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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
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Here are the empirical formulas of the compounds with the given compositions:
(a) 40.1% C, 6.6% H, 53.3% O
1. Assume a 100 g sample: This means we have 40.1 g C, 6.6 g H, and 53.3 g O.
2. Convert grams to moles:
Moles of C = 40.1 g / 12.01 g/mol = 3.34 mol C
Moles of H = 6.6 g / 1.008 g/mol = 6.55 mol H
Moles of O = 53.3 g / 16.00 g/mol = 3.33 mol O
3. Divide by the smallest number of moles:
C: 3.34 mol / 3.33 mol = 1.00
H: 6.55 mol / 3.33 mol = 1.97 ≈ 2.00
O: 3.33 mol / 3.33 mol = 1.00
4. Empirical formula: CH₂O
(b) 18.4% C, 21.5% N, 60.1% K
1. Assume a 100 g sample: We have 18.4 g C, 21.5 g N, and 60.1 g K.
2. Convert grams to moles:
Moles of C = 18.4 g / 12.01 g/mol = 1.53 mol C
Moles of N = 21.5 g / 14.01 g/mol = 1.53 mol N
Moles of K = 60.1 g / 39.10 g/mol = 1.54 mol K
3. Divide by the smallest number of moles:
C: 1.53 mol / 1.53 mol = 1.00
N: 1.53 mol / 1.53 mol = 1.00
K: 1.54 mol / 1.53 mol = 1.01 ≈ 1.00
4. Empirical formula: KCN
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