value (3.331 mol in this case):
C: 3.338 mol / 3.331 mol ≈ 1
H: 6.548 mol / 3.331 mol ≈ 2
O: 3.331 mol / 3.331 mol ≈ 1
Therefore, the empirical formula for (a) is CH2O.
You can follow the same steps for compound (b) to obtain the empirical formula KNO3.
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