A compound containing only C, H, and O was extracted from the bark of the sassafras tree. The combustion of 36.3mg produced 98.6 of CO2 and 20.2mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas
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What is the empirical formula of a compound containing C, H, O if combustion of 1.23g of the compound yields 1.8g CO2 and .74g of H2O
To determine the empirical and molecular formulas of the compound, we need to follow a series of steps.
Step 1: Calculate the number of moles of CO2 produced.
First, we need to convert the mass of CO2 produced (98.6 mg) to grams by dividing it by 1000:
98.6 mg ÷ 1000 = 0.0986 g
Next, we need to calculate the number of moles of CO2 by dividing the mass by the molar mass of CO2 (44.01 g/mol):
0.0986 g ÷ 44.01 g/mol ≈ 0.00224 mol
Step 2: Calculate the number of moles of H2O produced.
Similarly, we convert the mass of H2O produced (20.2 mg) to grams:
20.2 mg ÷ 1000 = 0.0202 g
Then, we calculate the number of moles of H2O by dividing the mass by the molar mass of H2O (18.02 g/mol):
0.0202 g ÷ 18.02 g/mol ≈ 0.00112 mol
Step 3: Determine the mole ratio of carbon and hydrogen in the compound.
Since all of the carbon in the compound is converted to CO2 and all of the hydrogen is converted to H2O during combustion, we can use the mole ratios to determine the number of moles of carbon and hydrogen in the compound.
From the balanced chemical equation for the combustion of the compound:
C:CO2 = 1:1
H:H2O = 2:1
Therefore, the number of moles of carbon in the compound is also 0.00224 mol, and the number of moles of hydrogen is double that, 0.00448 mol.
Step 4: Determine the number of moles of oxygen in the compound.
To find the number of moles of oxygen, subtract the sum of the moles of carbon and hydrogen from the total number of moles in the compound:
Total moles in the compound = 0.00224 mol (C) + 0.00448 mol (H) = 0.00672 mol
Moles of oxygen = Total moles in the compound - Moles of carbon - Moles of hydrogen:
0.00672 mol - 0.00224 mol - 0.00448 mol = 0.00000 mol
Step 5: Determine the empirical formula.
Now that we know the ratio of elements, we can determine the empirical formula by dividing the number of moles of each element by the smallest number of moles:
C: 0.00224 mol ÷ 0.00224 mol ≈ 1
H: 0.00448 mol ÷ 0.00224 mol ≈ 2
O: 0.00000 mol ÷ 0.00224 mol ≈ 0
The empirical formula is therefore CH2O.
Step 6: Determine the molecular formula.
To determine the molecular formula, we need to know the molar mass of the compound. Given that the molar mass is 162 g/mol, we need to find the ratio between the empirical formula mass and the molar mass.
Empirical formula mass (CH2O) = (12.01 g/mol x 1) + (1.01 g/mol x 2) + (16.00 g/mol x 1) = 30.03 g/mol
Molar mass ÷ Empirical formula mass = 162 g/mol ÷ 30.03 g/mol ≈ 5.39
Since the ratio is approximately 5.39, we multiply the subscripts in the empirical formula by this ratio to obtain the molecular formula:
Empirical formula: CH2O
Molecular formula: (CH2O) x 5.39 ≈ C5H10O5
Therefore, the empirical formula of the compound is CH2O, and its molecular formula is C5H10O5.