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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) 4 decimal point also do the 1mol
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Here are the empirical formulas for the compounds with the given compositions:
(a) 40.1% C, 6.6% H, 53.3% O
1. Assume 100g of the compound. This means we have 40.1g C, 6.6g H, and 53.3g O.
2. Convert grams to moles using the molar masses:
C: 40.1g / (12.011 g/mol) = 3.3401 mol
H: 6.6g / (1.008 g/mol) = 6.5536 mol
O: 53.3g / (15.9994 g/mol) = 3.3314 mol
3. Divide each mole value by the smallest mole value (3.3314 mol) to get the mole ratio:
C: 3.3401 mol / 3.3314 mol ≈ 1.0027
H: 6.5536 mol / 3.3314 mol ≈ 1.9679
O: 3.3314 mol / 3.3314 mol = 1
4. Round the mole ratios to the nearest whole number (if possible):
C: 1
H: 2
O: 1
Therefore, the empirical formula for compound (a) is CH₂O.
(b) 18.4% C, 21.5% N, 60.1% K
Following the same steps as above:
1. Assume 100g: 18.4g C, 21.5g N, 60.1g K
2. Convert to moles:
C: 18.4g / (12.011 g/mol) = 1.5320 mol
N: 21.5g / (14.007 g/mol) = 1.5350 mol
K: 60.1g / (39.0983 g/mol) = 1.5374 mol
3. Divide by the smallest mole value (1.5320 mol):
C: 1.5320 mol / 1.5320 mol = 1
N: 1.5350 mol / 1.5320 mol ≈ 1.0020
K: 1.5374 mol / 1.5320 mol ≈ 1.0035
4. Round to whole numbers:
C: 1
N: 1
K: 1
Therefore, the empirical formula for compound (b) is KCN.
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