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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 example 40.1g/(1mol/12.01)
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Here are the empirical formulas for the given compounds:
(a) 40.1% C, 6.6% H, 53.3% O
Assume a 100g sample, so you have 40.1g C, 6.6g H, and 53.3g O.
Convert grams to moles:
C: 40.1g / (12.01 g/mol) = 3.34 mol C
H: 6.6g / (1.008 g/mol) = 6.55 mol H
O: 53.3g / (16.00 g/mol) = 3.33 mol O
Divide each mole value by the smallest mole value (3.33 mol) to get the simplest whole-number ratio:
C: 3.34 mol / 3.33 mol = 1.003 ≈ 1
H: 6.55 mol / 3.33 mol = 1.97 ≈ 2
O: 3.33 mol / 3.33 mol = 1
Therefore, the empirical formula is CH₂O.
(b) 18.4% C, 21.5% N, 60.1% K
Assume a 100g sample, so you have 18.4g C, 21.5g N, and 60.1g K.
Convert grams to moles:
C: 18.4g / (12.01 g/mol) = 1.53 mol C
N: 21.5g / (14.01 g/mol) = 1.53 mol N
K: 60.1g / (39.10 g/mol) = 1.54 mol K
Divide each mole value by the smallest mole value (1.53 mol) to get the simplest whole-number ratio:
C: 1.53 mol / 1.53 mol = 1
N: 1.53 mol / 1.53 mol = 1
K: 1.54 mol / 1.53 mol = 1.01 ≈ 1
Therefore, the empirical formula is KCN.
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